Edition
MANUAL INTRODUCTION
TO
LINEAR
ALGEBRA
Third FOR INSTRUCTORS
Gilbert Strang
[e-mail]
Massachusetts Institute of Technology
http://web.mit.edu/18.06/www
http://math.mit.edu/gs
http://www.wellesleycambridge.com
Wellesley-Cambridge Press
Box 812060
Wellesley, Massachusetts 02482
Solutions to Exercises
Problem Set 1.1, page 6
1 Line through (1, 1, 1); plane; same plane!
3 v = (2, 2) and w = (1, 1).
4 3v + w = (7, 5) and v 3w = (1, 5) and cv + dw = (2c + d, c + 2d).
5 u + v = (2, 3, 1) and u + v + w = (0, 0, 0) and 2u + 2v + w = (add rst answers) = (2, 3, 1).
6 The components of every cv + dw add to zero. Choose c = 4 and d = 10 to get (4, 2, 6).
8 The other diagonal is v w (or else w v ). Adding diagonals gives 2v (or 2w ).
9 The fourth corner can be (4, 4) or (4, 0) or (2, 2).
10 i + j is the diagonal of the base.
1
11 Five more corners (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is ( 2 , 1 , 1 ). The
22
1
centers of the six faces are ( 1 , 1 , 0), ( 1 , 2 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), ( 1 , 1, 1 ).
22
2
22
22
2
2
2
2
12 A four-dimensional cube has 24 = 16 corners and 2 4 = 8 three-dimensional sides and 24
two-dimensional faces and 32 one-dimensional edges. See Worked Example 2.4 A.
13 sum = zero vector; sum = 4:00 vector; 1:00 is 60 from horizontal = (cos , sin ) = ( 1 ,
3
3
2
3
).
2
14 Sum = 12j since j = (0, 1) is added to every vector.
15 The point
3
v
4
+ 1 w is three-fourths of the way to v starting from w . The vector
4
1
v
4
+ 1 w is
4
1
halfway to u = 2 v + 1 w , and the vector v + w is 2u (the far corner of the parallelogram).
2
16 All combinations with c + d = 1 are on the line through v and w . The point V = v + 2w is
on that line beyond w .
17 The vectors cv + cw ll out the line passing through (0, 0) and u =
1
v
2
+ 1 w . It continues
2
beyond v + w and (0, 0). With c 0, half this line is removed and the ray starts at (0, 0).
18 The combinations with 0 c 1 and 0 d 1 ll the parallelogram with sides v and w .
19 With c 0 and d 0 we get the cone or wedge between v and w .
20 (a)
1
u
3
+ 1 v + 1 w is the center of the triangle between u , v and w ;
3
3
of the edge between u and w
1
u
2
+ 1 w is the center
2
(b) To ll in the triangle keep c 0, d 0, e 0, and
c + d + e = 1.
3
4
21 The sum is (v u ) + (w v ) + (u w ) = zero vector.
22 The vector
1
(u
2
+ v + w ) is outside the pyramid because c + d + e =
1
2
+
1
2
+
1
2
> 1.
23 All vectors are combinations of u , v , and w .
24 Vectors cv are in both planes.
25 (a) Choose u = v = w = any nonzero vector
(b) Choose u and v in dierent directions,
and w to be a combination like u + v .
26 The solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 The combinations of (1, 0, 0) and (0, 1, 0) ll the xy plane in xyz space.
28 An example is (a, b) = (3, 6) and (c, d) = (1, 2). The ratios a/c and b/d are equal. Then
ad = bc. Then (divide by bd) the ratios a/b and c/d are equal!
Problem Set 1.2, page 17
1 u v = 1.4, u w = 0, v w = 24 = w v .
2
u = 1 and v = 5 = w . Then 1.4 < (1)(5) and 24 < (5)(5).
3 Unit vectors v / v = ( 3 , 4 ) = (.6, .8) and w / w = ( 4 , 3 ) = (.8, .6). The cosine of is
55
55
v w = 24 . The vectors w , u , w make 0 , 90 , 180 angles with w .
v
w
25
4 u1 = v/ v =
could be
1 (3, 1)
10
and u 2 = w / w = 1 (2, 1, 2). U 1 =
3
1 (1, 3)
10
or
1 (1, 3).
10
U2
1
(1, 2, 0).
5
5 (a) v (v ) = 1
so = 90
(b) (v + w ) (v w ) = v v + w v v w w w = 1+(
)(
)1 = 0
(c) (v 2w ) (v + 2w ) = v v 4w w = 3
6 (a) cos =
1
(2)(1)
so = 60 or
(c) cos =
1+3
(2)(2)
=
1
2
3
radians
so = 60 or
3
(b) cos = 0 so = 90 or radians
2
(d) cos = 1/ 2 so = 135 or 34 .
7 All vectors w = (c, 2c); all vectors (x, y, z ) with x + y + z = 0 lie on a plane ; all vectors
perpendicular to (1, 1, 1) and (1, 2, 3) lie on a line.
8 (a) False
(b) True: u (cv + dw ) = cu v + du w = 0
(c) True
9 If v2 w2 /v1 w1 = 1 then v2 w2 = v1 w1 or v1 w1 + v2 w2 = 0.
10 Slopes
2
1
and 1 multiply to give 1: perpendicular.
2
11 v w < 0 means angle > 90 ; this is half of the plane.
12 (1, 1) perpendicular to (1, 5) c(1, 1) if 6 2c = 0 or c = 3; v (w cv ) = 0 if c = v w /v v .
13 v = (1, 0, 1), w = (0, 1, 0).
14 u = (1, 1, 0, 0), v = (0, 0, 1, 1), w = (1, 1, 1, 1).
15 1 (x + y ) = 5; cos = 2 16/ 10 10 = .8.
2
2
= 9 so v = 3; u = 1 v ; w = (1, 1, 0, . . . , 0).
3
2
2
2
17 cos = 1/ 2, cos = 0, cos = 1/ 2, cos2 + cos2 + cos2 = (v1 + v2 + v3 )/ v
16
v
2
= 1.
5
18
v
2
= 42 + 22 = 20, w
2
= (1)2 + 22 = 5, (3, 4)
2
= 25 = 20 + 5.
19 v w = (5, 0) also has (length)2 = 25. Choose v = (1, 1) and w = (0, 1) which are not
perpendicular; (length of v )2 + (length of w )2 = 12 + 12 + 12 but (length of v w )2 = 1.
20 (v + w ) (v + w ) = (v + w ) v +(v + w ) w = v (v + w )+ w (v + w ) = v v + v w + w v + w w =
v v + 2v w + w w . Notice v w = w v !
21 2v w 2 v
w leads to v + w
2
= v v + 2v w + w w v
2
+2 v
w+w
2
=
2
(v + w ).
22 Compare v v + w w with (v w ) (v w ) to nd that 2v w = 0. Divide by 2.
23 cos = w1 / w and sin = w2 / w . Then cos( a) = cos cos +sin sin = v1 w1 / v
v2 w2 / v
w = v w/ v
w+
w.
24 We know that (v w ) (v w ) = v v 2v w + w w . The Law of Cosines writes v
for v w . When < 90 this is positive and v v + w w is larger than v w
2
w cos
.
22
22
22
22
22
22
25 (a) v1 w1 + 2v1 w1 v2 w2 + v2 w2 v1 w1 + v1 w2 + v2 w1 + v2 w2 is true because the dierence is
22
22
v1 w2 + v2 w1 2v1 w1 v2 w2 which is (v1 w2 v2 w1 )2 0.
26 Example 6 gives |u1 ||U1 |
1
(u2
1
2
2
+ U1 ) and |u2 ||U2 |
1
(u2
2
2
2
+ U2 ). The whole line becomes
1
.96 (.6)(.8) + (.8)(.6) 2 (.62 + .82 ) + 1 (.82 + .62 ) = 1.
2
27 The cosine of is x/
x2 + y 2 , near side over hypotenuse. Then | cos |2 = x2 /(x2 + y 2 ) 1.
28 Try v = (1, 2, 3) and w = (3, 1, 2) with cos =
7
14
and = 120 . Write v w = xz + yz + xy
1
as 2 (x + y + z )2 1 (x2 + y 2 + z 2 ). If x + y + z = 0 this is 1 (x2 + y 2 + z 2 ), so v w / v
2
2
1
w = 2.
29 The length v w is between 2 and 8. The dot product v w is between 15 and 15.
30 The vectors w = (x, y ) with v w = x + 2y = 5 lie on a line in the xy plane. The shortest w
is (1, 2) in the direction of v .
31 Three vectors in the plane could make angles > 90 with each other: (1, 0), (1, 4), (1, 4).
Four vectors could not do this (360 total angle). How many can do this in R3 or Rn ?
Problem Set 1.3
1 (x, y, z ) = (2, 0, 0) and (0, 6, 0); n = (3, 1, 1); dot product (3, 1, 1) (2, 6, 0) = 0.
2 4x y 2z = 1 is parallel to every plane 4x y 2z = d and perpendicular to n = (4, 1, 2).
3 (a) True (assuming n = 0)
4 (a) x + 5y + 2z = 14
(b) False
(c) True.
(b) x + 5y + 2z = 30
(c) y = 0.
5 The plane changes to the symmetric plane on the other side of the origin.
6 x y z = 0.
7 x + 4y = 0; x + 4y = 14.
8 u = (2, 0, 0), v = (0, 2, 0), w = (0, 0, 2). Need c + d + e = 1.
6
9 x + 4y + z + 2t = 8.
10 x 4y + 2z = 0.
11 We choose v 0 = (6, 0, 0) and then in-plane vectors (3, 1, 0) and (1, 0, 1). The points on the
plane are v 0 + y (3, 1, 0) + z (1, 0, 1).
12 v 0 = (0, 0, 0); all vectors in the plane are combinations y (2, 1, 0) + z ( 1 , 0, 1).
2
13 v 0 = (0, 0, 0); all solutions are combinations y (1, 1, 0) + z (1, 0, 1).
14 Particular point (9, 0); solution (3, 1); points are (9, 0) + y (3, 1) = (3y + 9, y ).
15 v 0 = (24, 0, 0, 0); solutions (2, 1, 0, 0) and (3, 0, 1, 0) and (4, 0, 0, 1). Combine to get
(24 2y 3z 4t, y, z, t).
16 Choose v 0 = (0, 6, 0) with two zero components. Then set components to 1 to choose (1, 0, 0)
3
and (0, 3/2, 1). Combinations are (x, 6 2 z, z ).
17 Now |d|/ n = 12/ 56 = 12/2 14 = 6/ 14. Same answer because same plane.
18 (a) |d|/ n = 18/3 = 6 and v = (4, 4, 2)
(b) |d|/ n = 0 and v = 0
(c) |d|/ n = 6/ 2 and v = 3n = (3, 0, 3).
19 (a) Shortest distance is along perpendicular to line
(c) The distance to (5, 10) is 125.
20 (a) n = (a, b)
|c|/ a2 + b2 .
(b) t = c/(a2 + b2 )
(b) Need t + 4t = 25 or t = 5
(c) This distance to tn = (ca, cb)/(a2 + b2 ) is
21 Substitute x = 1 + t, y = 2t, z = 5 2t to nd (1 + t) + 2(2t) 2(5 2t) = 27 or 9 + 9t = 27
or t = 4. Then tn = 12.
22 Shortest distance in the direction of n ; w + tn lies on the plane when n w + tn n = d or
t = (d n w )/n n . The distance is |d n w |/ n (which is |d|/ n when w = 0).
23 The vectors (1, 2, 3) and (1, 1, 1) are perpendicular to the line. Set x = 0 to nd y = 16
and z = 14. Set y = 0 to nd x = 9/2 and z = 5/2. These particular points are (0, 16, 14)
and (9/2, 0, 5/2).
24 (a) n = (1, 1, 1, 1)
(d) v 0 = (1, 0, 0, 0)
(b) |d|/ n =
1
2
(c) dn /n n = ( 1 , 1 , 1 , 1 )
444
4
(e) (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)
(f) all points (1 y z + t, y, z, t).
25 n = (1, 1, 1) or any nonzero (c, c, c).
26 cos = (0, 1, 1) (1, 0, 1)/ 2 2 = 1 so = 60 .
2
Problem Set 2.1, page 29
1 The planes x = 2 and y = 3 and z = 4 are perpendicular to the x, y, z axes.
2 The vectors are i = (1, 0, 0) and j = (0, 1, 0) and k = (0, 0, 1) and b = (2, 3, 4) = 2i + 3j + 4k .
7
3 The planes are the same: 2y = 6 is y = 3, and 3z = 12 is z = 4. The solution is the same
intersection point. The columns are changed; but same combination x = x .
4 The solution is not changed; the second plane and row 2 of the matrix and all columns of the
matrix are changed.
5 If z = 2 then x + y = 0 and x y = z give the point (1, 1, 2). If z = 0 then x + y = 6 and
x y = 4 give the point (5, 1, 0). Halfway between is (3, 0, 1).
6 If x, y, z satisfy the rst two equations they also satisfy the third equation. The line L of
solutions contains v = (1, 1, 0) and w = ( 1 , 1, 1 ) and u =
2
2
1
v
2
+ 1 w and all combinations
2
cv + dw with c + d = 1.
7 Equation 1 + equation 2 equation 3 is now 0 = 4. Solution impossible.
8 Column 3 = Column 1; solutions (x, y, z ) = (1, 1, 0) or (0, 1, 1) and you can add any multiple
of (1, 0, 1); b = (4, 6, c) needs c = 10 for solvability.
9 Four planes in 4-dimensional space normally meet at a point. The solution to Ax = (3, 3, 3, 2)
is x = (0, 0, 1, 2) if A has columns (1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1).
10 Ax = (18, 5, 0), Ax = (3, 4, 5, 5).
11 Nine multiplications for Ax = (18, 5, 0).
12 (14, 22) and (0, 0) and (9, 7).
13 (z, y, x) and (0, 0, 0) and (3, 3, 6).
14 (a) x has n components, Ax has m components
(b) Planes in n-dimensional space, but the
columns are in m-dimensional space.
15 2x + 3y + z + 5t = 8 is Ax = b with the 1 by 4 matrix A = [ 2 3 1 5 ]. The solutions x ll
16
17
18
19
20
21
a 3D plane in 4 dimensions.
10
01
, P =
.
I=
01
10
01
1
0
, 180 rotation from R2 =
= I .
R=
1 0
0 1
010
001
P = 0 0 1 produces (y, z, x) and Q = 1 0 0 recovers (x, y, z ).
100
010
100
10
, E = 1 1 0 .
E=
1 1
001
100
100
E = 0 1 0 , E 1 = 0 1 0 , E v = (3, 4, 8), E 1 E v = (3, 4, 5).
101
1 0 1
10
00
5
0
, P 2 =
, P1 v = , P2 P1 v = .
P1 =
00
01
0
0
8
22 R =
1
2
22
.
2
2
x
23 The dot product [ 1 4 5 ] y = (1 by 3)(3 by 1) is zero for points (x, y, z ) on a plane in
z
three dimensions. The columns of A are one-dimensional vectors.
24 A = [ 1 2 ; 3 4 ] and x = [ 5 2 ] and b = [ 1 7 ] . r = b A x prints as zero.
25 A v = [ 3 4 5 ] and v v = 50; v A gives an error message.
26 ones(4, 4) ones(4, 1) = [ 4 4 4 4 ] ; B w = [ 10 10 10 10 ] .
27 The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) + 2(2, 1) =
4(column 1) + 2(column 2) = right side (0, 6).
28 The row picture shows 2 planes in 3-dimensional space. The column picture is in 2-dimensional
space. The solutions normally lie on a line.
29 The row picture shows four lines. The column picture is in four -dimensional space. No solution
unless the right side is a combination of the two columns.
.7
.65
. The components always add to 1. They are always positive.
30 u 2 = , u 3 =
.3
.35
31 u 7 , v 7 , w 7 are all close to (.6, .4). Their components still add to 1.
.8 .3
.6
.6
.8 .3
= = steady state s . No change when multiplied by
.
32
.2 .7
.4
.4
.2 .7
834
5+u
5u+v
5v
34 M = 1 5 9 = 5 u v
5
5 + u + v ; M3 (1, 1, 1) = (15, 15, 15);
672
5+v
5+uv
5u
M4 (1, 1, 1, 1) = (34, 34, 34, 34) because the numbers 1 to 16 add to 136 which is 4(34).
Problem Set 2.2, page 40
1 Multiply by l =
10
2
= 5 and subtract to nd 2x + 3y = 14 and 6y = 6.
2 y = 1 and then x = 2. Multiplying the right side by 4 will multiply (x, y ) by 4 to give the
solution (x, y ) = (8, 4).
3 Subtract 1 times equation 1 (or add
2
1
2
times equation 1). The new second equation is 3y = 3.
Then y = 1 and x = 5. If the right side changes sign, so does the solution: (x, y ) = (5, 1).
4 Subtract l =
c
a
times equation 1. The new second pivot multiplying y is d(cb/a) or (adbc)/a.
Then y = (ag cf )/(ad bc).
5 6x + 4y is 2 times 3x + 2y . There is no solution unless the right side is 2 10 = 20. Then all
points on the line 3x + 2y = 10 are solutions, including (0, 5) and (4, 1).
9
6 Singular system if b = 4, because 4x + 8y is 2 times 2x + 4y . Then g = 2 16 = 32 makes the
system solvable. The lines become the same : innitely many solutions like (8, 0) and (0, 4).
7 If a = 2 elimination must fail. The equations have no solution. If a = 0 elimination stops for
a row exchange. Then 3y = 3 gives y = 1 and 4x + 6y = 6 gives x = 3.
8 If k = 3 elimination must fail: no solution. If k = 3, elimination gives 0 = 0 in equation 2:
innitely many solutions. If k = 0 a row exchange is needed: one solution.
9 6x 4y is 2 times (3x 2y ). Therefore we need b2 = 2b1 . Then there will be innitely many
solutions.
10 The equation y = 1 comes from elimination. Then x = 4 and 5x 4y = c = 16.
11 2x + 3y + z = 8
y + 3z = 4
x=2
gives
12
2x 3y = 3
If a zero is at the start of row 2 or 3,
z=1
8z = 8
y=1
that avoids a row operation.
2x 3y = 3
y + z = 1 gives
2y 3z = 2
x=3
Subtract 2 row 1 from row 2
y + z = 1 and y = 1
Subtract 1 row 1 from row 3
5z = 0
Subtract 2 row 2 from row 3
z=0
13 Subtract 2 times row 1 from row 2 to reach (d 10)y z = 2. Equation (3) is y z = 3. If
d = 10 exchange rows 2 and 3. If d = 11 the system is singular; third pivot is missing.
14 The second pivot position will contain 2 b. If b = 2 we exchange with row 3. If b = 1
(singular case) the second equation is y z = 0. A solution is (1, 1, 1).
15
0x + 0y + 2z = 4
(a)
x + 2y + 2z = 5
0x + 3y + 4z = 4
(b)
x + 2y + 2z = 5
0x + 3y + 4z = 6
0x + 3y + 4z = 6
(exchange 1 and 2, then 2 and 3)
(rows 1 and 3 are not consistent)
16 If row 1 = row 2, then row 2 is zero after the rst step; exchange the zero row with row 3 and
there is no third pivot. If column 1 = column 2 there is no second pivot.
17 x + 2y + 3z = 0, 4x + 8y + 12z = 0, 5x + 10y + 15z = 0 has innitely many solutions.
18 Row 2 becomes 3y 4z = 5, then row 3 becomes (q + 4)z = t 5. If q = 4 the system is
singular no third pivot. Then if t = 5 the third equation is 0 = 0. Choosing z = 1 the
equation 3y 4z = 5 gives y = 3 and equation 1 gives x = 9.
19 (a) Another solution is
1
(x
2
+ X, y + Y, z + Z ).
(b) If 25 planes meet at two points, they
meet along the whole line through those two points.
20 Singular if row 3 is a combination of rows 1 and 2. From the end view, the three planes form
a triangle. This happens if rows 1 + 2 = row 3 on the left side but not the right side: for
example x + y + z = 0, x 2y z = 1, 2x y = 1. No parallel planes but still no solution.
21 Pivots 2,
345
,,
234
in the equations 2x + y = 0,
3
y
2
+ z = 0,
z = 3, y = 2, x = 1.
22 The solution is (1, 2, 3, 4) instead of (1, 2, 3, 4).
4
z
3
+ t = 0,
5
t
4
= 5. Solution t = 4,
10
23 The fth pivot is 6 . The nth pivot is (n+1) .
5
n
1
1
1
1
24 A = a a + 1
a + 1 for any a, b, c leads to U = 0
b b+c b+c+3
0
a2
if a = 2 or a = 0.
25 Elimination fails on
aa
1
1
1 .
3
1
0
26 a = 2 (equal columns), a = 4 (equal rows), a = 0 (zero column).
13
04
and
. A = [ 1 1 0 0; 1 0 1 0;
27 Solvable for s = 10 (add equations);
17
26
0 0 1 1; 0 1 0 1 ] and U = [ 1 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 0 ].
28 Elimination leaves the diagonal matrix diag(3, 2, 1). Then x = 1, y = 1, z = 4.
29 A(2, :) = A(2, :) 3 A(1, :) Subtracts 3 times row 1 from row 2.
30 The average pivots for rand(3) without row exchanges were
1
, 5, 10
2
in one experimentbut
pivots 2 and 3 can be arbitrarily large. Their averages are actually innite! With row exchanges
in MATLABs lu code, the averages .75 and .50 and .365 are much more stable (and should be
predictable, also for randn with normal instead of uniform probability distribution).
Problem Set 2.3, page 50
1
1 E21 = 5
0
0
1
0
0
0 ,
1
1
E32 = 0
0
0
1
7
0
0 ,
1
1
P = 0
0
0
0
1
0
0
1
1 1
0
0
0
0
0 = 0
0
1
0
1
1
0
0
0
1 .
0
= (1, 5, 35) but E21 E32 b = (1, 5, 0). Then row 3 feels no eect from row 1.
0
100
1
0
0
1
0
0
0 , 0 1 0 , 0
1
0
E21 , E31 , E32 4
1
0 .
1
201
0 2
1
M = E32 E31 E21
10 2
1
1
1
1
1
4 Elimination on column 4: b = 0 4 4 4 . Then back substitution in
0
0
2
10
2 E32 E21 b
10
3 4 1
00
1
U x = (1, 4, 10) gives z = 5, y = 2 , x = 1 . This solves Ax = (1, 0, 0).
2
5 Changing a33 from 7 to 11 will change the third pivot from 5 to 9. Changing a33 from 7 to 2
will change the pivot from 5 to no pivot.
6 If all columns are multiples of column 1, there is no second pivot.
1
7 To reverse E31 , add 7 times row 1 to row 3. The matrix is R31 = 0
7
0
1
0
0
0 .
1
8 The same R31 from Problem 7 is changed to I . Thus E31 R31 = R31 E31 = I .
11
1
0
0
9 M = 0 0 1 . After the exchange, E must act on the new row 3.
1 1 0
101
101
201
10 E13 = 0 1 0 ; 0 1 0 ; 0 1 0 .
001
101
101
122
11 A = 1 1 2 .
121
987
1
2
3
12 6 5 4 , 0
1 2 .
321
0
2 3
13 (a) E times the third column of B is the third column of EB
(b) E could add row 2 to
row 3 to give nonzeros.
14 E21 has l21 = 1 , E32 has l32 = 2 , E43 has l43 = 3 . Otherwise the E s match the identity
2
3
4
matrix.
1 4 7
1 4 7
1
0
15 A = 1 2 5 0 6 12 . E32 = 0
1
3
0 3
0 12 24
0 2
16 (a) X 2Y = 0 and X + Y = 33; X=22, Y=11
0
0 .
1
(b) 2m + c = 5 and 3m + c = 7; m = 2,
c = 1.
a+ b+ c= 4
17
a=2
a + 2b + 4c = 8 gives b = 1 .
a + 3b + 9c = 14
c=1
100
1
18 EF = a 1 0 , F E = a
bc1
b + ac
010
00
19 P Q = 0 0 1 , QP = 1 0
100
01
more).
0
0
c
22 (a)
a3j xj
(b) a21 a11
1
0
1
0
3
0 , F = 0
1
0
1
0
0
0 .
1
1
3c
2
2
2
2
0 , P = I, (P ) = I, I = I, (I ) = I (and many
0
20 (a) Each column is E times a column of B
1
0
are multiples of [ 1 2 4 ].
10
1
, F =
21 No. E =
11
0
1
2
0 , E = 2a
1
2b
1
1
, EF =
(b)
1
1
1
2
(c) x2 x1
1
0
1
1
, F E =
1
2
4
1
2
4
2
1
1
1
(d) (Ax )1 =
=
1
2
4
2
4
8
rows
.
a1j xj .
23 E (EA) subtracts 4 times row 1 from row 2. AE subtracts 2 times column 2 of A from column
1.
12
24 [ A b ] =
2
3
1
4
1
17
2
3
1
0 5
15
:
2x1 + 3x2 = 1
5x2 = 15
x1 =
5
x2 = 3.
25 The last equation becomes 0 = 3. Change the original 6 to 3. Then row 1 + row 2 = row 3.
1410
1
4
1
0
7
4
.
26 (a) Add two extra columns;
2701
0 1 2
1
2
1
27 (a) No solution if d = 0 and c = 0
(b) Innitely many solutions if d = 0 and c = 0. No
eect from a and b.
28 A = AI = A(BC ) = (AB )C = IC = C .
29 Given positive integers with ad bc = 1. Certainly c < a and b < d would be impossible. Also
c > a and b > d would be impossible with integers.This leaves row 1 < row 2 OR row 2 <
34
1 1
11
. Multiply by
to get
, then multiply
row 1. An example is M =
23
0
1
23
10
11
11
10
10
11
to get
. This shows that M =
.
twice by
1 1
01
01
11
11
01
1
0
00
1
0
00
1
1
1
0 0
1
0 0
and eventually M = inverse of Pascal =
30 E =
0 1
1 2
1 0
1 0
0
0 1 1
1
3 3 1
reduces Pascal to I .
Problem Set 2.4, page 59
1 BA = 3I is 5 by 5 AB = 5I is 3 by 3 ABD = 5D is 3 by 1. ABD: No A(B + C ): No.
2 (a) A (column 3 of B )
(b) (Row 1 of A) B
(c) (Row 3 of A)(column 4 of B )
(d) (Row 1 of C )D(column 1 of E ).
38
.
3 AB + AC = A(B + C ) =
69
4 A(BC ) = (AB )C = zero matrix
1 bn
2n 2n
and An =
.
5 An =
01
0
0
10 4
16
2
2
2
2
2
= A + AB + BA + B . But A + 2AB + B =
6 (A + B ) =
66
3
7 (a) True
(b) False
(c) True
2
0
.
(d) False.
8 Rows of DA are 3(row 1 of A) and 5(row 2 of A). Both rows of EA are row 2 of A. Columns
of AD are 3(column 1 of A) and 5(column 2 of A). Columns of AE are zero and column 1 of
A + column 2 of A.
a a+b
and E (AF ) equals (EA)F because matrix multiplication is associative.
9 AF =
c c+d
13
10 F A =
a+c
b+d
a+c
b+d
and then E (F A) =
. E (F A) is not F (EA) because
c
d
a + 2c b + 2d
multiplication is not commutative.
001
11 (a) B = 4I
(b) B = 0
(c) B = 0 1 0
(d) Every row of B is 1, 0, 0, . . .
100
a0
ab
= BA =
gives b = c = 0. Then AC = CA gives a = d : A = aI .
12 AB =
c0
00
13 (A B )2 = (B A)2 = A(A B ) B (A B ) = A2 AB BA + B 2 .
14 (a) True
(b) False
15 (a) mn (every entry)
(c) True
(d) False (take B = 0).
(b) mnp
(c) n3 (this is n2 dot products).
16 By linearity (AB )c agrees with A(B c ). Also for all other columns of C .
17 (a) Use only column 2 of B
(b) Use only row 2 of A
111
1 1
1
1/1 1/2 1/3
18 A = 1 2 2 , 1
1 1 , 2/1 2/2 2/3 .
123
1 1
1
3/1 3/2 3/3
(c)(d) Use row 2 of rst A.
19 Diagonal matrix, lower triangular, symmetric, all rows equal. Zero matrix.
20 (a) a11
(b) l31 = a31 /a11
040
0 0 4
, A3 =
0 0 0
000
0
0
21 A2 =
0
0
8t
0
A3 v = , A4 v = 0.
0
0
a31
(c) a32 ( a11 )a12
(d) a22 ( a21 )a12 .
a11
0008
2y
4z
0 0 0 0
2z
4t
, A4 = 0; then Av = , A2 v = ,
0 0 0 0
2t
0
0000
0
0
22 A = A2 = A3 = but AB =
.5 .5
and (AB )2 = 0.
.5 .5
01
1 1
11
00
2
has A = I ; BC =
=
;
23 A =
1 0
1 1
11
00
01
01
1 0
=
= ED.
DE =
10
1 0
01
010
001
01
has A2 = 0; A = 0 0 1 has A2 = 0 0 0 but A3 = 0.
24 A =
00
000
000
2n 2n 1
11
an an1 b
, An = 2n1
, An =
.
25 An =
1
2
3
0
1
11
0
0
14
1
0
3
26 2 3 3 0 + 4 1 2 1 = 6
2
1
6
27 (a) (Row 3 of A)(column
x
0
(b) x 0 x x = 0
0
0
28 A
3
6
6
0
0
0 + 4
0
1
0
8
2
0
;
3
4 = 10
1
7
1 of B = (Row of A)(column 2
)
3
xx
x
0
and x 0 0 x = 0
x x
00
x
0
; []B ; []
3
0
4 .
1
14
8
of B ) = 0
0x
0 x .
0x
29
30
31
32
x
1
Ax =
x2 = x1 (column 1) + x2 (column 2) + .
x3
100
100
10
E21 = 1 1 0 , E31 = 0 1 0 , E = E31 E21 = 1 1
001
4 0 1
4 0
2
01
11
, D =
, D cb /a =
.
In Problem 30, c =
8
53
13
A B
x
Ax By
real part
=
B
A
y
Bx + Ay
imaginary part.
0
2
0 , then EA = 0
1
0
33 A times X will be the identity matrix I .
3
3
1
34 The solution for b = 5 is 3x 1 + 5x 2 + 8x 3 = 8 ; A = 1
8
16
0
x1 , x2 , x3 .
0
1
1
0
1
1
1
0
1 .
3
0 to produce
1
35 S = D CA1 B is the Schur complement: block version of d (cb/a).
a+b a+b
a+c b+d
agrees with
when b = c and a = d.
36
c+d c+d
a+c b+d
37 If A is northwest and B is southeast then AB is upper triangular and BA is lower triangular. One reason: Row i of A can have n i + 1 nonzeros, with zeros after that. Column j
of B has j nonzeros, with zeros above that. If i > j then (row i of A) (column j of B ) = 0.
So AB is upper triangular.
Similarly BA is lower triangular. Problem 2.7.40 asks about inverses and transposes and
permutations of a northwest A and a southeast
01001
2011
1 0 1 0 0
0 2 0 1
2
38 A = 0 1 0 1 0 , A = 1 0 2 0
0 0 1 0 1
1 1 0 2
10010
0110
gives diameter 3.
B.
0
0
3
1
3
1 , A = 1
1
0
2
3
3
1
1
0
3
1
3
0
3
1
3
0
1
1
3
3
1
1 ,
3
0
A3 with A2
15
0
1
0
0 0 1
39 A = 0 0 0
0 0 0
100
so diameter 4.
0
0
1
0
0
0
0
0
0
2
0 , A = 0
1
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
3
1 , A = 1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
0
0
0
need also A4
Problem Set 2.5, Page 72
1 A1 =
0
1
4
1
3
0
1
2
0
7 4
, B 1 =
, C 1 =
.
1 1
5
3
2
001
1
1
2P
= P; P
= 1 0 0 . Always P 1 = transpose of P .
010
x
.5
5 2
1
0
t
.2
1
1
1
, =
so A =
. A =
and
3 =
10
y
.2
z
.1
2
1
0 1
0
1
xt
1 0
xt
.
any
yz
01
yz
0
1
and
4 x + 2y = 1, 3x + 6y = 0: impossible.
1 1
.
5 U =
0 1
6 (a) Multiply AB = AC by A1 to
nd B =
C
x
y
.
(b) B and C can be any matrices
x y
7 (a) In Ax = (1, 0, 0), equation 1 + equation 2 equation 3 is 0 = 1
must satisfy b1 + b2 = b3
(b) The right sides
(c) Row 3 becomes a row of zerosno third pivot.
8 (a) The vector x = (1, 1, 1) solves Ax = 0
(b) Elimination keeps columns 1+2 = column
3. When columns 1 and 2 end in zeros so does column 3: no third pivot.
9 If you exchange rows 1 and 2 of A, you exchange columns 1 and 2 of A1 .
0
0
0
1/5
3 2
0
0
0
4
0
1/4
0
3
0
0
, B 1 =
(invert each block).
10 A1 =
0
0
1/3
0
0
0
6 5
1/2
0
0
0
0
0 7
6
10
00
, B =
.
11 (a) A = I , B = I
(b) A =
00
01
12 C = AB gives C 1 = B 1 A1 so A1 = BC 1 .
13 M 1 = C 1 B 1 A1 so B 1 = CM 1 A.
16
14 B 1 = A1
1
0
1
1
1
1
= A1
1
0
1
: subtract column 2 of A1 from column 1.
15 If A has a column of zeros, so does BA. So BA = I is impossible. There is no A1 .
ab
d b
ad bc
0
=
= (ad bc)I . The inverse of one matrix is the
16
cd
c
a
0
ad bc
other divided by ad bc.
1
1
1
1
1
17
1 1
= 1
= E; 1 1
= L = E 1
1
1
1
1
1
1
1
1
0 1
1
111
after reversing the order and changing 1 to +1.
18 A2 B = I can be written as A(AB ) = I . Therefore A1 is AB .
19 The (1, 1) entry requires 4a 3b = 1; the (1, 2) entry requires 2b a = 0. Then b =
a=
2
.
5
For the 5 by 5 case 5a 4b = 1 and 2b a = 0 give b =
1
6
and a =
1
5
and
2
.
6
20 A ones(4, 1) is the zero vector so A cannot be invertible.
21 6 of
1
22
2
1
3
2
23 1
0
2
0
0
2
0
0
1
24 0
0
the 16 are invertible, including all four with three 1s.
310
1
3
1
0
1
0
7 3
= I A1 ;
701
0
1 2
1
0
1 2
1
310
1
0 8
3
= I A1 .
801
0
1
3 1
10100
2
1
0
1
0
0
2 1 0 1 0 0
3/2
1 1/2
1
0
12001
0
1
2
0
0
1
1
0
1
0
0
2
1
0
1
0
0
3/2
1 1/2
1
0 0 3/2
0 3/4
3/2 3/4
0
0 4/3
1/3 2/3
1
0 4/3
1/3 2/3
1
0
0
3/2
1
1/2
1
0
0
3/4 1/2
1/4
3/2
0 3/4
3/2 z 3/4 0
1
0 1/2
1 1/2 .
0 4/3
1/3 2/3
1
0
0
1
1/4 1/2
3/4
ab100
1 a 0 1 0 b
1 0 0 1 a ac b
1 c 0 1 0 0 1 0 0 1 c 0 1 0 0
1
c .
01001
00100
1
0010
0
1
3 1 1
1
0
1
1
1
25 A = 4 1
3 1 ; B 1 = 0 so B does not exist.
1 1
3
1
0
10
12
1 1
10
10
1
A =
. Then
A =
. Multiply by D =
26
2 1
02
0
1
2 1
02
0
3 1
= A1 .
to reach I . Here D1 E12 E21 =
1 1/2
0
1/2
17
1
0
0
2 1
0
1
27 A1 = 2
1 3 (notice the pattern); A = 1
2 1 .
0
0
1
0 1
1
2201
2
0 1
1
1
0 1/2 1/2
= I A1 .
28
0210
0
2
1
0
0
1
1/2
0
29 (a) True (AB has a row of zeros)
(c) True (inverse of A1
(b) False (matrix of all 1s)
is A) (d) True (inverse of A2 is (A1 )2 ).
30 Not invertible for c = 7 (equal columns), c = 2 (equal rows), c = 0 (zero column).
a
0 b
1
31 Elimination produces the pivots a and a b and a b. A1 =
a
a
0 .
a(a b)
0 a
a
1100
0 1 1 0
1
. The 5 by 5 A1 also has 1s on the diagonal and superdiagonal.
32 A =
0 0 1 1
0001
33 x = (2, 2, 2, 1).
34 x = (1, 1, . . . , 1) has P x = Qx so (P Q)x = 0.
I0
A1
0
D I
and
and
.
35
C I
D1 CA1 D1
I0
36 If AC = CA, multiply left and right by A1 to nd CA1 = A1 C . If also BC = CB , then
(using the associative law!!),
(AB )C = A(BC ) = A(CB ) = (AC )B = (CA)B = C (AB ).
37 A can be invertible but B is always singular. Each row of B will add to zero, from 0 + 1 + 2 3,
so the vector x = (1, 1, 1, 1) will give B x = 0. I thought A would be invertible as long as you
put the 3s on its main diagonal, but thats wrong:
3
0
Ax =
1
1
0
1
3
1
2
3
2
0
2
1
2 1
= 0
0 1
1
3
but
0
3
A=
2
1
1
2
0
1
3
0
2
3
3
2
1
0
is invertible
38 AD = pascal(4, 1) is its own inverse.
39 hilb(6) is not the exact Hilbert matrix because fractions are rounded o.
40 The three Pascal matrices have S = LU = LLT and then inv(S ) = inv(LT )inv(L). Note that
the triangular L is abs(pascal(n, 1)) in MATLAB.
41 For Ax = b with A = ones(4, 4) = singular matrix and b = ones(4, 1) in its column space,
MATLAB will pick the shortest solution x = (1, 1, 1, 1)/4. Any vector in the nullspace of A
could be added to this particular solution.
42 If AC = I for square matrices then C = A1 (it is proved in 2I that CA = I will also be true).
The same will be true for C . But a square matrix has only one inverse so C = C .
18
43 M M 1 =
(In U V ) (In + U (Im V U )1 V )
=
In U V + U (Im V U )1 V U V U (Im V U )1 V
=
In U V + U (Im V U )(Im V U )1 V = In
(formulas 1, 2, 4 are similar)
Problem Set 2.6, page 84
1
2
21
31
= 1; L =
= 1 and
1
0
1
1
32
times U x = c is Ax = b :
= 2 (and
33
x
5
= .
2
y
7
1
1
1
= 1): reverse the steps to recover x + 3y + 6z = 11 from U x = c :
1 times (x + y + z = 5) + 2 times (y + 2z = 2) + 1 times (z = 2) gives x + 3y + 6z = 11.
10
c1
5
5
11
x1
5
3
= ; c = . U x = c is
= ; x = .
3 Lc = b is
11
c2
7
2
01
x2
2
2
5
1
c = 7 ; c = 2 . U x =
1
11
2
1
4 Lc = 1
1
1
2
1
5 EA = 0
3
1
0
2
3
1
This is 0
0
0
2
0
1
1
2
c
1
1
1
1
0
1
2
1
2
0
1
1
2
1
3
0
1
0
0
1
U.
1 1
0 U = E21 E32 U = LU .
1
1
2
0
1
2 .
5
2
4
0 U = LU .
1
1
1
b
1
2 = U ; A = LU = 0
5
3
0
1
1
1
1
0 = U . Then A = 2
2
3
1
0
4
8 E = E32 E31 E21 =
=A
1
5
5
2 x = 2 ; x = 2 .
1
2
2
1
1
2 3 = U . Then A = 2
0 6
0
1
1
1
3
1
2
2 = 0
5
0
4
1
1
2 1 A = 0
1
001
0
7 E32 E31 E21 A =
L
0
1
5
1
0
1
6
1
6 0 1
0 2
1
1
1
a
1
1
= a
1
1
ac b c
1
. This is
.
1
9 2 by 2: d = 0 not allowed; 1
1
1
1
2
0
1
2 = l
1
m
1
n
1
d
e
f
g
h
i
d = 1, e = 1, then l = 1
f = 0 is not allowed
no pivot in row 2
10 c = 2 leads to zero in the second pivot position: exchange rows and the matrix will be OK.
c = 1 leads to zero in the third pivot position. In this case the matrix is singular.
19
2
11 A = 0
0
12
13
14
15
16
4
8
2
; A = LU has U = A (pivots on the diagonal);
9 has L = I and D =
3
07
7
124
A = LDU has U = D1 A = 0 1 3 with 1s on the diagonal.
001
2
4
10
24
10
20
12
=
=
= LDU ; notice U is LT
A=
4 11
21
03
21
03
01
1
1
4
0
1
1
1
4
0
T
A = 4
0 4
0
1
4 = 4
1
4
1 1 = LDL .
0 1
1
0
0
4
0 1
1
4
0
0
1
aaaa
1
a
a
a
a
a=0
a b b b 1 1
b a b a b a
b=a
=
. Need
a b c c 1 1 1
c b c b
c=b
abcd
1111
dc
d=c
arrr
1
a
r
r
r
a=0
a b s s 1 1
b r s r s r
b=r
=
. Need
a b c t 1 1 1
c s t s
c=s
abcd
1111
dt
d=t
10
2
2
24
2
5
c = gives c = . Then
x = gives x = .
41
11
3
01
3
3
2
4
2
times x is b = .
Check that A = LU =
8 17
11
100
4
4
111
4
3
1 1 0 c = 5 gives c = 1 . Then 0 1 1 x = 1 gives x = 0 .
111
6
1
001
1
1
3
17 (a) L goes to I
(b) I goes to L1
(c) LU goes to U .
18 (a) Multiply LDU = L1 D1 U1 by inverses to get L1 LD = D1 U1 U 1 . The left side is lower
1
triangular, the right side is upper triangular both sides are diagonal.
(b) Since L, U, L1 , U1 have diagonals of 1s we get D = D1 . Then
1
110
a
a
0
19 1 1
1 1 = LIU ; a a + b
b = (same L)
011
1
0
b
b+c
L1 L is I and U1 U 1 is I .
1
a
(same U ).
b
c
20 A tridiagonal T has 2 nonzeros in the pivot row and only one nonzero below the pivot (so 1
operation to
12
2 3
T =
0 1
00
nd the multiplier and 1 to nd new pivot!). T = bidiagonal L
the
00
1
200
1
0
0 1 1 0
2
1 0
1
U =
. Reverse steps by L =
0
0 1
2 3
0 3 3
34
0
001
0
0
timesU :
00
0 0
.
1 0
11
20
21 For A, L has the 3 lower zeros but U may not have the upper zero. For B , L has the bottom
left zero and U has the upper right zero. One zero in A and two zeros in B are lled in.
xxx
100
22 x x x = 1 0 0
(s are all known after the rst pivot is used).
xxx
1
0
531
420
200
111
23 3 3 1 2 2 0 2 2 0 = L. Then A = U L with U = 0 1 1 .
111
111
111
001
11005
1
1
0
0
5
2 1 1 0 8
0 1
1
0 2
1 1
x2
2
. Solve
= for x2 = 3
24
0 1 3 2 8
0
1
1
0
4
11
x3
4
00112
0
0
1
1
2
and x3 = 1 in the middle. Then x1 = 2 backward and x4 = 1 forward.
25 The 2 by 2 upper submatrix B has the rst two pivots 2, 7. Reason: Elimination on A starts
in the upper left corner with elimination on B .
26 The rst three pivots for M are still 2, 7, 6. To be sure that 9 is the fourth pivot, put zeros in
the
1
1
27 1
1
1
rest of row 4 and column
1
1
1
1
1
2
3
4
5 1
3
6 10 15 = 1
4 10 20 35 1
5 15 35 70
1
4.
1
2
1
3
3
1
4
6
4
1
1
1
1
1
1
2
3
1
3
1
1
4
6 .
4
1
Pascals triangle in L and U .
MATLABs lu code will wreck
the pattern. chol does no row
exchanges for symmetric
matrices with positive pivots.
28 c = 6 and also c = 7 will make LU impossible (c = 6 needs a row exchange).
32 inv(A) b should take 3 times as long as A\b (n3 for A1 vs n3 /3 multiplications for LU ).
34 The upper triangular part triu(A) should be about three times faster to invert.
35 Each new right side costs only n2 steps compared to n3 /3 for full elimination A\b .
36 This L comes from the 1, 2, 1 tridiagonal A = LDLT . (Row i of L) (Column j of L1 ) =
j
+ (1)
i1
1i
i
j
i
= 0 for i > j so LL1 = I . Then L1 leads to A1 = (L1 )T D1 L1 .
The 1, 2, 1 matrix has inverse A1 = j (n i + 1)/(n + 1) for i j (reverse for i j ).
ij
Problem Set 2.7, page 95
T
1 A =
1
c2
1
9
0
3
0
c
c 1
1
, A
1
=
3
0
1/3
1 T
, (A
T 1
) = (A )
=
1
3
0
1/3
= (A1 )T .
2 In case AB = BA, transpose both sides: AT commutes with B T .
3
(AB )1
T
= (B 1 A1 )T = (A1 )T (B 1 )T .
; AT = A and A1 =
21
4 A=
0
1
0
0
has A2 = 0. But the diagonal entries of AT A are dot products of columns of A
with themselves. If AT A = 0, zero dot products zero columns A = zero matrix.
2
T
T
5 (a) x Ay = a22 = 5
(b) x A = 4 5 6
(c) Ay = .
5
AT C T
; M T = M needs AT = A, B T = C, DT = D.
6 MT =
B T DT
7 (a) False (needs A = AT )
(b) False
(c) True
(d) False.
8 The 1 in column 1 has n choices; then the 1 in column 2 has n 1 choices; . . . (n! choices
overall).
0
9 P1 P2 = 0
1
1
0
1
1 0
0
0
0
0
0
0
1
0
1 = P2 P1 .
0
10 (3, 1, 2, 4), (2, 3, 1, 4) keep 4 in position; 6 more keeping 1 or 2 or 3 in position; (2, 1, 4, 3) and
(3, 4, 1, 2)
0
11 P = 0
1
exchanging 2 pairs.
10
0 1 ;
00
1
0
No AP is lower triangular (this is a column exchange); P1 = 0
0
T
T
T
T
0
1
0
0
1 , P2 = 0
0
1
T
0
1
0
1
0 .
0
T
12 (P x ) (P y ) = x P P y = x y because P P = I ; In general P x y = x P y = x P y :
0
0
1
0
13 P = 0
1
1
0
0
0
1
0
0
1
1
1
0
1 2 1 = 2 0
0
3
2
3
1
0
1
1 1 .
0
2
1
0
0
0
1
1 or its transpose; P =
0
0
0
P
for the same P had P 4 = P .
14 There are n! permutation matrices of order n. Eventually two powers of P must be the same:
P r = P s and P rs = I . Certainly r s n!
0
P
01
2
is 5 by 5 with P2 =
and P3 = 0
P=
P3
10
1
E0
= P T with E
15 (a) P T (row 4) = row 1
(b) P =
0E
1
0
1 .
00
01
moves all rows.
=
10
0
16 A2 B 2 and ABA are symmetric if A and B are symmetric.
11
01
11
1
has D =
17 (a) A =
(b) A =
(c) A =
11
11
10
0
0
1
.
22
18 (a) 5 + 4 + 3 + 2 + 1 = 15 independent entries if A = AT
15 in LDL
T
(b) L has 10 and D has 5: total
T
(c) Zero diagonal if A = A, leaving 4 + 3 + 2 + 1 = 10.
19 (a) The transpose of RT AR is RT AT RTT = RT AR = n by n
(b) (RT R)jj = (column j
of R) (column j of R) = length squared of column j .
13
10
1
0
13
1b
10
1
0
1b
=
;
=
.
20
32
31
0 7
01
bc
b1
0 c b2
01
5 7
d b2 e bc
,
.
21 Lower right 2 by 2 matrix is
7 32
e bc f c2
1
0
1
1
1
1
2
0
01
1
22 1 0
A = 0 1
1
1
1
1 ;
0 1 A = 1 1
1
231
1
10
201
1
001
23 A = 1 0 0 = P and L = U = I ; exchanges rows 12 then rows 23.
010
1
012
1
2
1
1
24
0 3 8 = 0
1
1
3
8 . If we wait to exchange, then
1
211
0 1/3 1
2/3
1
1
211
A = L 1 P 1 U1 = 3 1
1
0 1 2 .
002
1
1
23
01
and P
; no more elimination
25 abs(A(1, 1)) = 0 and abs(A(2, 1)) > tol; A
01
10
234
so L = I and U = new A. abs(A(1, 1)) = 0 and abs(A(2, 1)) > tol; A 0 0 1 and
056
010
234
010
abs(A(2, 2)) = 0
P 1 0 0 ;
; A 0 5 6 , L = I , P 0 0 1 .
abs(A(3, 2)) > tol
001
100
001
110
26 abs(A(1, 1)) = 0 so nd abs(A(2, 1)) > tol; exchange rows to A = 0 1 2 and P =
254
010
110
100
1 0 0 ; eliminate to A = 0 1 2 and L = 0 1 0 , same P ; abs(A(2, 2)) > tol
001
034
201
1
1
0
100
so eliminate to A = 0
1
2 = nal U and L = 0 1 0 .
231
0
0 2
27 No solution
23
1
28 L1 = 1
2
1
0
1
shows the elimination steps as actually done (L is aected by P ).
29 One way to decide even vs. odd is to count all pairs that P has in the wrong order. Then
P is even or odd when that count is even or odd. Hard step: show that an exchange always
reverses that count! Then 3 or 5 exchanges will leave that count odd.
1
100
1
T
30 E21 = 3 1
and E21 AE21 = 0 2 4 is still symmetric; E32 =
1
1
049
4
1
T
T
and E32 E21 AE21 E32 = D. Elimination from both sides gives the symmetric LDLT directly.
yBC
yBC + yBS
1
0
1
31 Total currents are AT y = 1
1
0 yCS = yBC + yCS .
0 1 1
yBS
yCS yBS
Either way (Ax )T y = x T (AT y ) = xB yBC + xB yBS xC yBC + xC yCS xS yCS xS yBS .
700
1
50
x1
1
40
2
6820
1 truck
3 =
32 Inputs 40 1000 = Ax ; AT y =
x2
50 1000 50
188000
1 plane
2
50
3000
33 Ax y is the cost of inputs while x AT y is the value of outputs.
34 P 3 = I so three rotations for 360 ; P rotates around (1, 1, 1) by 120 .
12
10
12
=
= EH
35
49
21
25
36 L(U T )1 = triangular times triangular. The transpose of U T DU is U T DT U TT = U T DU
again.
37 These are groups: Lower triangular with diagonal 1s, diagonal invertible D, permutations P ,
orthogonal
012
1 2 3
38
2 3 0
301
matrices with QT = Q1 .
3
0
(I dont know any rules for constructions like this)
1
2
39 Reordering the rows and/or columns of
ab
cd
will move the entry a.
40 Certainly B T is northwest. B 2 is a full matrix! B 1 is southeast:
1 1 1
10
=
rows of B are in reverse order from a lower triangular L, so B = P L. Then B
0
1
1 1
1
=L
. The
1
P 1
has the columns in reverse order from L1 . So B 1 is southeast. Northwest times southeast is
upper triangular! B = P L and C = P U give BC = (P LP )U = upper times upper.
41 The i, j entry of P AP is the n i + 1, n j + 1 entry of A. The main diagonal reverses order.
24
Problem Set 3.1, Page 107
1 x + y = y + x and x + (y + z ) = (x + y ) + z and (c1 + c2 )x = c1 x + c2 x .
2 The only broken rule is 1 times x equals x .
3 (a) cx may not be in our set: not closed under scalar multiplication. Also no 0 and no x
(b) c(x + y ) is the usual (xy )c , while cx + cy is the usual (xc )(y c ). Those are equal. With
c = 3, x = 2, y = 1 they equal 8. This is 3(2 + 1)!! The zero vector is the number 1.
00
1 1
2
2
1
; A =
and A =
. The smallest
4 The zero vector in M is
2
00
1 1
2
2
subspace containing A consists of all matrices cA.
5 (a) One possibility: The matrices cA form a subspace not containing B
subspace must contain A B = I
(b) Yes: the
(c) All matrices whose main diagonal is all zero.
6 h(x) = 3f (x) 4g (x) = 3x2 20x.
7 Rule 8 is broken: If cf (x) is dened to be the usual f (cx) then (c1 + c2 )f = f ((c1 + c2 )x) is
dierent from c1 f + c2 f = usual f (c1 x) + f (c2 x).
8 If (f + g )(x) is the usual f (g (x)) then (g + f )x is g (f (x)) which is dierent. In Rule 2 both
sides are f (g (h(x))). Rule 4 is broken because there might be no inverse function f 1 (x) such
that f (f 1 (x)) = x. If the inverse function exists it will be the vector f .
9 (a) The vectors with integer components allow addition, but not multiplication by
1
2
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is
allowed but not all vector additions.
10 Only (a) (d) (e) are subspaces.
ab
a
11 (a) All matrices
(b) All matrices
00
0
a
0
(c) All diagonal matrices.
12 The sum of (4, 0, 0) and (0, 4, 0) is not on the plane.
13 P0 has the equation x + y 2z = 0; (2, 0, 1) and (0, 2, 1) and their sum (2, 2, 2) are in P0 .
14 (a) The subspaces of R2 are R2 itself, lines through (0, 0), and (0, 0) itself
4
(b) The sub-
4
spaces of R are R itself, three-dimensional planes n v = 0, two-dimensional subspaces
(n 1 v = 0 and n 2 v = 0), one-dimensional lines through (0, 0, 0, 0), and (0, 0, 0, 0) alone.
15 (a) Two planes through (0, 0, 0) probably intersect in a line through (0, 0, 0)
and line probably intersect in the point (0, 0, 0)
(b) The plane
(c) Suppose x is in S T and y is in
S T . Both vectors are in both subspaces, so x + y and cx are in both subspaces.
16 The smallest subspace containing P and L is either P or R3 .
10
00
+
is not singular.
17 (a) The zero matrix is not invertible
(b)
00
01
18 (a) True
(b) True
(b) False.
19 The column space of A is the x axis = all vectors (x, 0, 0). The column space of B is the xy
plane = all vectors (x, y, 0). The column space of C is the line of vectors (x, 2x, 0).
25
20 (a) Solution only if b2 = 2b1 and b3 = b1
(b) Solution only if b3 = b1 .
21 A combination of the columns of C is also a combination of the columns of A (same column
space; B has a dierent column space).
22 (a) Every b
(b) Solvable only if b3 = 0
(c) Solvable only if b3 = b2 .
23 The extra
column b
enlarges the column space unless b is already in the column space of A:
101
(larger column space)
101
(b already in column space)
[A b ] =
001
(no solution to Ax = b )
011
(Ax = b has a solution)
24 The column space of AB is contained in (possibly equal to) the column space of A. If B = 0
and A = 0 then AB = 0 has a smaller column space than A.
25 The solution to Az = b + b is z = x + y . If b and b are in the column space so is b + b .
26 The column space of any invertible 5 by 5 matrix is R5 . The equation Ax = b is always
solvable (by x = A1 b ) so every b is in the column space.
27 (a) False
11
28 A = 1 0
01
(b) True
0
1
0 or 1
0
0
(c) True
(d) False.
12
120
0 1 ; A = 2 4 0 (columns on 1 line).
11
360
29 Every b is in the column space so that space is R9 .
Problem Set 3.2, Page 118
1
1 (a) U = 0
0
2
2
4
6
0
1
2
0
0
0
Free variables x2 , x4 , x5
3
Pivot variables x1 , x3
0
2
(b) U = 0
0
4
2
4
Free x3
4
Pivot x1 , x2
0
0
2 (a) Free variables x2 , x4 , x5 and solutions (2, 1, 0, 0, 0), (0, 0, 2, 1, 0), (0, 0, 3, 0, 1)
(b) Free variable x3 : solution (1, 1, 1).
3 The complete solutions are (2x2 , x2 , 2x4 3x5 , x4 , x5 ) and (2x3 , x3 , x3 ).
The nullspace contains only 0 when there are no free variables.
12000
1
0 1
4 R = 0 0 1 2 3 , R = 0
1
1 , R has the same nullspace as U and A.
00000
0
0
0
1 3
5
10
1 3 5
1 3 5
10
1
3
5
=
;
=
.
5
2 6 10
21
000
2 6 7
21
0
0 3
6 (a) Special solutions (3, 1, 0) and (5, 0, 1)
(b) (3, 1, 0). Total count of pivot and free is n.
7 (a) Nullspace of A is the plane x + 3y + 5z = 0; it contains all vectors (3y + 5z, y, z )
(b) The line through (3, 1, 0) has equations x + 3y + 5z = 0 and 2x + 6y + 7z = 0.
1 3 5
1 3
0
10
with I = [ 1 ]; R =
with I =
.
8 R=
0
0
0
0
0
1
01
9 (a) False
(b) True
(c) True (only n columns)
(d) True (only m rows).
26
10 (a) Impossible above diagonal
1
(b) A = invertible = 1
1
(d) A = 2I, U = 2I, R = I .
0111111
11
0 0 0 1 1 1 1 0 0
11
0 0 0 0 1 0 0 0 0
0000000
00
11011100
0
0 0 1 1 1 1 0 0 0
,
12
0 0 0 0 0 0 1 0 0
00000001
0
1
1
1
1
1
1
1
1
0
0
0
1
0
0
0
0
1
1
1
1
1
1
0
0
1
1
0
0
1
0
1
1
0
0
0
1
1
1
0
0
0
0
0
0
1
2
1
0
1
1
2
0
0 0
0 0
00
1
1
.
1
0
1
(c) A = 1
1
0
1
1
1
0
0
0
1
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
13 If column 4 is all zero then x4 is a free variable. Its special solution is (0, 0, 0, 1, 0).
14 If column 1 = column 5 then x5 is a free variable. Its special solution is (1, 0, 0, 0, 1).
15 There are n r special solutions. The nullspace contains only x = 0 when r = n. The column
space is Rm when r = m.
16 The nullspace contains only x = 0 when A has 5 pivots. Also the column space is R5 , because
we can solve Ax = b and every b is in the column space.
17 A = [ 1 3 1 ]; y and z are free; special solutions (3, 1, 0) and (1, 0, 1).
18 Fill in 12 then 3 then 1.
19 If LU x = 0, multiply by L1 to nd U x = 0. Then U and LU have the same nullspace.
20 Column 5 is sure to have no pivot since it is a combination of earlier columns. With 4 pivots in
the other columns, the special solution is s = (1, 0, 1, 0, 1). The nullspace contains all multiples
of s (a line in R5 ).
21 Free variables
1
22 A = 0
0
1
23 A = 1
5
0
1
0
0
3
1
1
0
x3 , x4 : A =
0 1
0 4
0 3 .
1 2
1/2
2 .
3
2
3
2
1
.
24 This construction is impossible: 2 pivot columns, 2 free variables, only 3 columns.
1 1
0
0
25 A = 1
0 1
0 .
1
0
0 1
01
.
26 A =
00
27
27 If nullspace = column space (r pivots) then n r = r. If n = 3 then 3 = 2r is impossible.
28 If A times every column of B is
zero, the column space of B is contained in the nullspace of
11
1
1
, B =
.
A: A =
11
1 1
29 R is most likely to be I ; R is most likely to be I with fourth row of zeros.
01
10
T
shows that (a)(b)(c) are all false. Notice rref(A ) =
.
30 A =
00
00
1 0 0 2
31 Three pivots (4 columns and 1 special solution); R = 0 1 0 1 (add any zero rows).
0010
100
, R = I .
32 Any zero rows come after these rows: R = [ 1 2 3 ], R =
010
10
10
11
01
00
,
,
,
,
33 (a)
(b) All 8 matrices are Rs !
01
00
00
00
00
34 One reason: A and A have the same nullspace (and also the same column space).
Problem Set 3.3, page 128
1 (a) and (c) are correct; (d) is false because R might happen to have 1s in nonpivot columns.
1111
1
0 1 2
1 1
1 1
2 R = 0 0 0 0 r = 1; R = 0
1
2
3 r = 2; R = 0
0
0
0 r = 1
0000
0
0
0
0
0
0
0
0
120
RA
0
Zero row in the upper
3 RA = 0 0 1
RB = RA RA
RC
0
RA
000
R moves all the way to the bottom.
0I
I
. The nullspace matrix is N = .
4 If all pivot variables come last then R =
00
0
5 I think this is true.
6 A and AT have the same rank r. But pivcol (the column number) is 2 for A and 1 for AT :
010
A = 0 0 0 .
000
2 3
1
0
4 5
7 The special solutions are the columns of N =
and
N = 0 2 .
1
0
0
1
0
1
28
1
8 A = 2
4
2
4
8
4
2
3
6
a
3 3/2 , M =
c
6
3
8 , B = 1
16
2
b
bc/a
.
9 If A has rank 1, the column space is a line in Rm . The nullspace is a plane in Rn (given by
one equation). The column space of AT is a line in Rn .
10 u = (3, 1, 4), v = (1, 2, 2); u = (2, 1), v = (1, 1, 3, 2).
11 A rank one matrix has one pivot. The second row of U is zero.
13
10
.
12 S =
and
S= 1
and
S=
14
01
13 P has rank r (the same as A) because elimination produces the same pivot columns.
14 The rank of RT is also r, and the example matrix A has rank 2:
1
P = 2
2
3
6
7
P
T
=
1
2
2
3
6
7
T
S =
1
2
3
7
S=
1
3
2
7
.
15 Rank(AB ) = 1; rank(AM ) = 1 except AM = 0 if c = 1/2.
16 (uv T )(w z T ) = u (v T w )z T has rank one unless v T w = 0.
17 (a) By matrix multiplication, each column of AB is A times the corresponding column of B .
So a combination of columns of B turns into a combination of columns of AB .
(b) The rank of B is r = 1. Multiplying by A cannot increase this rank. The rank stays the
same for A1 = I and it drops to zero for A2 = 0 or A2 = [ 1 1; 1 1 ].
18 If we know that rank(B T AT ) rank(AT ), then since rank stays the same for transposes, we
have rank(AB ) rank(A).
19 We are given AB = I which has rank n. Then rank(AB ) rank(A) forces rank(A) = n.
20 Certainly A and B have at most rank 2. Then their product AB has at most rank 2. Since BA
is 3 by 3, it cannot be I even if AB = I :
A=
1
0
0
1
0
0
1
B = 0
0
0
1
0
AB = I
and
BA = I.
21 (a) A and B will both have the same nullspace and row space as R (same R for both matrices).
(b) A equals an invertible matrix times B , when they share the same R. A key fact!
1
3
0
2 1
10
1
3
0
2 1
(nonzero rows of R).
22 A = 0
0
1
4 3 = 0 1
0
0
1
4 3
1
3
1
6 4
11
29
0
1
4
0
8
1
23 A = (pivot columns)(nonzero rows of R) = 1
1
1
B = 1
1
0
1
4
0
8
1
0
0
1
1
1
0
0
0
1
= 1
1
1
1
0
1
1
= 1
1
1
0
1
0
1
1
1
0
1
1
1
0
1
1
0
0
0
1
0 + 0
0
0
0
0
0
0
0
4
0
0
0
8
0
0
1
0
0 + 0
0
0
0
0
4 .
8
0
0
0
4 .
8
24 The m by n matrix Z has r ones at the start of its main diagonal. Otherwise Z is all zeros.
is decided by the rank which is the same for A and AT .
2
1022
0 has x2 , x3 , x4 free. If c = 1, R = 0 1 0 0 has x3 , x4 free.
0
0000
1 2 2
2 2
1
0
0
0
0
(c = 1) and N =
(c = 1)
in N =
0
1
1
0
0
0
0
1
0
1
1
1 2
and x1 free; if c = 2, R =
and x2 free; R = I if c = 1, 2
0
0
0
2
1
in N = (c = 1) or N = (c = 2) or N = 2 by 0 empty matrix.
1
0
I
= ; N = empty.
I
25 Y = Z because the form
112
26 If c = 1, R = 0 0 0
000
Special solutions
If c = 1, R =
0
0
Special solutions
I
27 N = ;
I
N
Problem Set 3.4, page 136
2
1 2
2
4
6
4
5
7
6
3
5
2
b1
2
b2 0
b3
0
4
6
4
1
1
2
1
1
2
b1
2
b2 b1 0
b3 b1
0
4
6
4
b1
1
1
2
b2 b1
0
0
0
b3 + b2 2b1
Ax = b has a solution when b3 + b2 2b1 = 0; the column space contains all combinations
of (2, 2, 2) and (4, 5, 3) which is the plane b3 + b2 2b1 = 0 (!); the nullspace contains all
combinations of s 1 = (1, 1, 1, 0) and s 2 = (2, 2, 0, 1); x complete = x p + c1 s 1 + c2 s 2 ;
1
[ R d ] = 0
0
2
1
3
b1
0
1
2
1
1
2
0
0
0
2
1
3
4
1 gives the particular solution x p = (4, 1, 0, 0).
0
b1
1
1/2
3/2
5
2 6 3 9 b2 0 0 0 b2 3b1 Then [ R d ] = 0 0
0
0 Ax = b
4 2 6 b3
0 0 0 b3 2b1
00
0
0
has a solution when b2 3b1 = 0 and b3 2b1 = 0; the column space is the line through (2, 6, 4)
30
which is the intersection of the planes b2 3b1 = 0 and b3 2b1 = 0; the nullspace contains all
combinations of s 1 = (1/2, 1, 0) and s 2 = (3/2, 0, 1); particular solution x p = (5, 0, 0) and
complete solution x p + c1 s 1 + c2 s 2 .
2
3
3x
= + x2 1 .
complete 0
1
0
1/2
3
0
0
1
0
+ x2 + x4 .
4x
=
complete 1/2
0
2
0
0
1
5b1 2b2
2
5 Solvable if 2b1 + b2 = b3 . Then x = b2 2b1 + x3 0 .
0
1
6 (a) Solvable if b2 = 2b1 and 3b1 3b3 + b4 = 0. Then x =
(b) Solvable if b2 = 2b1 and
1
7 3
2
3
1
8
2
4
0
b1
1
3
b2 0 1
b3
0 2
5b1 2b3
(no free variables)
b3 2b1
5b1 2b3
1
3b1 3b3 + b4 = 0. Then x = b3 2b1 + x3 1 .
0
1
1
b2
row 3 2 (row 2) + 4 (row 1)
1
b2 3b1
is the zero row
2
b3 2b1
[ 0 0 0 b3 2b2 + 4b1 ]
8 (a) Every b is in the column space:
1
00
12
9 L[ U c ] = 2
1 0 0 0
3 1 1
00
independent rows. (b) Need b3 = 2b2 . Row 3 2 row 2 = 0.
3 5 b1
= [ A b ];
2 2 b2 2b1
0 0 b3 + b2 5b1
x p = (9, 0, 3, 0) so 9(1, 2, 3) + 3(3, 8, 7) = (0, 6, 6) is exactly Ax p = b .
1
0 1
2
x = .
10
0
1 1
4
11 A 1 by 3 system has at least two free variables.
12 (a) x 1 x 2 and 0 solve Ax = 0
(b) 2x 1 2x 2 solves Ax = 0; 2x 1 x 2 solves Ax = b .
13 (a) The particular solution x p always
is
multiplied by 1
(b) Any solution can be the
par
33
x
6
1
2
= . Then is shorter (length 2) than
ticular solution
(c)
33
y
6
1
0
(d) The homogeneous solution in the nullspace is x n = 0 when A is invertible.
14 If column 5 has no pivot, x5 is a free variable. The zero vector is not the only solution to
Ax = 0. If Ax = b has a solution, it has innitely many solutions.
15 If row 3 of U has no pivot, that is a zero row. U x = c is solvable only if c3 = 0. Ax = b might
not be solvable, because U may have other zero rows.
31
16 The largest rank is 3. Then there is a pivot in every row. The solution always exists. The
column space is R3 . An example is A = [ I F ] for any 3 by 2 matrix F .
17 The largest rank is 4. There is a pivot in every column. The solution is unique. The nullspace
contains only the zero vector. An example is A = [ I ; G ] for any 4 by 2 matrix G.
18 Rank = 3; rank = 3 unless q = 2 (then rank = 2).
19 All ranks = 2.
1
0
3
4
1
0
1
0
0
1
0
1
0
; A = 2 1 0 0
2 2
3 .
21
0 3
0
1
031
0
0 11 5
x
4
1
1
x
4
1
21 (a) y = 0 + y 1 + z 0
(b) y = 0 + z 0 .
z
0
0
1
z
0
1
20 A =
22 If Ax 1 = b and Ax 2 = b then we can add x 1 x 2 to any solution of Ax = B . But there will
be no solution to Ax = B if B is not in the column space.
23 For A, q = 3 gives rank 1, every other q gives rank 2. For B , q = 6 gives rank 1, every other q
gives rank 2.
1
24 (a)
(b) [ 1 1 ]
1
(c) [ 0 ] or any r < m, r < n
25 (a) r < m, always r n
(b) r = m, r < n
1
0 2
26 R = 0
1
2 , R = I .
0
0
0
(d) Invertible.
(c) r < m, r = n
(d) r = m = n.
27 R has n pivots equal to 1. Zeros above and below pivots make R = I .
2
1
1230
1200
1235
1 2 0 1
; x n = 1 ;
x p = 0 .
28
0040
0010
0048
001 2
0
2
The pivot columns contain I so 1 and 2 go into x p .
0
1
0
0 1
1000
29 R = 0 0 1 0 and x n = 1 ; 0
0
1
2 : no solution because of row 3.
0000
0
0
0
0
5
4
2
1023 2
102
32
1 0 2 0 4
3
0
30 1 3 2 0 5 0 3 0 3 3 0 1 0 0
3 ; x p = and x n = x3 .
0
1
2 0 4 9 10
000
36
0001
2
2
0
11
31 A = 0 2 ; B cannot exist since 2 equations in 3 unknowns cannot have a unique solution.
03
32
1
1
32 A = LU =
2
1
10
.
36 A =
30
0
0
1
0
2
1
2
0
0
1
3
0 0 1
0 0
0
1
0
0
1
7
7
2
and x = 2 + x3 2 and then no solution.
0
0
1
0
Problem Set 3.5, page 150
1
1 0
0
1
1
0
1
c1
1 c2 = 0 gives c3 = c2 = c1 = 0. But v 1 + v 2 4v 3 + v 4 = 0 (dependent).
1
c3
2 v 1 , v 2 , v 3 are independent. All six vectors are on the plane (1, 1, 1, 1) v = 0 so no four of
these six vectors can be independent.
3 If a = 0 then column 1 = 0; if d = 0 then b(column 1) a(column 2) = 0; if f = 0 then all
columns end in zero (all are perpendicular to (0, 0, 1), all in the xy plane, must be dependent).
abc
x
0
4 U x = 0 d e y = 0 gives z = 0 then y = 0 then x = 0.
00f
z
0
123
1
2
3
1
2
3
5 (a) 3 1 2 0 5 7 0 5
7 : invertible independent columns
231
0 1 5
0
0 18/5
1
2 3
1
2 3
1
2 3
1
0
(b) 3
1
2 0
7 7 0
7 7 ; A 1 = 0 , columns add to 0.
2 3
1
0 7
7
0
0
0
1
0
6 Columns 1, 2, 4 are independent. Also 1, 3, 4 and 2, 3, 4 and others (but not 1, 2, 3). Same
column numbers (not same columns!) for A.
7 The sum v 1 v 2 + v 3 = 0 because (w 2 w 3 ) (w 1 w 3 ) + (w 1 w 2 ) = 0.
8 If c1 (w 2 + w 3 )+ c2 (w 1 + w 3 )+ c3 (w 1 + w 2 ) = 0 then (c2 + c3 )w 1 +(c1 + c3 )w 2 +(c1 + c2 )w 3 = 0.
Since the w s are independent this requires c2 + c3 = 0, c1 + c3 = 0, c1 + c2 = 0. The only
solution is c1 = c2 = c3 = 0. Only this combination of v 1 , v 2 , v 3 gives zero.
9 (a) The four vectors are the columns of a 3 by 4 matrix A. There is a nonzero solution to
Ax = 0 because there is at least one free variable
(b) dependent if [ v 1 v 2 ] has rank 0 or 1
(c) 0v 1 + 3(0, 0, 0) = 0.
10 The plane is the nullspace of A = [ 1 2 3 1 ]. Three free variables give three solutions
(x, y, z, t) = (2, 1, 0, 0) and (3, 0, 1, 0) and (1, 0, 0, 1).
11 (a) Line in R3
(b) Plane in R3
(c) Plane in R3
(d) All of R3 .
33
12 b is in the column space when there is a solution to Ax = b ; c is in the row space when there
is a solution to AT y = c . False. The zero vector is always in the row space.
13 All dimensions are 2. The row spaces of A and U are the same.
14 The dimension of S is
(a) zero when x = 0
(b) one when x = (1, 1, 1, 1)
(c) three
when x = (1, 1, 1, 1) because all rearrangements of this x are perpendicular to (1, 1, 1, 1)
(d) four when the xs are not equal and dont add to zero. No x gives dim S = 2.
15 v = 1 (v + w ) + 1 (v w ) and w = 1 (v + w ) 1 (v w ). The two pairs span the same space.
2
2
2
2
They are a basis when v and w are independent.
16 The n independent vectors span a space of dimension n. They are a basis for that space. If
they are the columns of A then m is not less than n (m n).
17 These bases are not unique!
(a) (1, 1, 1, 1)
(c) (1, 1, 1, 0), (1, 1, 0, 1)
(b) (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)
(d) (1, 0)(0, 1); (1, 0, 1, 0, 0), (0, 1, 0, 1, 0), (1, 0, 0, 0, 1).
18 Any bases for R2 ; (row 1 and row 2) or (row 1 and row 1 + row 2).
19 (a) The 6 vectors might not span R4
(b) The 6 vectors are not independent
(c) Any four might be a basis.
20 Independent columns rank n. Columns span Rm rank m. Columns are basis for Rm
rank = m = n.
21 One basis is (2, 1, 0), (3, 0, 1). The vector (2, 1, 0) is a basis for the intersection with the xy
plane. The normal vector (1, 2, 3) is a basis for the line perpendicular to the plane.
22 (a) The only solution is x = 0 because the columns are independent
(b) Ax = b is solvable
5
because the columns span R .
23 (a) True
(b) False because the basis vectors may not be in S.
24 Columns 1 and 2 are bases for the (dierent) column spaces; rows 1 and 2 are bases for the
(equal) row spaces; (1, 1, 1) is a basis for the (equal) nullspaces.
25 (a) False for [ 1 1 ]
(b) False
(c) True: Both dimensions = 2 if A is invertible, dimen-
sions = 0 if A = 0, otherwise dimensions = 1
(d) False, columns may be dependent.
26 Rank 2 if c = 0 and d = 2; rank 2 except when c = d or c = d.
000
100
000
000
010
001
27 (a) 0 0 0 , 0 1 0 , 0 0 0 (b) Add 1 0 0 , 0 0 0 , 0 0 1
000
000
001
000
100
010
010
001
0
00
T
(c) 1 0 0 , 0 0 0 , 0
0 1 are a basis for all A = A .
000
1 0 0
0 1 0
100
100
110
101
100
28 I , 0 1 0 , 0 2 0 , 0 1 0 , 0 1 0 , 0 1 1 ; echelon matrices do
002
001
001
001
001
not form a subspace; they span the upper triangular matrices (not every U is echelon).
34
1
0
0
0
1
0
0
,
0
0
0
1
1 1
,
29
1 0 0
0 1
1
1
30
+ 1
1
1
1
31 (a) All 3 by 3 matrices
1
2
0
1
,
32
0
0
0
0
0
(b) Upper triangular matrices
0
2
0
0
0
0
,
,
0
0
1
2
0
1
33 (a) y (x) = constant C
1
1
;
0 1
1
1 +
(b) y (x) = 3x
1
1
1
0 1
and
.
0
1
0
1
1
1
1
+
=0
1 1
1
1
1
(c) All multiples cI .
0
0
.
0
2
(c) y (x) = 3x + C = y p + y n .
34 y (0) = 0 requires A + B + C = 0. One basis is cos x cos 2x and cos x cos 3x.
35 (a) y (x) = e2x
(b) y = x (one basis vector in each case).
36 y1 (x), y2 (x), y3 (x) can be x, 2x, 3x (dim 1) or x, 2x, x2 (dim 2) or x, x2 , x3 (dim 3).
37 Basis 1, x, x2 , x3 ; basis x 1, x2 1, x3 1.
38 Basis for S: (1, 0, 1, 0), (0, 1, 0, 0), (1, 0, 0, 1); basis for T: (1, 1, 0, 0) and (0, 0, 2, 1); S T
has dimension 1.
39 See Solution 30 for I = combination of ve other P s. Check the (1, 1) entry, then (3, 2), then
(3, 3), then (1, 2) to show that those ve P s are independent.
Four conditions on the 9 entries make all row sums and column sums equal: row sum 1 = row
sum 2 = row sum 3 = column sum 1 = column sum 2 (= column sum 3 is automatic).
40 The subspace of matrices that have AS = SA has dimension three.
41 (a) No, dont span
(b) No, dependent
(c) Yes, a basis
(d) No, dependent
42 If the 5 by 5 matrix [ A b ] is invertible, b is not a combination of the columns of A. If [ A b ]
is singular, and the 4 columns of A are independent, b is a combination of those columns.
Problem Set 3.6, page 161
1 (a) Row and column space dimensions = 5, nullspace dimension = 4, left nullspace dimension
= 2 sum = 16 = m + n
(b) Column space is R3 ; left nullspace contains only 0.
2 A: Row space (1, 2, 4); nullspace (2, 1, 0) and (4, 0, 1); column space (1, 2); left nullspace
(2, 1). B : Row space (1, 2, 4) and (2, 5, 8); column space (1, 2) and (2, 5);
nullspace (4, 0, 1); left nullspace basis is empty.
3 Row space (0, 1, 2, 3, 4) and (0, 0, 0, 1, 2); column space (1, 1, 0) and (3, 4, 1); nullspace basis
(1, 0, 0, 0, 0), (0, 2, 1, 0, 0), (0, 2, 0, 2, 1); left nullspace (1, 1, 1).
10
9 3
4 (a) 1 0
(b) Impossible: r + (n r) must be 3
(c) [ 1 1 ]
(d)
3
1
01
(e) Impossible: Row space = column space requires m = n. Then m r = n r.
35
5 A=
1
1
1
2
1
0
, B = 1 2 1 .
6 A: Row space (0, 3, 3, 3) and (0, 1, 0, 1); column space (3, 0, 1) and (3, 0, 0);
nullspace (1, 0, 0, 0) and (0, 1, 0, 1); left nullspace (0, 1, 0). B : Row space (1), column space
(1, 4, 5), nullspace: empty basis, left nullspace (4, 1, 0) and (5, 0, 1).
7 Invertible A: row space basis = column space basis = (1, 0, 0), (0, 1, 0), (0, 0, 1); nullspace basis
and left nullspace basis are empty. Matrix B : row space basis (1, 0, 0, 1, 0, 0), (0, 1, 0, 0, 1, 0)
and (0, 0, 1, 0, 0, 1); column space basis (1, 0, 0), (0, 1, 0), (0, 0, 1); nullspace basis
(1, 0, 0, 1, 0, 0) and (0, 1, 0, 0, 1, 0) and (0, 0, 1, 0, 0, 1); left nullspace basis is empty.
8 Row space dimensions 3, 3, 0; column space dimensions 3, 3, 0; nullspace dimensions 2, 3, 2; left
nullspace dimensions 0, 2, 3.
9 (a) Same row space and nullspace. Therefore rank (dimension of row space) is the same
(b) Same column space and left nullspace. Same rank (dimension of column space).
10 Most likely rank = 3, nullspace and left nullspace contain only (0, 0, 0). When the matrix is 3
by 5: Most likely rank = 3 and dimension of nullspace is 2.
11 (a) No solution means that r < m. Always r n. Cant compare m and n
(b) If m r > 0, the left nullspace contains a nonzero vector.
11
221
101
= 2 4 0 ; r + (n r) = n = 3 but 2 + 2 is 4.
12 0 2
120
10
101
13 (a) False
(b) True
(c) False (choose A and B same size and invertible).
14 Row space basis (1, 2, 3, 4), (0, 1, 2, 3), (0, 0, 1, 2); nullspace basis (0, 1, 2, 1); column space
basis (1, 0, 0), (0, 1, 0), (0, 0, 1); left nullspace has empty basis.
15 Row space and nullspace stay the same; (2, 1, 3, 4) is in the new column space.
16 If Av = 0 and v is a row of A then v v = 0.
17 Row space = yz plane; column space = xy plane; nullspace = x axis; left nullspace = z axis.
For I + A: Row space = column space = R3 , nullspaces contain only zero vector.
18 Row 3 2 row 2 + row 1 = zero row so the vectors c(1, 2, 1) are in the left nullspace. The
same vectors happen to be in the nullspace.
19 Elimination leads to 0 = b3 b2 b1 so (1, 1, 1) is in the left nullspace. Elimination leads
to b3 2b1 = 0 and b4 + b2 4b1 = 0, so (2, 0, 1, 0) and (4, 1, 0, 1) are in the left nullspace.
20 (a) All combinations of (1, 2, 0, 0) and ( 1 , 0, 3, 1)
4
21 (a) u and w
dependent
12
1
22 2 2
0
41
(b) v and z
(d) The rank
1
00
= 2
11
4
(b) One
(c) (1, 2, 3), (0, 1, 4).
(c) rank < 2 if u and w are dependent or v and z are
T
of uv + wz T is 2.
22
2 2 .
11
36
23 Row space basis (3, 0, 3), (1, 1, 2); column space basis (1, 4, 2), (2, 5, 7); rank is only 2.
24 AT y = d puts d in the row space of A; unique solution if the left nullspace (nullspace of AT )
contains only y = 0.
25 (a) True (same rank)
unsymmetric)
(b) False A = [ 1 0 ]
(c) False (A can be invertible and also
(d) True.
26 The rows of AB = C are combinations of the rows of B . So rank C rank B . Also rank C
rank A. (The columns of C are combinations of the columns of A).
27 Choose d = bc/a. Then the row space has basis (a, b) and the nullspace has basis (b, a).
28 Both ranks are 2; if p = 0, rows 1 and 2 are a basis for the row space. N(B T ) has six vectors
with 1 and 1 separated by a zero; N(C T ) has (1, 0, 0, 0, 0, 0, 0, 1) and (0, 1, 0, 0, 0, 0, 1, 0)
and columns 3, 4, 5, 6 of I ; N(C ) is a challenge.
29 a11 = 1, a12 = 0, a13 = 1, a22 = 0, a32 = 1, a31 = 0, a23 = 1, a33 = 0, a21 = 1 (not unique).
Problem Set 4.1, page 171
1 Both nullspace vectors are orthogonal to the row space vector in R3 . Column space is perpendicular to the nullspace of AT in R2 .
2 The nullspace is Z (only zero vector) so x n = 0. and row space = R2 . Plane line in R3 .
1
2 3
2
1
1
1
3 (a) 2 3
1 (b) Impossible, 3 not orthogonal to 1 (c) 1 in C (A) and 0
3
5 2
5
1
1
0
1 1
1 1
in N (AT ) is impossible: not perpendicular (d) This asks for A2 = 0; take A =
(e) (1, 1, 1) will be in the nullspace and row space; no such matrix.
4 If AB = 0, the columns of B are in the nullspace of A. The rows of A are in the left nullspace
of B . If rank = 2, all four subspaces would have dimension 2 which is impossible for 3 by 3.
5 (a) If Ax = b has a solution and AT y = 0, then y is perpendicular to b . (Ax )T y = b T y = 0.
(b) c is in the row space, x is in the nullspace: c T x = y T Ax = y T 0 = 0.
6 Multiply the equations by y1 = 1, y2 = 1, y3 = 1. They add to 0 = 1 so no solution:
y = (1, 1, 1) is in the left nullspace. Cant have 0 = (y T A)x = y T b = 1.
7 Multiply by y = (1, 1, 1), then x1 x2 = 1 plus x2 x3 = 1 minus x1 x3 = 1 is 0 = 1.
8 x = x r + x n , where x r is in the row space and x n is in the nullspace. Then Ax n = 0 and
Ax = Ax r + Ax n = Ax r . All vectors Ax are combinations of the columns of A.
9 Ax is always in the column space of A. If AT Ax = 0 then Ax is also in the nullspace of AT .
Perpendicular to itself, so Ax = 0.
10 (a) For a symmetric matrix the column space and row space are the same
(b) x is in the
nullspace and z is in the column space = row space: so these eigenvectors have x T z = 0.
37
11 The nullspace of A is spanned by (2, 1), the row space is spanned by (1, 2). The nullspace of
B is spanned by (0, 1), the row space is spanned by (1, 0).
12 x splits into x r + x n = (1, 1) + (1, 1) = (2, 0).
13 V T W = zero matrix makes each basis vector for V orthogonal to each basis vector for W .
Then every v in V is orthogonal to every w in W (they are combinations of the basis vectors).
14 Ax = B x means that [ A B ]
x
x
= 0. Three homogeneous equations in four unknowns
always have a nonzero solution. Here x = (3, 1) and x = (1, 0) and Ax = B x = (5, 6, 5) is in
both column spaces. Two planes in R3 must intersect in a line at least!
15 A p-dimensional and a q -dimensional subspace of Rn share at least a line if p + q > n.
16 AT y = 0 (Ax )T y = x T AT y = 0. Then y Ax and N (AT ) C (A).
17 If S is the subspace of R3 containing only the zero vector, then S is R3 . If S is spanned
by (1, 1, 1), then S is spanned by (1, 1, 0) and (1, 0, 1). If S is spanned by (2, 0, 0) and
(0, 0, 3), then S is spanned by (0, 1, 0).
151
. Therefore S is a subspace even if S is not.
18 S is the nullspace of A =
222
19 L is the 2-dimensional subspace (a plane ) in R3 perpendicular to L. Then (L ) is a 1dimensional subspace (a line ) perpendicular to L . In fact (L ) is L.
20 If V is the whole space R4 , then V contains only the zero vector. Then (V ) = R4 = V .
21 For example (5, 0, 1, 1) and (0, 1, 1, 0) span S = nullspace of A = [ 1 2 2 3; 1 3 3 2 ].
22 (1, 1, 1, 1) is a basis for P . A = [ 1 1 1 1 ] has the plane P as its nullspace.
23 x in V is perpendicular to any vector in V . Since V contains all the vectors in S , x is also
perpendicular to any vector in S . So every x in V is also in S .
24 Column 1 of A1 is orthogonal to the space spanned by the 2nd, 3rd, . . ., nth rows of A.
25 If the columns of A are unit vectors, all mutually perpendicular, then AT A = I .
2
2 1
T
T
26 A = 1
2
2 , A A = 9I is diagonal : (A A)ij = (column i of A)(column j of A).
2 1
2
27 The lines 3x + y = b1 and 6x + 2y = b2 are parallel. They are the same line if b2 = 2b1 . In that
case (b1 , b2 ) is perpendicular to (2, 1). The nullspace is the line 3x + y = 0. One particular
vector in the nullspace is (1, 3).
28 (a) (1, 1, 0) is in both planes. Normal vectors are perpendicular, but planes still intersect!
(b) Need three orthogonal vectors to span the whole orthogonal complement.
(c) Lines
1
29 A = 2
3
can meet without being orthogonal.
23
1
1 1
1 0 , B = 2 1
0 ; v can not be in the nullspace and row space, or in
01
3
0 1
the left nullspace and column space. These spaces are orthogonal and v T v = 0.
38
30 When AB = 0, the column space of B is contained in the nullspace of A. So rank(B ) 4
rank (A) = (dimension of the nullspace A).
31 null(N ) produces a basis for the row space of A (perpendicular to N (A)).
Problem Set 4.2, page 181
1 (a) a T b /a T a = 5/3; p = (5/3, 5/3, 5/3); e = (2/3, 1/3, 1/3)
(b) a T b /a T a = 1; p = (1, 3, 1); e = (0, 0, 0).
2 (a) p = (cos , 0)
(b) p = (0, 0) since a T b = 0.
111
5
131
1
1
1
1
2
3 P1 = 1 1 1 and P1 b = 5 and P1 = P1 . P2 =
3 9 3 and P2 b = 3 .
3
3
11
111
5
131
1
1 1
10
. P1 P2 = 0 and P1 + P2 is not a projection matrix.
, P 2 = 1
4 P1 =
2 1
00
1
1 2 2
4
4 2
1
1
5 P1 = 2
4
4 , P2 = 4
4 2 . P1 P2 = zero matrix because a 1 a 2 .
9
9
2
4
4
2 2
1
6 p 1 = ( 1 , 2 , 2 ) and p 2 = ( 4 , 4 , 2 ) and p 3 = ( 4 , 2 , 4 ). Then p 1 + p 2 + p 3 = (1, 0, 0) = b .
9
9
9
99
9
9
99
1 2 2
4
4 2
4 2
4
1
1
1
7 P1 + P2 + P3 = 2
4
4 + 4
4 2 + 2
1 2 = I .
9
9
9
2
4
4
2 2
1
4 2
4
8 p 1 = (1, 0) and p 2 = (0.6, 1.2). Then p 1 + p 2 = b .
9 Since A is invertible, P = A(AT A)1 AT = AA1 (AT )1 AT = I : project onto all of R2 .
0.2 0.4
0.2
10
0.2
, P 2 a 1 =
, P 1 =
, P1 P2 a 1 =
. No, P1 P2 = (P1 P2 )2 .
10 P2 =
0.4 0.8
0.4
00
0
11 (a) p = A(AT A)1 AT b = (2, 3, 0) and e = (0, 0, 4)
(b) p
100
0.5 0.5
12 P1 = 0 1 0 = projection on xy plane. P2 = 0.5 0.5
000
0
0
1000
0 1 0 0
.
13 p = (1, 2, 3, 0). P = square matrix =
0 0 1 0
0000
= (4, 4, 6) and e = (0, 0, 0).
0
0 .
1
14 The projection of this onto the column space of A is b itself, but P is not necessarily I .
b
5
8 4
1
P=
8 17
2 and p = (0, 2, 4).
21
4
2 20
15 The column space of 2A is the same as the column space of A. x for 2A is half of x for A.
39
16
1
(1, 2, 1)
2
+ 3 (1, 0, 1) = (2, 1, 1). Therefore b is in the plane. Projection shows P b = b .
2
17 P 2 = P and therefore (I P )2 = (I P )(I P ) = I P I IP + P 2 = I P . When P
projects onto the column space of A then I P projects onto the left nullspace of A.
18 (a) I P is the projection matrix onto (1, 1) in the perpendicular direction to (1, 1)
(b) I P is the projection matrix onto the plane x + y + z = 0 perpendicular to (1, 1, 1).
5/6
1/6
1/3
19 For any choice, say (1, 1, 0) and (2, 0, 1), the matrix P is 1/6
5/6 1/3 .
1/3 1/3
1/3
1
1/6 1/6 1/3
5/6
1/6
1/3
20 e = 1 , Q = ee T /e T e = 1/6
1/6
1/3 , P = I Q = 1/6
5/6 1/3 .
2
1/3
1/3
2/3
1/3 1/3
1/3
21
A(AT A)1 AT
2
= A(AT A)1 (AT A)(AT A)1 AT = A(AT A)1 AT . Therefore P 2 = P . P b is
always in the column space (where P projects). Therefore its projection P (P b ) is P b .
22 P T = A(AT A)1 AT
T
= A (AT A)1
T
AT = A(AT A)1 AT = P . (AT A is symmetric.)
23 If A is invertible then its column space is all of Rn . So P = I and e = 0.
24 The nullspace of AT is orthogonal to the column space C (A). So if AT b = 0, the projection
of b onto C (A) should be p = 0. Check P b = A(AT A)1 AT b = A(AT A)1 0 = 0.
25 The column space of P will be S (n-dimensional). Then r = dimension of column space = n.
26 A1 exists since the rank is r = m. Multiply A2 = A by A1 to get A = I .
27 Ax is in the nullspace of AT . But Ax is always in the column space of A. To be in both of
those perpendicular spaces, Ax must be zero. So A and AT A have the same nullspace.
28 P 2 = P = P T give P T P = P . Then the (2, 2) entry of P equals the (2, 2) entry of P T P which
is the length squared of column 2.
29 Set A = B T . Then A has independent columns. By 4G, AT A = BB T is invertible.
3
1 9 12
aa T
. We cant
30 (a) The column space is the line through a = so PC = T =
aa
25 12 25
4
use (AT A)1 because A has dependent columns.
T
(b) The row space is the line through
T
v = (1, 2, 2) and PR = v v /v v . Always PC A = A and APR = A and then PC APR = A !
Problem Set 4.3, page 192
1
1
1 A=
1
1
0
8
1
4
T
and b = give A A =
8
3
8
4
20
0
8
26
T
and A b =
36
112
.
1
3
5
1
AT Ax = AT b gives x = and p = Ax = and e = b p = . E = e
5
13
4
17
3
1
2
= 44.
40
1
0
1
2
1
1
0
1
8
5
1 C
1
= . Change the right side to p = ; x = exactly solves Ax = b .
8
13
3 D
4
4
20
17
3 p = A(AT A)1 AT b = (1, 5, 13, 17). e = (1, 3, 5, 3). e is indeed perpendicular to both
columns of A. The shortest distance e is 44.
4 E = (C + 0D)2 + (C + 1D 8)2 + (C + 3D 8)2 + (C + 4D 20)2 .
Then E/C = 2C +
2(C +D 8)+2(C +3D 8)+2(C +4D 20) = 0 and E/D 12(C +D 8)+32(C D 8)+
=
+3
4
8
C
36
=
.
4 2(C + 4D 20) = 0. These normal equations are again
8 26
D
112
5 E = (C 0)2 + (C 8)2 + (C 8)2 + (C 20)2 . AT = [ 1 1 1 1 ], AT A = [ 4 ] and AT b = [ 36 ]
and (AT A)1 AT b = 9 = best height C . Errors e = (9, 1, 1, 11).
6 x = a T b /a T a = 9 and projection p = (9, 9, 9, 9); e T a = (9, 1, 1, 11)T (1, 1, 1, 1) = 0 and
e = 204.
7 A = [ 0 1 3 4 ]T , AT A = [ 26 ] and AT b = [ 112 ]. Best D = 112/26 = 56/13.
8 x = 56/13, p = (56/13)(0, 1, 3, 4). C = 9, D = 56/13 dont match (C, D) = (1, 4); the
columns of A were not perpendicular so we cant project separately to nd C = 1 and D = 4.
10
0
0
C
4
8
26
C
36
1 1
1 8
D = . AT Ax = 8 26
9 Closest parabola:
92 D = 112 .
1 3
8
9
E
26 92 338
E
400
1 4 16
20
10
0
0
C
0
C
0
1 1
D
1
1 D 8
1 47
= . Then =
. Exact cubic so p = b , e = 0.
10
3 28
1 3
E
9 27 E 8
1 4 16 64
F
20
F
5
11 (a) The best line is x = 1 + 4t, which goes through the center point (t, b ) = (2, 9)
(b) From the rst equation: C m + D
m
i=1 ti
=
m
i=1
bi . Divide by m to get C + Dt = b .
12 (a) a T a = m, a T b = b1 + + bm . Therefore x is the mean of the bs
e
2
=
m
i=1 (bi
2
x)
(b)e = b
11
1
T
(c) p = (3, 3, 3), e = (2, 1, 3), p e = 0. P = 1 1
3
11
x a.
1
1 .
1
13 (AT A)1 AT (b Ax ) = x x . Errors b Ax = (1, 1, 1) add to 0, so the x x add to 0.
14 (x x )(x x )T = (AT A)1 AT (b Ax )(b Ax )T A(AT A)1 . Average (b Ax )(b Ax )T = 2 I
gives the covariance matrix (AT A)1 AT 2 A(AT A)1 which simplies to 2 (AT A)1 .
15 Problem 14 gives the expected error (x x)2 as 2 (AT A)1 = 2 /m. By taking m measurements, the variance drops from 2 to 2 /m.
16
1
9
1
b10 +
x9 =
(b1 + + b10 ).
10
10
10
41
1 1
7
C
9
3
17 1
= 7 . The solution x = comes from
1
D
4
2
1
2
21
2
6
C
D
=
35
42
.
18 p = Ax = (5, 13, 17) gives the heights of the closest line. The error is b p = (2, 6, 4).
19 If b = e then b is perpendicular to the column space of A. Projection p = 0.
20 If b = Ax = (5, 13, 17) then error e = 0 since b is in the column space of A.
21 e is in N(AT ); p is in C(A); x is in C(AT ); N(A) = {0} = zero vector.
5
0
C
5
=
. Solution: C = 1, D = 1.
22 The least squares equation is
0 10
D
10
23 The square of the distance between points on two lines is E = (y x)2 + (3y x)2 + (1 + x)2 .
Set
1
E/x
2
= (y x) (3y x) + (x + 1) = 0 and
1
E/y
2
= (y x) + 3(3y x) = 0.
The solution is x = 5/7, y = 2/7; E = 2/7, and the minimal distance is
24 e is orthogonal to p ;
e
2
25 The derivatives of Ax b
2/7.
= e T (b p ) = e T b = b T b b T p .
2
are zero when x = (AT A)1 AT b .
26 Direct approach to 3 points on a line: Equal slopes (b2 b1 )/(t2 t1 ) = (b3 b2 )/(t3 t2 ).
Linear algebra approach: If y is orthogonal to the columns (1, 1, 1) and (t1 , t2 , t3 ) and b is in
the column space then y T b = 0. This y = (t2 t3 , t3 t1 , t1 t2 ) is in the left nullspace. Then
y T b = 0 is the same equal slopes condition written as (b2 b1 )(t3 t2 ) = (b3 b2 )(t2 t1 ).
1
1
0
0
C
400
8
C
2
1
1
0
1
D = has AT A = 0 2 0 , AT b = 2 , D = 1 . At
27
1 1
3
0
E
002
3
E
3/2
1
0 1
4
3
x, y = 0, 0 the best plane 2 x 2 y has height C = 2 which is the average of 0, 1, 3, 4.
Problem Set 4.4, page 203
1 (a) Independent
(b) Independent and orthogonal
(c) Independent and orthonormal.
For orthonormal, (a) becomes (1, 0), (0, 1) and (b) is (.6, .8), (.8, .6).
5/9
10
T
22
1
122
but QQT = 2/9
2 q 1 = ( 3 , 3 , 3 ). q 2 = ( 3 , 3 , 3 ). Q Q =
01
4/9
3 (a) AT A = 16I
(b) AT A is
10
1
T
4 (a) Q = 0 1 , QQ 0
00
0
2/9 = 4/9
8/9
2/9
2/9 .
5/9
diagonal with entries 1, 4, 9.
00
1 0 (b) (1, 0) and (0, 0) are orthogonal, not independent
00
1
1
(c) ( 2 , 1 , 1 , 1 ), ( 1 , 2 , 1 , 1 ), ( 1 , 1 , 1 , 1 ), ( 1 , 1 , 1 , 1 ).
222
2
2
2
2
22
2
222
2
1
1
1
5 Orthogonal vectors are (1, 1, 0) and (1, 1, 1). Orthonormal are ( 2 , 2 , 0), ( 3 ,
1
1
, ).
3
3
42
6 If Q1 and Q2 are orthogonal matrices then (Q1 Q2 )T Q1 Q2 = QT QT Q1 Q2 = QT Q2 = I which
2
1
2
means that Q1 Q2 is orthogonal also.
1
0
7 The least squares solution to Q Qx = Q b is x = Q b . This is 0 if Q = and b = .
0
1
T
T
T
8 If q 1 and q 2 are orthonormal vectors in R5 then (q T b )q 1 + (q T b )q 2 is closest to b .
1
2
100
T
9 (a) P = QQ = 0 1 0
(b) (QQT )(QQT ) = Q(QT Q)QT = QQT .
000
10 (a) If q 1 , q 2 , q 3 are orthonormal then the dot product of q 1 with c1 q 1 + c2 q 2 + c3 q 3 = 0 gives
(b) Qx = 0 QT Qx = 0 x = 0.
c1 = 0. Similarly c2 = c3 = 0 independent
11 (a) Two orthonormal vectors are
1
(1, 3, 4, 5, 7)
10
1
(7, 3, 4, 5, 1)
10
and
(b) The closest
T
vector in the plane is the projection QQ (1, 0, 0, 0, 0) = (0.5, 0.18, 0.24, 0.4, 0).
12 (a) a T b = a T (x1 a 1 + x2 a 2 + x3 a 3 ) = x1 (a T a 1 ) = x1
1
1
1
(b) a T b = a T (x1 a 1 + x2 a 2 + x3 a 3 ) = x1 (a T a 1 ). Therefore x1 = a T b /a T a 1
1
1
1
1
1
(c) x1 is the rst component of A1 times b .
a Tb
13 The multiple to subtract is a T b /a T a . Then B = b a T a a = (4, 0) 2 (1, 1) = (2, 2).
14
a
q Tb
1/ 2
1/ 2
222
1
= q1 q2
=
14
= QR.
10
0
B
1/ 2 1/ 2
022
15 (a) q 1 =
1
(1, 2, 2),
3
1
(2, 1, 2),
3
q2 =
q3 =
1
(2, 2, 1)
3
(b)
The nullspace of AT
(c) x = (AT A)1 AT (1, 2, 7) = (1, 2).
contains q 3
16 The projection p = (a T b /a T a )a = 14a /49 = 2a /7 is closest to b ; q 1 = a / a
= a /7 is
(4, 5, 2, 2)/7. B = b p = (1, 4, 4, 4)/7 has B = 1 so q 2 = B .
17 p = (a T b /a T a )a = (3, 3, 3) and e = (2, 0, 2). q 1 = (1, 1, 1)/ 3 and q 2 = (1, 0, 1)/ 2.
18 A = a = (1, 1, 0, 0); B = b p = ( 1 , 1 , 1, 0); C = c p A p B = ( 1 , 1 , 1 , 1). Notice the
22
333
pattern in those orthogonal vectors A, B , C .
19 If A = QR then AT A = RT R = lower times upper triangular. Pivots of AT A are 3 and 8.
20 (a) True
(b) True. Qx = x1 q 1 + x2 q 2 . Qx
2
= x2 + x2 because q 1 q 2 = 0.
1
2
21 The orthonormal vectors are q 1 = (1, 1, 1, 1)/2 and q 2 = (5, 1, 1, 5)/ 52. Then b =
(4, 3, 3, 0) projects to p = (7, 3, 1, 3)/2. Check that b p = (1, 3, 7, 3)/2 is
orthogonal to both q 1 and q 2 .
22 A = (1, 1, 2), B = (1, 1, 0), C = (1, 1, 1).
1
0
0
1
23 q 1 = 0 , q 2 = 0 , q 3 = 1 . A = 0
0
1
0
0
Not yet orthonormal.
00
124
0 1 0 3 6 .
10
005
24 (a) One basis for this subspace is v 1 = (1, 1, 0, 0),
(b) (1, 1, 1, 1)
(c) b 2 =
1
(2, 1, 1,1)
22
2
and b 1 =
v 2 = (1, 0, 1, 0),
( 1 , 1 , 1 , 3 ).
2222
v 3 = (1, 0, 0, 1)
43
11
1 2 1 1 5 3
1 1 1 1 2
=
=
25
. Singular
51
501
21
20
11
2
11
1
The Gram-Schmidt process breaks down when A is singular and ad bc = 0.
2
1
2
0
.
T
26 (q T C )q 2 = BT c B because q 2 = B and the extra q 1 in C is orthogonal to q 2 .
2
BB
B
27 When a and b are not orthogonal, the projections onto these lines do not add to the projection
onto their plane.
28 q 1 = 1 (2, 2, 1), q 2 = 1 (2, 1, 2), q 3 = 1 (1, 2, 2).
3
3
3
29 There are mn multiplications in (11) and
1
m2 n
2
multiplications in each part of (12).
30 The columns of the wavelet matrix W are orthonormal. Then W 1 = W T . See Section 7.3
for more about wavelets.
31 (a) c =
1
2
(b) Change all signs in rows 2, 3, 4; then in columns 2, 3, 4.
32 p 1 = 1 (1, 1, 1, 1) and p 2 = (0, 0, 1, 1).
2
33 Q1 =
1
0
0
1
reects across x axis, Q2 = 0
1
0
0
0
1
0
1 across plane y + z = 0.
0
34 (a) Qu = (I 2uu T )u = u 2uu T u . This is u , provided that u T u equals 1
(b) Qv = (I 2uu T )v = u 2uu T v = u , provided that u T v = 0.
35 No solution
36 Orthogonal and lower triangular 1 on the main diagonal, 0 elsewhere.
Problem Set 5.1, page 213
1 det(2A) = 8 and det(A) = (1)4 det A =
1
2
and det(A2 ) =
and det(A1 ) = 2.
1
4
2 det( 1 A) = ( 1 )3 det A = 1 and det(A) = (1)3 det A = 1; det(A2 ) = 1; det(A1 ) = 1.
2
2
8
3 (a) False: 2 by 2 I
(b) True
(c) False: 2 by 2 I
(d) False (but trace = 0).
4 Exchange rows 1 and 3. Exchange rows 1 and 4, then 2 and 3.
5 |J5 | = 1, |J6 | = 1, |J7 | = 1. The determinants are 1, 1, 1, 1 repeating, so |J101 | = 1.
6 Multiply the zero row by t. The determinant is multiplied by t but the matrix is the same
det = 0.
7 det(Q) = 1 for rotation, det(Q) = 1 for reection (1 2 sin2 2 cos2 = 1).
8 QT Q = I |Q|2 = 1 |Q| = 1; Qn stays orthogonal so cant blow up. Same for Q1 .
9 det A = 1, det B = 2, det C = 0.
10 If the entries in every row add to zero, then (1, 1, . . . , 1) is in the nullspace: singular A has
det = 0. (The columns add to the zero column so they are linearly dependent.) If every row
adds to one, then rows of A I add to zero (not necessarily det A = 1).
11 CD = DC |CD| = (1)n |DC | and not |DC |. If n is even we can have |CD| = 0.
44
12 det(A1 ) = det
d
adbc
b
adbc
c
adbc
a
adbc
=
ad bc
1
=
.
(ad bc)2
ad bc
13 Pivots 1, 1, 1 give det = 1; pivots 1, 2, 3/2 give det = 3.
14 det(A) = 24 and det(A) = 5.
15 det = 0 and det = 1 2t2 + t4 = (1 t2 )2 .
16 A singular rank one matrix has det = 0; Also det K = 0.
17 Any 3 by 3 skew-symmetric K has det(K T ) = det(K ) = (1)3 det(K ). This is det(K ).
But also det(K T ) = det(K ), so we must have det(K ) = 0.
1
18
a
a2
1
b
b2 = 0
1
c
2
c
1
0
a2
a
ba
ca
19 det(U ) = 6, det(U 1 ) =
2
2
c a
1
,
6
1
b+a
1
b2 a2 = (b a)(c a)
c+a
= (b a)(c a)(c b).
det(U 2 ) = 36, det(U ) = ad, det(U 2 ) = a2 d2 . If ad = 0 then
det(U 1 ) = 1/ad.
a Lc b Ld
= (ad bc)(1 Ll).
20 det
c la d lb
21 Rules 5 and 3 give Rule 2. (Since Rules 4 and 3 give 5, they also give Rule 2.)
22 det(A) = 3, det(A1 ) =
1
,
3
det(A I ) = 2 4 + 3. Then = 1 and = 3 give
det(A I ) = 0. Note to instructor : If you discuss this exercise, you can explain that this is
the reason determinants come before eigenvalues. Identify 1 and 3 as the eigenvalues.
18
7
3 1
2
2
1
1
, det(A ) = 100, A
, det(A1 ) =
23 det(A) = 10, A =
= 10
14 11
2
4
1
.
10
det(A I ) = 2 7 + 10 = 0 when = 2 or = 5.
24 det(L) = 1, det(U ) = 6, det(A) = 6, det(U 1 L1 ) = 1 , and det(U 1 L1 A) = 1.
6
25 Row 2 = 2 times row 1 so det A = 0.
26 Row 3 row 2 = row 2 row 1 so A is singular.
27 det A = abc, det B = abcd, det C = a(b a)(c b).
28 (a) True : det(AB ) = det(A)det(B ) = 0
(c) False : A = 2I and B = I
(b) False : may exchange rows
(d) True : det(AB ) = det(A)det(B ) = det(BA).
29 A is rectangular so det(AT A) = (det AT )(det A): these are not dened.
d
b
f /a f /c
d b
1
= adbc adbc =
= A1 .
30
c
a
ad bc c
f /b f /d
a
adbc
adbc
31 The Hilbert determinants are 1, .08, 4.6 104 , 1.6 107 , 3.7 1012 , 5.4 1018 , 4.8 1025 ,
2.7 1033 , 9.7 1043 , 2.2 1053 . Pivots are ratios of determinants so 10th pivot is near
1010 .
32 Typical determinants of rand(n) are 106 , 1025 , 1079 , 10218 for n = 50, 100, 200, 400). Using
randn(n) with normal bell-shaped probabilities these are 1031 , 1078 , 10186 , Inf 21024 . MATLAB computes 1.999999999999999 21023 1.8 10308 but one more 9 gives Inf!
45
33 n=5; p=(n 1)^2; A0=ones(n); maxdet=0; for k=0:2^p 1
Asub=rem(oor(k. * 2.^( p + 1:0)),2); A=A0; A(2:n,2:n)=1 2 * reshape(Asub,n 1,n 1);
if abs(det(A))>maxdet, maxdet=abs(det(A)); maxA=A; end end
Output: maxA = 1
1
1
1
1
1
1
1 1 1
1
1 1
1 1
1 1
1
1 1
1 1 1 1
1
maxdet = 48.
34 Reduce B to [ row 3 : row 2; row 1 ]. Then det B = 6.
Problem Set 5.2, page 225
1 det A = 1 + 18 + 12 9 4 6 = 12, rows are independent; det B = 0, rows are dependent;
det C = 1, independent rows.
2 det A = 2, independent; det B = 0, dependent; det C = (2)(0), dependent.
3 Each of the 6 terms in det A is zero; the rank is at most 2; column 2 has no pivot.
4 (a) The last three rows must be dependent
(b) In each of the 120 terms: Choices from
the last 3 rows must use 3 columns; at least one choice will be zero.
5 a11 a23 a32 a44 gives 1, a14 a23 a32 a41 gives +1 so det A = 0; det B = 2 4 4 2 1 4 4 1 = 48.
6 Four zeros in a row guarantee det = 0; A = I has 12 zeros.
7 (a) If a11 = a22 = a33 = 0 then 4 terms are sure zeros
(b) 15 terms are certainly zero.
8 5!/2 = 60 permutation matrices have det = +1. Put row 5 of I at the top (4 exchanges).
9 Some term a1 a2 an is not zero! Move rows 1, 2, . . ., n into rows , , . . ., . Then
these nonzero as will be on the main diagonal.
10 To get +1 for the even permutations the matrix needs an even number of 1s. For the odd P s
the matrix needs an odd number of 1s. So six 1s and det = 6 are impossible: max(det) = 4.
11 det(I + Peven ) = 16 or 4 or 0 (16 comes from I + I ).
0
42 35
6 3
. C = 0 21
12 C =
14 . det B = 1(0) + 2(42) + 3(35) = 21.
1
2
3
6 3
321
400
13 C = 2 4 2 and AC T = 0 4 0 . Therefore A1 = 1 C T .
4
123
004
1 1
1 1
1 1
= 2|B3 | |B2 |.
14 |B4 | = 2 det 1
= 2|B3 | det
2 1 + det 1
2
1
2
1
2
1 1
15 (a) C1 = 0, C2 = 1, C3 = 0, C4 = 1
(b) Cn = Cn2 by cofactors of row 1 then
cofactors of column 1. Therefore C10 = C8 = C6 = C4 = 1.
46
16 Must choose 1s from column 2 then column 1, column 4 then column 3, and so on. Therefore
n must be even to have det An = 0. The number of row exchanges is
1
n
2
so Cn = (1)n/2 .
17 The 1, 1 cofactor is En1 . The 1, 2 cofactor has a single 1 in its rst column, with cofactor
En2 . Signs give En = En1 En2 . Then 1, 0, 1, 1, 0, 1 repeats by sixes; E100 = 1.
18 The 1, 1 cofactor is Fn1 . The 1, 2 cofactor has a 1 in column 1, with cofactor Fn2 . Multiply
by (1)1+2 and also (1) from the 1, 2 entry to nd Fn = Fn1 + Fn2 (so Fibonacci).
19 |Bn | = |An | |An1 | = (n + 1) n = 1.
20 Since x, x2 , x3 are all in the same row, they are never multiplied in det V4 . The determinant
is zero at x = a or b or c, so det V has factors (x a)(x b)(x c). Multiply by the cofactor
V3 . Any Vandermonde matrix Vij = (ci )j 1 has det V = product of all (cl ck ) for l > k.
21 G2 = 1, G3 = 2, G4 = 3, and Gn = (1)n1 (n 1) = (product of the n eigenvalues!)
22 S1 = 3, S2 = 8, S3 = 21. The rule looks like every second number in Fibonaccis sequence
. . . 3, 5, 8, 13, 21, 34, 55, . . . so the guess is S4 = 55. Following the solution to Problem 32
with 3s instead of 2s conrms S4 = 81 + 1 9 9 9 = 55.
23 The problem asks us to show that F2n+2 = 3F2n F2n2 . Keep using the Fibonacci rule:
F2n+2 = F2n+1 + F2n = F2n + F2n1 + F2n = F2n + (F2n F2n2 ) + F2n = 3F2n F2n2 .
24 Changing 3 to 2 in the corner reduces the determinant F2n+2 by 1 times the cofactor of
that corner entry. This cofactor is the determinant of Sn1 (one size smaller) which is F2n .
Therefore changing 3 to 2 changes the determinant to F2n+2 F2n which is F2n+1 .
25 (a) If we choose an entry from B we must choose an entry from the zero block; result zero.
This leaves a pair of entries from A times a pair fromD leading to (det A)(det D)
10
00
01
00
, B =
, C =
, D =
.
(b) and (c) Take A =
00
10
00
01
3
26 (a) All Ls have det = 1; det Uk = det Ak = 2, 6, 6 for k = 1, 2, 3
(b) Pivots 2, 2 , 1 .
3
I
0
AB
= 1 and det
= |A| times |D CA1 B | which
27 Problem 25 gives det
1
CA
I
CD
is |AD ACA1 B |. If AC = CA this is |AD CAA1 B | = det(AD CB ).
28 If A is a row and B is a column then det M = det AB = dot product of A and B . If A is a
column and B is a row then AB has rank 1 and det M = det AB = 0 (unless m = n = 1).
29 (a) det A = a11 A11 + + a1n A1n . The derivative with respect to a11 is the cofactor A11 .
30 Row 1 2 row 2 + row 3 = 0 so the matrix is singular.
31 There are ve nonzero products, all 1s with a plus or minus sign. Here are the (row, column)
numbers and the signs: + (1, 1)(2, 2)(3, 3)(4, 4) + (1, 2)(2, 1)(3, 4)(4, 3) (1, 2)(2, 1)(3, 3)(4, 4)
(1, 1)(2, 2)(3, 4)(4, 3) (1, 1)(2, 3)(3, 2)(4, 4). Total 1 + 1 1 1 1 = 1.
32 The 5 products in solution 31 change to 16 + 1 4 4 4 since A has 2s and 1s:
(2)(2)(2)(2) + (1)(1)(1)(1) (1)(1)(2)(2) (2)(2)(1)(1) (2)(1)(1)(2).
47
33 det P = 1 because the cofactor of P14 = 1 in row one has sign (1)1+4 . The big formula for
det P has only one term (1 1 1 1) with minus sign because three exchanges take 4, 1, 2, 3
0I
01
= det
is not right.
into 1, 2, 3, 4; det(P 2 ) = (det P )(det P ) = +1 so det
I0
10
34 With a11 = 1, the 1, 2, 1 matrix has det = 1 and inverse (A1 )ij = n + 1 max(i, j ).
35 With a11 = 2, the 1, 2, 1 matrix has det = n + 1 and (n + 1)(A1 )ij = i(n j + 1) for i j
and symmetrically (n + 1)(A1 )ij = j (n i + 1) for i j .
36 Subtracting 1 from the n, n entry subtracts its cofactor Cnn from the determinant. That
cofactor is Cnn = 1 (smaller Pascal matrix). Subtracting 1 from 1 leaves 0.
Problem Set 5.3, page 240
1 (a) det A = 3, det B1 = 6, det B2 = 3 so x1 = 6/3 = 2 and x2 = 3/3 = 1
(b) |A| = 4, |B1 | = 3, |B2 | = 2, |B3 | = 1. Therefore x1 = 3/4 and x2 = 1/2 and x3 = 1/4.
2 (a) y = c/(ad bc)
(b) y = (f g id)/D.
3 (a) x1 = 3/0 and x2 = 2/0: no solution
4 (a) x1 = det [ b a 2 a 3 ]/ det A, if det A = 0
(b) x1 = 0/0 and x2 = 0/0: undetermined.
(b) The determinant is linear in column 1
so we get x1 |a 1 a 2 a 3 | + x2 |a 2 a 2 a 3 | + x3 |a 3 a 2 a 3 |. The last two determinants are zero.
5 If the rst column in A is also the right side b then det A = det B1 . Both B2 and B3 are
singular since a column is repeated. Therefore x1 = |B1 |/|A| = 1 and x2 = x3 = 0.
1 2
0
321
3
1
1
0
2 1 2 . The inverse of a symmetric matrix is symmetric.
(b)
6 (a)
0
3
4
0 4
1
123
3
11
1
7 If all cofactors = 0 then A would be the zero matrix if it existed; cannot exist. A =
11
has no zero cofactors but it is not invertible.
6 3
0
300
T
8 C= 3
1 1 and AC = 0 3 0 . Therefore det A = 3. Cofactor of 100 is 0.
6
2
1
003
9 If we know the cofactors and det A = 1 then C T = A1 and det A1 = 1. The inverse of A1
is A, so A is the cofactor matrix for C .
10 Take the determinant of AC T = (det A)I . The left side gives det AC T = (det A)(det C ) while
the right side gives (det A)n . Divide by det A to reach det C = (det A)n1 .
1
11 We nd det A = (det C ) n1 with n = 4. Then det A1 is 1/ det A. Construct A1 using the
cofactors. Invert to nd A.
12 The cofactors of A are integers. Division by det A = 1 gives integer entries in A1 .
48
13 Both det A and det A1 are integers since the matrices contain only integers. But det A1 =
1/ det A
0
14 A = 1
2
so det A = 1 or 1.
13
1
2
1
1
1T
0 1 has cofactor matrix C = 3 6
2 and A = 5 C .
10
1
3 1
(b) C12 = C21 , C31 = C13 , C32 = C23 make S 1 symmetric.
15 (a) C21 = C31 = C32 = 0
16 For n = 5 the matrix C contains 25 cofactors and each 4 by 4 cofactor contains 24 terms and
each term needs 3 multiplications: total 1800 multiplications vs. 125 for Gauss-Jordan.
32
14
17 (a) Area
311
131
113
18 Volume =
19 (a) Area
20
21
23
= 10
22
13
(c) 5.
= 20. Area of faces = length of cross product
1 211
341
2 051
=4=
(b) 5
=5
(b) 5 + new triangle area
211
1
051
2 1 0 1
ijk
311
131
= 2i 2j +8k is 6 2.
= 5 + 7 = 12.
because the transpose has the same determinant. See #23.
21 The edges of the hypercube have length
1 + 1 + 1 + 1 = 2. The volume det H is 24 = 16.
(H/2 has orthonormal columns. Then det(H/2) = 1 leads again to det H = 16.)
22 The maximum volume is L1 L2 L3 L4 reached when the four edges are orthogonal in R4 . With
entries 1 and 1 all lengths are 1 + 1 + 1 + 1 = 2. The maximum determinant is 24 = 16,
achieved by Hadamard above. For a 3 by 3 matrix, det A = ( 3)3 cant be achieved.
23 TO SEND IN EMAIL
aT
a Ta
T
24 AT A = b a b c = 0
cT
0
0
0
0
0
cTc
1
25 The box has height 4. The volume is 4 = det 0
2
b Tb
n1
n
26 The n-dimensional cube has 2 corners, n2
0
1
3
det AT A
=
(a
det A
has
=
a
0
b
b
c )2
c
0 ; i j = k and (k w ) = 4.
4
edges and 2n (n 1)-dimensional faces. Those
are coecients of (2 + x)n in Worked Example 2.4A. The cube whose edges are the rows of
2I has volume 2n .
27 The pyramid has volume 1/6. The 4-dimensional pyramid has volume 1/24 = 1/4!.
28 J = r. The columns are orthogonal and their lengths are 1 and r.
29 J =
sin cos cos cos sin sin
sin sin cos sin sin cos
cos
sin
0
r/x
r/y
/x
30
/y
=
= 2 sin , needed for triple integrals inside spheres.
cos
sin
( sin )/r
(cos )/r
=
1
.
r
31 The triangle with corners (0, 0), (6, 0), (1, 4) has area 24. Rotated by = 600 area is
the
unchanged. The determinant of the rotation matrix is J =
32 Base area 10, height 2, volume 20.
cos sin
sin cos
=
1
2
3
2
23
1
2
= 1.
49
2
4
0
33 V = det 1 3 0 = 20.
122
u
u2 u3
1
v2 v 3
v
u2 1
34 v1 v2 v3 = u1
w2 w3
w1
w1 w2 w3
v3
w3
+ u3
v1
v2
w1
w2
= u (v w ).
35 (w u ) v = (v w ) u = (u v ) w : Cyclic = even permutation of (u , v , w ).
36 S = (2, 1, 1). The area is P Q P S = (2, 2, 1) = 3. The other four corners could
be (0, 0, 0), (0, 0, 2), (1, 2, 2), (1, 1, 0). The volume of the tilted box is |det| = 1.
xyz
37 If (1, 1, 0), (1, 2, 1), (x, y, z ) are in a plane the volume is det 1 1 0 = x y + z = 0.
121
xyz
38 det 3 2 1 = 0 = 7x 5y + z ; plane contains the two vectors.
123
39 Doubling each row multiplies the volume by 2n .
Then 2 det A = det(2A) only if n = 1.
Problem Set 6.1, page 253
1 A and A2 and A all have the same eigenvectors. The eigenvalues are 1 and 0.5 for A, 1 and
0.25 for A2 , 1 and 0 for A . Therefore A2 is halfway between A and A .
Exchanging the rows of A changes the eigenvalues to 1 and 0.5 (it is still a Markov matrix
with eigenvalue 1, and the trace is now 0.2 + 0.3so the other eigenvalue is 0.5).
Singular matrices stay singular during elimination, so = 0 does not change.
2 1 = 1 and 2 = 5 with eigenvectors x 1 = (2, 1) and x 2 = (1, 1). The matrix A + I has
the same eigenvectors, with eigenvalues increased by 1 to 0 and 6.
3 A has 1 = 4 and 2 = 1 (check trace and determinant) with x 1 = (1, 2) and x 2 = (2, 1).
A1 has the same eigenvectors as A, with eigenvalues 1/1 = 1/4 and 1/2 = 1.
4 A has 1 = 3 and 2 = 2 (check trace and determinant) with x 1 = (3, 2) and x 2 = (1, 1).
A2 has the same eigenvectors as A, with eigenvalues 2 = 9 and 2 = 4.
1
2
5 A and B have 1 = 1 and 2 = 1. A + B has 1 = 1, 2 = 3. Eigenvalues of A + B are not
equal to eigenvalues of A plus eigenvalues of B .
6 A and B have 1 = 1 and 2 = 1. AB and BA have =
1
(3
2
5). Eigenvalues of AB are
not equal to eigenvalues of A times eigenvalues of B . Eigenvalues of AB and BA are equal.
7 The eigenvalues of U are the pivots. The eigenvalues of L are all 1s. The eigenvalues of A are
not the same as the pivots.
8 (a) Multiply Ax to see x which reveals
(b) Solve (A I )x = 0 to nd x .
50
9 (a) Multiply by A: A(Ax ) = A(x ) = Ax gives A2 x = 2 x
A1 Ax = A1 x = A1 x gives A1 x =
1
x
(b) Multiply by A1 :
(c) Add I x = x : (A + I )x = ( + 1)x .
10 A has 1 = 1 and 2 = .4 with x 1 = (1, 2) and x 2 = (1, 1). A has 1 = 1 and 2 = 0 (same
eigenvectors). A100 has 1 = 1 and 2 = (.4)100 which is near zero. So A100 is very near A .
11 M = (A 2 I )(A 1 I ) = zero matrix so the columns of A 1 I are in the nullspace of
A 2 I . This Cayley-Hamilton Theorem M = 0 in Problem 6.2.35 has a short proof: by
Problem 9 = M has eigenvalues (1 2 )(1 1 ) = 0 and (2 2 )(2 1 ) = 0. Same
x 1, x 2.
12 P has = 1, 0, 1 with eigenvectors (1, 2, 0), (2, 1, 0), (0, 0, 1). Add the rst and last vectors:
(1, 2, 1) also has = 1. P 100 = P so P 100 gives the same answers.
13 (a) P u = (uu T )u = u (u T u ) = u so = 1
(b) P v = (uu T )v = u (u T v ) = 0 so = 0
(c) x 1 = (1, 1, 0, 0), x 2 = (3, 0, 1, 0), x 3 = (5, 0, 0, 1) are eigenvectors with = 0.
14 The eigenvectors are x 1 = (1, i) and x 2 = (1, i).
15 = 1 (1 i 3); the three eigenvalues are 1, 1, 1.
2
16 Set = 0 to nd det A = (1 )(2 ) (n ).
17 If A has 1 = 3 and 2 = 4 then det(A I ) = ( 3)( 4) = 2 7 + 12. Always
1 = 1 (a + d + (a d)2 + 4bc) and 2 = 1 (a + d
). Their sum is a + d.
2
2
40
32
22
,
,
.
18
05
1 6
3 7
19 (a) rank = 2
(b) det(B T B ) = 0
(d) eigenvalues of (B + I )1 are 1, 1 , 1 .
23
0
1
has trace 11 and determinant 28.
20 A =
28 11
21 a = 0, b = 9, c = 0 multiply 1, , 2 in det(A I ) = 9 3 : A = companion matrix.
10
11
T
and
: dierent eigenvectors.
22 (A I ) has the same determinant as (A I ) .
10
00
23 = 1 (for Markov),
00
01
,
,
24
10
00
0 (for singular), 1 (so sum of eigenvalues = trace = 1 ).
2
2
1 1
. Always A2 = zero matrix if = 0, 0 (Cayley-Hamilton 6.2.35).
1 1
25 = 0, 0, 6 with x 1 = (0, 2, 1), x 2 = (1, 2, 0), x 3 = (1, 2, 1).
26 Ax = c1 1 x 1 + + cn n x n equals B x = c1 1 x 1 + + cn n x n for all x . So A = B .
27 = 1, 2, 5, 7.
28 rank(A) = 1 with = 0, 0, 0, 4; rank(C ) = 2 with = 0, 0, 2, 2.
29 B has = 1, 1, 1, 3 so det B = 3. The 5 by 5 matrix A has = 0, 0, 0, 0, 5 and
B = A I has = 1, 1, 1, 1, 4.
30 (A) = 1, 4, 6; (B ) = 2, 3, 3; (C ) = 0, 0, 6.
ab
1
a+b
1
=
= (a + b) ; 2 = d b to produce trace = a + d.
31
cd
1
c+d
1
51
32 Eigenvector (1, 3, 4) for A with = 11 and eigenvector (3, 1, 4) for P AP .
33 (a) u is a basis for the nullspace, v and w give a basis for the column space
11
(b) x = (0, 3 , 5 ) is a particular solution. Add any cu from the nullspace
(c) If Ax = u had a solution, u would be in the column space, giving dimension 3.
34 With 1 = e2i/3 and 2 = e2i/3 , the determinant is 1 2 = 1 and the trace is 1 + 2 = 1:
e2i/3 + e2i/3 = cos
A=
1 1
1 0
2
2
2
2
+ i sin
+ cos
i sin
= 1. Also 3 = 3 = 1.
1
2
3
3
3
3
has this trace 1 and determinant 1. Then A3 = I and every (M 1 AM )3 = I .
Choosing 1 = 2 = 1 leads to I or else to a matrix like A =
11
01
that has A3 = I .
35 det(P I ) = 0 gives the equation 3 = 1. This reects the fact that P 3 = I . The solutions
of 3 = 1 are = 1 (real) and = e2i/3 , = e2i/3 (complex conjugates). The real
eigenvector x 1 = (1, 1, 1) is not changed by the permutation P . The complex eigenvectors are
x 2 = (1, e2i/3 , e4i/3 ) and x 3 = (1, e2i/3 , e4i/3 ) = x 2 .
Problem Set 6.2, page 266
1
1
0
2
3
=
1
0
1
1
1
0
0
3
1 1
0
1
;
1
2
1
1
=
2
1
1
2
0
0
2
0 3
1
3
3
1
3
.
1
3
2 If A = S S 1 then A3 = S 3 S 1 and A1 = S 1 S 1 .
11
20
1 1
23
=
.
3 A=
01
05
0
1
05
4 If A = S S 1 then the eigenvalue matrix for A + 2I is + 2I and the eigenvector matrix is
still S . A + 2I = S ( + 2I )S 1 = S S 1 + S (2I )S 1 = A + 2I .
5 (a) False: dont know s
(b) True
(c) True
(d) False: need eigenvectors of S !.
6 A is a diagonal matrix. If S is triangular, then S 1 is triangular, so S S 1 is also triangular.
7 The columns of S are nonzero multiples of (2, 1) and (0, 1) in either order. Same for A1 .
ab
for any a and b.
8
ba
21
32
53
2
3
4
, A =
, A =
; F20 = 6765.
9 A =
11
21
32
.5 .5
has 1 = 1, 2 = 1 with x 1 = (1, 1), x 2 = (1, 2)
10 (a) A =
2
10
2
1
2
1
n
1
1
1
0
3
3 3
3
(b) An =
A =
n
1
1
2
1
1 2
0 (.5)
3
3
3
3
2
1
2
Gk+1
G1
1
= Ak 3 3 = 3 .
(c)
2
1
2
Gk
G0
0
3
3
3
52
1 2
1 0
1 2
1
=
.
11 A = S S 1 =
1 2 1
10
1
0 2
1
1
1 2
k 0
1 2
1
1
1
=
.
S k S 1 =
1 2 1
1
0 k
1
1
0
(k k )/(1 2 )
2
1
2
1
1
12 The equation for the s is 2 1 = 0 or 2 = +1. Multiply by k to get k+2 = k+1 + k .
13 Direct computation gives L0 , . . . , L10 as 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123. My calculator gives
10 = (1.618 . . .)10 = 122.991 . . ..
1
14 The rule Fk+2 = Fk+1 + Fk produces the pattern: even, odd, odd, even, odd, odd, . . .
15 (a) True
(b) False
(c) False (might have 2 or 3 independent eigenvectors).
16 (a) False: dont know
(b) True: missing an eigenvector
(c) True.
83
94
10 5
(or other), A =
, A =
; only eigenvectors are (c, c).
17 A =
3 2
4 1
5 0
18 The rank of A 3I is one. Changing any entry except a12 = 1 makes A diagonalizable.
19 S k S 1 approaches zero if and only if every || < 1; B k 0.
1
1
10
1
1
10
2 2
and S =
; k
and S k S 1
20 =
: steady state.
1
1
0 .2
1 1
00
2
2
.9 0
3 3
3
3
3
3
6
, S =
; B 10 = (.9)10 , B 10 = (.3)10 , B 10 =
21 =
0 .3
1
1
1
1
1
1
0
sum of those two.
k
21
0
11
1 1 1 3 0 1 1
1 1 1 3
=
.
22
and Ak =
21
21
12
1
01
1 1
1
01
1 1
23 B k =
1
1
0 1
3
0
0
2
k
1
1
0 1
=
3k
3k 2k
0
2k
.
24 det A = (det S )(det )(det S 1 ) = det = 1 n . This works when A is diagonalizable.
25 trace AB = (aq + bs) + (cr + dt) = (qa + rc) + (sb + td) = trace BA. Proof for diagonalizable
case: the trace of S S 1 is the trace of (S 1 )S = which is the sum of the s.
10
.
26 AB BA = I : impossible since trace AB trace BA = zero trace I . E =
11
A0
S0
0
S 1
0
1
=
.
27 If A = S S
then B =
0 2A
0S
0 2
0
S 1
28 The As form a subspace since cA and A1 + A2 have the same S . When S = I the As give
the subspace of diagonal matrices. Dimension 4.
29 If A has columns x 1 , . . . , x n then A2 = A means every Ax i = x i . All vectors in the column
space are eigenvectors with = 1. Always the nullspace has = 0. Dimensions of those spaces
add to n by the Fundamental Theorem so A is diagonalizable (n independent eigenvectors).
30 Two problems: The nullspace and column space can overlap, so x could be in both. There
may not be r independent eigenvectors in the column space.
53
1
21
has R2 = A. B would have = 9 and = 1 so its trace is
31 R = S S =
12
01
1
0
.
can have 1 = i and i, and real square root
not real. Note
1 0
0 1
32 AT = A gives x T AB x = (Ax )T (B x ) Ax
gives x T BAx = (B x )T Ax Ax
Bx
by the Schwarz inequality. B T = B
B x . Add these to get Heisenberg when AB BA = I .
33 The factorizations of A and B into S S 1 are the same. So A = B .
34 A = S 1 S 1 and B = S 2 S 1 . Diagonal matrices always give 1 2 = 2 1 . Then AB = BA
from S 1 S 1 S 2 S 1 = S 1 2 S 1 = S 2 1 S 1 = S 2 S 1 S 1 S 1 = BA.
35 If A = S S 1 then the product (A 1 I ) (A n I ) equals S ( 1 I ) ( n I )S 1 .
The factor j I is zero in row j . The product is zero in all rows = zero matrix.
11
21
2
has A =
and A2 A I = zero matrix conrms Cayley-Hamilton.
36 A =
10
11
37 (A aI )(A dI ) =
0b
0 da
ad b
00
=
00
00
38 (a) The eigenvectors for = 0 always span the nullspace
(b) The eigenvectors for = 0
span the column space if there are r independent eigenvectors: then algebraic multiplicity =
geometric multiplicity for each nonzero .
39 The eigenvalues 2, 1, 0 and their eigenvectors are in and S . Then Ak = S k S 1 is
2
1
0
2k
4
1
1
422
1 1 1
k
2k
1
(1)
1 1
2 2 2 =
2 1 1 +
1
1
(1)k
1
1
6
3
6
1 1 1
0k
0
1 1
211
1
1
1
Check k = 1! The (2, 2) entry of A4 is 24 /6 + (1)4 /3 = 18/6 = 3. The 4-step paths that
begin and end at node 2 are 2 to 1 to 1 to 1 to 2, 2 to 1 to 2 to 1 to 2, and 2 to 1 to 3 to 1 to
2. Harder to nd the eleven 4-step paths that start and end at node 1.
Notice the column times row multiplication above. Since A = AT the eigenvectors in the
columns of S are orthogonal. They are in the rows of S 1 divided by their length squared.
40 B has the same eigenvectors (1, 0) and (0, 1)as A, so B is also diagonal. The 4 equations
a
b
a 2b
00
=
have coecient matrix with rank 2.
AB BA =
2c 2d
c 2d
00
41 AB = BA always has the solution B = A. (In case A = 0 every B is a solution.)
42 B has = i and i, so B 4 has 4 = 1 and 1; C has = (1 3i)/2 = exp(i/3) so 3 = 1
and 1. Then C 3 = I and C 1024 = C .
Problem Set 6.3, page 279
1
1
1
1
1 u 1 = e4t , u 2 = et . If u (0) = (5, 2), then u (t) = 3e4t + 2et .
0
1
0
1
54
2 z (t) = 2et ; then dy/dt = 4y 6et with y (0) = 5 gives y (t) = 3e4t + 2et as in Problem 1.
y
01
y
. Then = 1 (5 41).
3 =
2
y
45
y
6 2
2
1
has 1 = 5, x 1 = , 2 = 2, x 2 = ; rabbits r(t) = 20e5t + 10e2t ,
4
2
1
1
2
w(t) = 10e5t + 20e2t . The ratio of rabbits to wolves approaches 20/10; e5t dominates.
5 (v + w)/dt = dv/dt + dw/dt = (w v ) + (v w) = 0, so thetotal v + w is constant. A =
d
1
1
1
1
v (1) = 20 + 10e2
has 1 = 0 and 2 = 2 with x 1 = and x 2 = ;
.
1 1
1
1
w(1) = 20 10e2
6 1 = 0 and 2 = 2. Now v (t) = 20 + 10e2t as t .
01
1t
+ zeros =
.
7 eAt = I + t
00
01
01
has trace 6, det 9, = 3 and 3 with only one independent eigenvector (1, 3).
8 A=
9 6
m0
y
b k
y
=
.
9 my + by + ky = 0 is
01
y
1
0
y
10 When A is skew-symmetric, u (t) = eAt u (0) = u (0) . So eAt is an orthogonal matrix.
1
1
1
1
1
cos t
.
11 (a) = 1 + 1 . Then u (t) = 1 eit + 1 eit =
2
2
2
2
0
i
i
i
i
sin t
12 y (t) = cos t starts at y (0) = 1 and y (0) = 0.
4
1
0
4
13 u p = A1 b = 4 and u(t) = ce2t + 4; u p = and u (t) = c1 e2t + c2 e3t + .
2
0
1
2
14 Substituting u = ect v gives cect v = Aect v ect b or (A cI )v = b or v = (A cI )1 b =
particular solution. If c is an eigenvalue then A cI is not invertible.
1
0
10
11
,
,
. In each case eAt blows up.
15
0 1
01
1 1
16 d/dt(eAt ) = A + A2 t + 1 A3 t2 + 1 A4 t3 + = A(I + At + 1 A2 t2 + 1 A3 t3 + ) = AeAt .
2
6
2
6
1 t
0 1
. Derivative =
= B.
17 eBt = I + Bt =
0
1
0
0
18 The solution at time t + T is also eA(t+T ) u (0). Thus eAt times eAT equals eA(t+T ) .
11
1
1
10
1
1
1
1
et 0
1
1
et et 1
At
=
; e =
=
.
19
00
0 1
00
0 1
0 1
01
0 1
0
1
10
et 1 et 1
2
At
2
3
t
1
1
+
.
20 If A = A then e = I + At + 2 At + 6 At + = I +(e 1)A =
01
0
0
e e1
1 1
e e2
e0
A
B
AB
BA
A+ B
, e =
, e e = e e =
=e
.
21 e =
=
0
1
0
1
0
1
01
1
11
11
3 0 0
et 1 (e3t et )
2
2
At
=
.
22 A =
, then e =
0
e3t
03
20
01
1 1
2
55
23 A2 = A so A3 = A and by Problem 20 eAt = I + (et 1)A =
24 (a) The inverse of eAt is eAt
et
3(et 1)
0
1
.
(b) If Ax = x then eAt x = et x and et = 0.
25 x(t) = e4t and y (t) = e4t is growing solution. The correct matrix for the exchanged
a
2 2
and it does have the same eigenvalues as the original matrix.
unknown u = (y, x) is
4
0
Problem Set 6.4, page 290
1
3
1 A = 3
6
T
3
3
6
0 1 2
3+1
5
2
T
T
1
1
0 3 = 2 (A + A ) + 2 (A A ) = symmetric + skew-symmetric.
3
0
T
2 (A CA) = AT C T (AT )T = AT CA. When A is 6 by 3, C is 6 by 6 and AT CA is 3 by 3.
3 = 0, 2, 1 with unit eigenvectors (0, 1, 1)/ 2 and (2, 1, 1)/ 6 and (1, 1, 1)/ 3.
2
1 1
.
4 Q=
5 2 1
2
1
2
1
5 Q = 2 2 1 .
3
1 2
2
.8
.6
.8
.6
or
or exchange columns.
6 Q=
.6
.8
.6
.8
12
has = 1 and 3
7 (a)
(b) The pivots have the same signs as the s
21
(c) trace = 1 + 2 = 2, so A cant have two negative eigenvalues.
01
. If A is symmetric then A3 =
8 If A3 = 0 then all 3 = 0 so all = 0 as in A =
00
Q3 QT = 0 gives = 0 and the only symmetric possibility is A = Q 0 QT = zero matrix.
9 If is complex then is also an eigenvalue (Ax = x ). Always + is real. The trace is real
so the third eigenvalue must be real.
10 If x is not real then = x T Ax /x T x is not necessarily real. Cant assume real eigenvectors!
1
1
1
1
31
.64 .48
.36 .48
2
2
2 2 9 12
= 2
+ 25
11
= 0
+ 4
;
1
1
1
13
12 16
.48
.36
.48 .64
1
2
2
2
2
xT
1
= I;
12 [ x 1 x 2 ] is an orthogonal matrix so P1 + P2 = x 1 x T + x 2 x T = [ x 1 x 2 ]
1
2
xT
2
P1 P2 = x 1 (x T x 2 )x T = 0. Second
1
2
0
3
13 = ib and ib; A = 3
0
0 4
2
proof: P1 P2 = P1 (I P1 ) = P1 P1 = 0 since P1 = P1 .
0
3
4 has det(A I ) = 25 = 0 and = 0, 5i, 5i.
0
56
14 Skew-symmetric and orthogonal; = i, i, i, i to have trace zero.
15 A has = 0, 0 and only one independent eigenvector x = (i, 1).
16 (a) If Az = y and AT y = z then B [ y ; z ] = [ Az ; AT y ] = [ y ; z ]. So is
also an eigenvalue of B .
(b) AT Az = AT (y ) = 2 z . The eigenvalues of AT A are 0
(c) = 1, 1, 1, 1;
x 1 = (1, 0, 1, 0), x 2 = (0, 1, 0, 1), x 3 = (1, 0, 1, 0), x 4 = (0, 1, 0, 1).
17 The eigenvalues of B are 0, 2, 2 with x 1 = (1, 1, 0), x 2 = (1, 1, 2), x 3 = (1, 1, 2).
18 y is in the nullspace of A and x is in the column space. A = AT has column space = row
space, and this is perpendicular to the nullspace. Then y T x = 0. If Ax = x and Ay = y
then shift by : (A I )x = ( )x and (A I )y = 0 and again x y .
10
1
101
19 B has eigenvectors in S = 0 1
0 0 1 0 ; independent but not perpendicular.
0 0 1+d
002
20 = 5 and 5 have the same signs as the pivots 3 and 25/3.
12
(b) True (c) True. A1 = Q1 QT is also symmetric (d) False.
21 (a) False. A =
01
22 If AT = A then AT A = AAT = A2 . If A is orthogonal then AT A = AAT = I .
a1
1
1
is normal only if a = d. Then x = is perpendicular to .
A=
1 d
i
i
01
T
has 1 = i
23 A and A have the same s but the order of the x s can change. A =
1 0
and 2 = i with x 1 = (1, i) for A but x 1 = (1, i) for AT .
24 A is invertible, orthogonal, permutation, diagonalizable, Markov; B is projection, diagonalizable, Markov. QR, S S 1 , QQT possible for A; S S 1 and QQT possible for B .
25 Symmetry gives QQT when b = 1; repeated and no S when b = 1; singular if b = 0.
26 Orthogonal and symmetric requires || = 1 and real, so = 1. Then = I or
every
A
cos sin
1
0
cos sin
cos 2
sin 2
=
= reection.
A = QQT =
sin
cos
0 1
sin cos
sin 2 cos 2
27 Eigenvectors (1, 0) and (1, 1) give a 45 angle even with AT very close to A.
28 The roots of 2 + b + c = 0 dier by b2 4c. For det(A + tB I ) we have b = 3 8t
and c = 2 + 16t t2 . The minimum of b2 4c is 1/17 at t = 2/17. Then 2 1 = 1/ 17.
29 We get good eigenvectors for the symmetric part 1 (P + P T ) which MATLAB would recognize
2
as symmetric. But the projection matrix P = A(AT A)1 AT = product of 3 matrices is not
recognized as exactly symmetric.
Problem Set 6.5, page 302
1 A4 has two positive eigenvalues because a = 1 and ac b2 = 1; x T A1 x is zero for x = (1, 1)
and x T A1 x < 0 for x = (6, 5).
57
2
Positive denite
for 3 < b < 3
Positive denite
for c > 8
2
1
0
b
1
1
0
2
1
1
0
2
0
b
1
0
1
0
1
b
=
= LDLT ;
9 b2
b1
0 9 b2
01
4
10
2
0
12
=
= LDLT .
c8
21
0 c8
01
2
3 f (x, y ) = x + 4xy + 9y = (x + 2y )2 + 5y 2 ; f (x, y ) = x2 + 6xy + 9y 2 = (x + 3y )2 .
4 x2 + 4xy + 3y 2 = (x + 2y )2 y 2 is negative at x = 2, y = 1.
01
01
x
produces f (x, y ) = [ x y ]
= 2xy . A has = 1 and 1.
5 A=
10
10
y
6 x T AT Ax = (Ax )T (Ax ) = 0 only if Ax = 0. Since A has independent columns this only
happens when x = 0.
7 AT A =
8 A=
3
6
4
9 A = 4
8
2
10 A = 1
0
1
2
2
13
6
16
=
4
8
1
0
21
3
0
5
2
3
3
are positive denite; AT A = 3 5 4 is singular.
56
345
0
12
. Pivots outside squares, and L inside.
4
01
and AT A =
6
4 8 has only one
8 16
1
0
2 1 has pivots 2,
1
2
pivot = 4, rank A = 1, eigenvalues are 24, 0, 0, det A = 0.
1
0
34
, ; A = 1
2 1 is singular; A 1 = 0 .
23
1 1
2
1
0
2 1 1
11 |A1 | = 2, |A2 | = 6, |A3 | = 30. The pivots are 2/1, 6/2, 30/6.
12 A is positive denite for c > 1; determinants c, c2 1, c3 + 2 3c > 0. B is never positive
denite (determinants d 4 and 4d + 12 are never both positive).
1
5
has a + c > 2b but ac < b2 , so not positive denite.
13 A =
5 10
14 The eigenvalues of A1 are positive because they are 1/(A). And the entries of A1 pass the
determinant tests. And x T A1 x = (A1 x )T A(A1 x ) > 0 for all x = 0.
15 Since x T Ax > 0 and x T B x > 0 we have x T (A + B )x = x T Ax + x T B x > 0 for all x = 0.
Then A + B is a positive denite matrix.
16 x T Ax is not positive when (x1 , x2 , x3 ) = (0, 1, 0) because of the zero on the diagonal.
17 If ajj were smaller than all the eigenvalues, A ajj I would have positive eigenvalues (so positive
denite). But A ajj I has a zero in the (j, j ) position; impossible by Problem 16.
18 If Ax = x then x T Ax = x T x . If A is positive denite this leads to = x T Ax /x T x > 0
(ratio of positive numbers).
19 All cross terms are x T x j = 0 because symmetric matrices have orthogonal eigenvectors.
i
20 (a) The determinant is positive, all > 0
(b) All projection matrices except I are singular
(c) The diagonal entries of D are its eigenvalues
(d) I has det = 1 when n is even.
58
21 A is positive denite when s > 8; B is positive denite when t > 5 (check determinants).
40
31
1 1 1 9
1 1 1 2 1
QT =
.
22 R =
=
; R = Q
21
2 1 1
1
1
12
02
13
23 1 = 1/a2 and 2 = 1/b2 so a = 1/ 1 and b = 1/ 2 . The ellipse 9x2 + 16y 2 = 1 has axes
with half-lengths a =
1
3
and b = 1 .
4
24 The ellipse x2 + xy + y 2 = 1 has axes with half-lengths a = 1/ 1 = 2 and b = 2/3.
93
20
; C =
.
25 A =
35
43
300
10
0
26 C = L D = 0 1 0 and C = 1 1
0 have square roots of the pivots from D.
022
11
5
b
27 ax2 + 2bxy + cy 2 = a(x + a y )2 +
acb2 2
y;
a
2x2 + 8xy + 10y 2 = 2(x + 2y )2 + 2y 2 .
28 det A = 10; = 2 and 5; x 1 = (cos , sin ), x 2 = ( sin , cos ); the s are positive.
6x2 2x
is positive denite if x = 0; f1 = ( 1 x2 + y )2 = 0 on the curve 1 x2 + y = 0;
29 A1 =
2
2
2x
2
6x 1
61
=
is indenite and (0, 1) is a saddle point.
A2 =
10
10
30 ax2 + 2bxy + cy 2 has a saddle point if ac < b2 . The matrix is indenite ( < 0 and > 0).
31 If c > 9 the graph of z is a bowl, if c < 9 the graph has a saddle point. When c = 9 the graph
of z = (2x + 3y )2 is a trough staying at zero on the line 2x + 3y = 0.
32 Orthogonal matrices, exponentials eAt , matrices with det = 1 are groups. Examples of subgroups are orthogonal matrices with det = 1, exponentials eAn for integer n.
Problem Set 6.6, page 310
1 C = (M N )1 A(M N ) so if B is similar to A and C is similar to B , then A is similar to C .
2 B = (F G1 )1 A(F G1 ). If C is similar to A and also to B then A
1
10
01
01
01
1
0
0
=
; M =
; M =
3
10
10
01
10
0 1
1
4 A has no repeated so it
11
00
1
,
,
5
00
11
1
is similar to B .
1
gives B = M 1 AM .
0
can be diagonalized: S 1 AS = makes A similar to .
0
01
10
01
,
are similar;
by itself and
by itself.
0
01
01
10
6 Eight families of similar matrices: 6 matrices have = 0, 1; 3 matrices have = 1, 1 and 3 have
1
= 0, 0 (two families each!); one has = 1, 1; one has = 2, 0; two have = 2 (1 5).
7 (a) (M 1 AM )(M 1 x ) = M 1 (Ax ) = M 1 0 = 0
(b) The nullspaces of A and of M 1 AM
have the same dimension. Dierent vectors and dierent bases.
59
8
0
1
0
0
and
2
9 A =
10 J 2 =
0
2
0
0
1
2
0
1
c2
2c
0
c2
have the same line of eigenvectors and the same eigenvalues 0, 0.
3
, A =
1
3
0
1
, J 3 =
k
, every A =
c3
3c2
0
c3
, J k =
1
k
0
1
0
. A =
ck
kck1
0
ck
1
0
0
1
1
and A
; J 0 = I , J 1 =
=
1 1
0
1
c1 c2
c1
0
.
.
11 w(t) = w(0) + tx(0) + 1 t2 y (0) + 1 t3 z (0) e5t .
2
6
m21
m22
m23
m24
0
0
0
0
0
0
1
= MK =
12 If M JM = K then JM =
m41 m42 m43 m44
0
0
0
0
0
0
That means m21 = m22 = m23 = m24 = 0 and M is not invertible.
m12
m13
m22
m23
m32
m33
m42
m43
13 (1) Choose Mi = reverse diagonal matrix to get Mi1 Ji Mi = MiT in each block
those blocks Mi on its block diagonal to get
M0 1 JM0
T
=J .
T
(3) A
= (M
0
0
.
0
0
(2) M0 has
1 T
) J T M T is
(M 1 )T M0 1 JM0 M T = (M M0 M T )1 A(M M0 M T ), and AT is similar to A.
14 Every matrix M JM 1 will be similar to J .
15 det(M 1 AM I ) = det(M 1 AM M 1 IM ) = det(M 1 (A I )M ) = det(A I ).
ab
dc
ba
cd
01
is similar to
;
is similar to
. I is not similar to
.
16
cd
ba
dc
ab
10
17 (a) True: One has = 0, the other doesnt
(b) False. Diagonalize a nonsymmetric
01
0 1
and
are similar
matrix and is symmetric
(c) False:
(d) True:
1 0
1
0
All eigenvalues of A + I are increased by 1, so dierent from the eigenvalues of A.
18 AB = B 1 (BA)B so AB is similar to BA. Also AB x = x leads to BA(B x ) = (B x ).
19 Diagonals 6 by 6 and 4 by 4; AB has all the same eigenvalues as BA plus 6 4 zeros.
20 (a) A = M 1 BM A2 = (M 1 BM )( 1 BM ) = M 1 B 2 M
M
to B = A (but it could be!)
(d)
3
1
0
3
(c)
3
0
(b) A may not be similar
1
30
is diagonalizable to
because 1 = 2
4
04
has only one eigenvector, so not diagonalizable
(e) P AP T is similar to A.
21 J 2 has three 1s down the second superdiagonal, and two
independent eigenvectors for = 0.
010
J3
01
with J3 = 0 0 1 and J2 =
.
Its 5 by 5 Jordan form is
J2
00
000
Note to professors: You could list all 3 by 3 and 4 by 4 Jordan J s:
60
a
0
0
a
0
b
0
1
a
0
a
1
0
a
1
0
0 , 0 a 0 , 0 a 1 with 3, 2, 1 eigenvectors; diag(a, b, c, d) and
00b
00a
c
a1
a1
a1
a1
a1
a
with 4, 3, 2, 1 eigenvectors.
,
,
,
b
b 1
a 1
a
a
b
b
c
Problem Set 6.7, page 318
1/ 17
4/ 17
has
A A=
= 85, v 1 = , v 2 =
.
20 80
4/ 17
1/ 17
2/ 5
17 34
1/ 5
2
has 1 = 85, u1 = , u2 =
(a) AAT =
.
34 68
2/ 5
1/ 5
17
14
1/ 17
1/ 5
= = 85 = 1 u1 .
(b) Av 1 =
28
4/ 17
2 17
2/ 5
1/ 17
2/ 5
1/ 5
u1 = for the column space, v 1 = for the row space, u2 =
for
2/ 5
4/ 17
1/ 5
4/ 17
the nullspace, v 2 =
for the left nullspace.
1/ 17
21
T
T
2
2
has eigenvalues 1 = 3 + 5 and 2 = 3 5 .
A A = AA =
2
2
11
1
2
3
4
T
5
20
2
1
Since A = AT the eigenvectors of AT A are the same as for A. Since 2 =
1 5
2
is negative,
1 = 1 but 2 = 2 . The eigenvectors are the same as in Section 6.2 for A, except for the
1 / 1 + 2
2 / 1 + 2
2
1
.
and u2 = v 2 =
eect of this minus sign: u1 = v 1 =
1/ 1 + 2
1/ 1 + 2
2
1
6 A proof that eigshow nds the SVD for 2 by 2 matrices. Starting at the orthogonal pair
V 1 = (1, 0), V 2 = (0, 1) the demo nds AV 1 and AV 2 at angle . After a 90 turn by the
mouse to V 2 , V 1 the demo nds AV 2 and AV 1 at angle . Somewhere between, the
constantly orthogonal v 1 , v 2 must have produced Av 1 and Av 2 at angle = /2. Those are
the orthogonal directions for u 1 and u 2 .
21
1/ 2
1/ 2
2
2
T
has 1 = 3 with u1 = and 2 = 1 with u2 =
7 AAT =
. A A =
12
1/ 2
1/ 2
1/ 3
1/ 2
110
1/ 6
2
2
1 2 1 has 1 = 3 with v 1 = 2/ 6 , 2 = 1 with v 2 = 0 ; and v 3 = 1/ 3 .
1/ 3
1/ 2
1/ 6
011
110
300
= u1 u2
v1 v2 v3 T .
Then
011
0
10
8 A = U V T since all j = 1.
61
9 A = 12 U V T .
10 A = W W T is the same as A = U V T .
11 Multiply U V T using columns (of U ) times rows (of V T ).
2
2
12 Since AT = A we have 1 = 2 and 2 = 2 . But 2 is negative, so 1 = 3 and 2 = 2. The
2
1
unit eigenvectors of A are the same u1 = v 1 as for AT A = AAT and u2 = v 2 (notice sign
change because 2 = 2 ).
13 Suppose the SVD of R is R = U V T . Then multiply by Q. So the SVD of this A is (QU )V T .
14 The smallest change in A is to set its smallest singular value 2 to zero.
15 (a) If A changes to 4A, multiply by 4.
(b) AT = V T U T . And if A1 exists, it is square
and equal to (V T )1 1 U 1 .
16 The singular values of A + I are not j + 1. They come from eigenvalues of (A + I )T (A + I ).
17 This simulates the random walk used by Google on billions of sites to solve Ap = p . It is
like the power method of 9.3 except that it follows the links in one walk where the vectors
p k = Ak p 0 averages over all walks.
Problem Set 7.1, page 325
1 With w = 0 linearity gives T (v + 0) = T (v ) + T (0). Thus T (0) = 0. With c = 1 linearity
gives T (0) = T (0). Thus T (0) = 0.
2 T (cv + dw ) = cT (v ) + dT (w ); add eT (u ).
3 (d) is not linear.
4 (a) S (T (v )) = v
(b) S (T (v 1 ) + T (v 2 )) = S (T (v 1 )) + S (T (v 2 )).
5 Choose v = (1, 1) and w = (1, 0). Then T (v ) + T (w ) = v + w but T (v + w ) = (0, 0).
6 (b) and (c) are linear
7 (a) T (T (v )) = v
(b) T (T (v )) = v + (2, 2)
8 (a) Range R2 , kernel {0}
kernel R2
(d) satises T (cv ) = cT (v ).
(c) T (T (v )) = v
(b) Range R2 , kernel {(0, 0, v3 )}
(d) T (T (v )) = T (v ).
(c) Range {0},
(d) Range = multiples of (1, 1), kernel = multiples of (1, 1).
9 T (T (v )) = (v3 , v1 , v2 ); T 3 (v ) = v ; T 100 (v ) = T (v ).
10 (a) T (1, 0) = 0
(b) (0, 0, 1) is not in the range
(c) T (0, 1) = 0.
11 V = Rn , W = Rm ; the outputs ll the column space; v is in the kernel if Av = 0.
12 T (v ) = (4, 4); (2, 2); (2, 2); if v = (a, b) = b(1, 1) +
ab
(2, 0)
2
then T (v ) = b(2, 2) + (0, 0).
13 Associative gives A(M1 + M2 ) = AM1 + AM2 . Distributive over cs gives A(cM ) = c(AM ).
14 A is invertible. Multiply AM = 0 and AM = B by A1 to get M = 0 and M = A1 B .
2
2
00
=
.
15 A is not invertible. AM = I is impossible. A
1 1
00
62
16 No matrix A gives A
0
0
0
1
=
. To professors: The matrix space has dimension 4.
10
00
Linear transformations come from 4 by 4 matrices. Those in Problems 1315 were special.
17 (a) True
(b) True
0
18 T (I ) = 0 but M =
0
b
0
(c) True
(d) False.
= T (M ); these ll the range. M =
a
0
c
d
in the kernel.
19 If v = 0 is a column of B and u T = 0 is a row of A, choose M = uv T .
20 T (T 1 (M )) = M so T 1 (M ) = A1 M B 1 .
21 (a) Horizontal lines stay horizontal, vertical lines stay vertical
(c) Vertical lines stay vertical.
a0
with d > 0
23 (a) A =
(b) A = 3I
0d
(b) House squashes onto a
line
24 (a) ad bc = 0
(b) ad bc > 0
(c) A =
(c) |ad bc| = 1.
cos sin
sin
cos
.
If vectors to two corners transform
to themselves then by linearity T = I . (Fails if one corner is (0, 0).)
25 Rotate the house by 180 and shift one unit to the right.
27 This emphasizes that circles are transformed to ellipses (gure in Section 6.7).
30 Squeezed by 10 in y direction; attened onto 45 line; rotated by 45 and stretched by
2;
ipped over and skewed so squares become parallelograms.
Problem Set 7.2, page 337
0
0
1 S v 1 = S v 2 = 0, S v 3 = 2v 1 , S v 4 = 6v 2 ; B =
0
0
0
2
0
0
0
0
0
0
0
6
.
0
0
2 All functions v (x) = a + bx; all vectors (a, b, 0, 0).
3 A2 = B when T 2 = S and output basis = input basis.
4 Third derivative has 6 in the (1, 4) position; fourth derivative of cubic is zero.
011
5 A = 1 0 0 .
011
6 T (v 1 + v 2 + v 3 ) = 2w 1 + w 2 + 2w 3 ; A times (1, 1, 1) gives (2, 1, 2).
7 v = c(v 2 v 3 ) gives T (v ) = 0; nullspace is (0, c, c); solutions are (1, 0, 0) + any (0, c, c).
8 (1, 0, 0) is not in the column space; w 1 is not in the range.
9 We dont know T (w ) unless the w s are the same as the v s. In that case the matrix is A2 .
10 Rank = 2 = dimension of the range of T .
63
1
11 A = 1 1 0 ; for output 0 choose input v = v 1 v 2 .
111
0
1
0
0
1
1
1
1
12 A = 1
1
0 so T (w 1 ) = v 1 v 2 , T (w 2 ) = v 2 v 3 , T (w 3 ) = v 3 ; the only
0 1
1
solution to T (v ) = 0 is v = 0.
1
0
0
13 (c) is wrong because w 1 is not generally in the input space.
14 (a) T (v 1 ) = v 2 , T (v 2 ) = v 1
then T = I .
21
15 (a)
(b)
53
rs
16 (a) M =
tu
10
2
17 M N =
12
5
(b) T (v 1 ) = v 1 , T (v 2 ) = 0
3 1
5
2
= inverse of (a)
(b) N =
1
3
1
a
b
c
d
2
1
(c) A must be 2A .
6
3
1
3 1
=
7
(c) If T 2 = I and T 2 = T
3
(c) ad = bc.
.
18 Permutation matrix; positive diagonal matrix.
19 (a, b) = (cos , sin ). Minus sign from Q1 = QT .
11
; (a, b) = (5, 4) = rst column of M 1 .
20 M =
45
21 w 2 (x) = 1 x2 ; w 3 (x) = 1 (x2 x); y = 4w 1 + 5w 2 + 6w 3 .
2
0
1
0
1
1
1
22 w s to v s: .5
0 .5 . v s to w s: inverse matrix = 1
0
0 .
.5 1
.5
1 1
1
1 a a2
A
4
23 1 b b2 B = 5 ; Vandermonde determinant = (b a)(c a)(c b); a, b, c must
1 c c2
C
6
be distinct.
24 The matrix M with these nine entries must be invertible.
25 a 2 = r12 q 1 + r22 q 2 gives a 2 as a combination of the q s. So the change of basis matrix is R.
26 Row 2 of A is l21 (row 1 of U ) + l22 (row 2 of U ). The change of basis matrix is always invertible.
27 The matrix is .
28 If T is not invertible then T (v 1 ), . . ., T (v n ) will not be a basis. Then we couldnt choose
w i = T (v i ).
03
29 (a)
00
(b)
1
0
0
0
.
30 T (x, y ) = (x, y ) and then S (x, y ) = (x, y ). Thus ST = I .
64
31 S (T (v )) = (1, 2) but S (v ) = (2, 1) and T (S (v )) = (1, 2).
cos 2( ) sin 2( )
rotates by 2( ).
32
sin 2( )
cos 2( )
33 False, because the v s might not be linearly independent.
Problem Set 7.3, page 345
1 Multiply by W 1
1
4
1
4
1
4
1
4
1
1
1 1
4
4
4
4
=
. Then e = 1 w1 + 1 w2 + 1 w3 and v = w3 + w4 .
4
4
2
1
1
0
0
2 2
1
0
0
1
2
2
2 The last step writes 6, 6, 2, 2 as the overall average 4, 4, 4, 4 plus the dierence 2, 2, 2, 2.
Therefore c1 = 4 and c2 = 2 and c3 = 1 and c4 = 1.
3 The wavelet basis is (1, 1, 1, 1, 1, 1, 1, 1) and the long wavelet and two medium wavelets (1, 1,
1, 1, 0, 0, 0, 0) and
1
1
0
2
2
1
1
2 2 0
4 W2 1 =
01
0
0
00
(0, 0, 0, 0, 1, 1, 1, 1) and 4 short wavelets with a single pair 1, 1.
1
1
0
0
0
2
2
1
1
0
0
02
2
1
.
and W1 =
1
0
0
2 1 0
2
1
0
0 1 1
2
2
5 The Hadamard matrix H has orthogonal columns of length 2. So the inverse is H T /4 = H/4.
6 If V b = W c then b = V 1 W c. The change of basis matrix is V 1 W .
7 The transpose of W W 1 = I is (W 1 )T W T = I . So the matrix W T (which has the ws in its
rows) is the inverse to the matrix that has the w s in its columns.
Problem Set 7.4, page 353
1 AT A =
2 AAT =
10
20
5
15
1 1
1 2
has = 50 and 0, v 1 =
, v2 =
; 1 = 50.
52
5 1
40
15
1 1
1 3
has = 50 and 0, u 1 =
, u2 =
.
10 3
10 1
45
20
3 Orthonormal bases: v 1 for row space, v 2 for nullspace, u 1 for column space, u 2 for N(AT ).
4 The matrices with those four subspaces are multiples cA.
1 7 1 1 10 20
5 A = QH =
. H is semidenite because A is singular.
50 1
50 20 40
7
1/ 50 0
13
.2 .4
.1 .3
UT = 1
; A+ A =
, AA+ =
.
6 A+ = V
50
0
0
26
.4 .8
.3 .9
65
7 AT A =
10
8
8
has = 18 and 2, v 1 =
1
2
1
, v 2 =
1
1
2
1
, 1 =
1
18 and 2 = 2.
10
18 0
1
0
has u 1 = , u 2 = .
8 AAT =
02
0
1
T
v1
T
T
T
T
9 1 u 1 2 u 2
= 1 u 1 v 1 + 2 u 2 v 2 . In general this is 1 u 1 v 1 + + r u r v r .
T
v2
1
18
0
1 1
and K =
10 Q = U V T =
.
2 1 1
0
2
11 A+ is A1 because A is invertible.
9 12 0
.6
.8
0
T
12 A A = 12 16 0 has = 25, 0, 0 and v 1 = .8 , v 2 = .6 , v 3 = 0 .
0
00
0
0
1
AAT = [ 25 ] and 1 = 5.
.2
.12
.36
13 A = [ 1 ] [ 5 0 0 ]V T and A+ = V 0 [ 1 ] = .16 ; AA+ = [ 1 ]; A+ A = .48
0
0
0
.48
.64
0
0
0
0
+
14 Zero matrix; = 0; A = 0 is 3 by 2.
15 If det A = 0 then rank(A) < n; thus rank(A+ ) < n and det A+ = 0.
16 A must be symmetric and positive denite.
17 (a) AT A is singular
(b) AT Ax + = AT b
(c) (I AA+ ) projects onto N(AT ).
18 x + in the row space of A is perpendicular to x x + in the nullspace of AT A = nullspace of
A. The right triangle has c2 = a2 + b2 .
19 AA+ p = p , AA+ e = 0, A+ Ax r = x r , A+ Ax n = 0.
20 A+ = 1 [ .6 .8 ] = [ .12 .16 ] and A+ A = [ 1 ] and AA+ =
5
21 L is determined by
21 .
.36
.48
.48
.64
.
Each eigenvector in S is determined by one number. The counts are
1 + 3 for LU , 1 + 2 + 1 for LDU , 1 + 3 for QR, 1 + 2 + 1 for U V T , 2 + 2 + 0 for S S 1 .
22 The counts are 1 + 2 + 0 because A is symmetric.
23 Column times row multiplication gives A = U V T =
i u i v T and also A+ = V + U T =
i
i 1 v i u T . Multiplying A+ A and using orthogonality of each u i to all other u j leaves the
i
projection matrix A+ A:
A+ A =
1v i v T . Similarly AA+ =
i
1u i u T from V V T = I .
i
24 The columns of U are a basis for the column space of A. So are the rst r columns of U .
Those r columns must have the form U M1 for some r by r invertible matrix M1 . Similarly
the columns of V and the rst r columns of V are bases for the row space of A. So V = V M2 .
Keep only the r by r invertible corner r of (the rest is all zero). Then A = U V T has the
T
T
required form A = U M1 r M2 V T with an invertible M = M1 r M2 in the middle.
0
A
u
u
= . That block matrix connects to AT A and AAT .
25
T
A
0
v
v
66
Problem Set 8.1, page ???
1 Det AT C0 A0 is by direct calculation. Set c4 = 0 to
0
111
c1
10
1
T
1
1
2 (A1 C1 A1 ) = 0 1 1
1 1
c2
1
11
001
c3
nd det AT C1 A1 = c1 c2 c3 .
1
1
0
c + c1 + c1 c1 + c1
2
3
2
3
1
1
1
1
c2 + c3
c2 + c1
0 =
3
1
c1
c1
3
3
c1
3
c1
3
c1
3
3 The rows of the free-free matrix in equation (9) add to [ 0 0 0 ] so the right side needs
f1 + f2 + f3 = 0. For f = (1, 0, 1) elimination gives c2 u1 c2 u2 = 1, c3 u2 c3 u3 = 1, and
0 = 0. Then u particular = (c1 c1 , c1 , 0). Add any multiple of u nullspace = (1, 1, 1).
2
3
3
d
dx
du
dx
du
du
(0) c(1) (1) = 0 so we need
dx
dx
4
5
dy
= f (x) gives y (x) = C
dx
c(x)
dx = c(0)
1
f (x) dx = 0.
x
1
f (t) dt. Then y (1) = 0 gives C =
0
f (t) dt and y (x) =
0
f (t) dt. If f (x) = 1 then y (x) = 1 x.
x
6 Multiply AT C1 A1 as columns of AT times cs times rows of A1 . The rst element matrix
1
1
c1 E1 = [ 1 0 0 ]T c1 [ 1 0 0 ] has c1 in the top left corner.
7 For 5 springs and 4 masses, the 5 by 4 A has all aii = 1 and ai+1,i = 1. With C =
diag(c1 , c2 , c3 , c4 , c5 ) we get K = AT CA, symmetric tridiagonal with Kii = ci + ci+1 and
Ki+1,i = ci+1 . With C = I this K is the 1, 2, 1 matrix and K (2, 3, 3, 2) = (1, 1, 1, 1).
8 The solution to u = 1 with u(0) = u(1) = 0 is u(x) = 1 (x x2 ). At x = 1 , 2 , 3 ,
2
555
4
5
this u(x)
2
equals u = 2, 3, 3, 2 (discrete solution in Problem 7) times (x) = 1/25.
9 u = mg has complete solution u(x) = A + Bx 1 mgx2 . From u(0) = 0 we get A = 0. From
2
1
u (1) = 0 we get B = mg . Then u(x) = 2 mg (2xx2 ) at x = 1 , 2 ,
33
3
3
equals mg/6, 4mg/9, mg/2.
This u(x) is not proportional to the discrete u at the meshpoints.
10 The graphs of 100 points are discrete parabolas starting at (0, 0): symmetric around 50 in
the xed-xed case, ending with slope zero in the xed-free case.
11 Forward vs. backward dierences for du/dx have a big eect on the discrete u , because that
term has the large coecient 10 (and with 100 or 1000 we would have a real boundary layer
= near discontinuity at x = 1). The computed values are u = 0, .01, .03, .04, .05, .06, .07, .11, 0
versus u = 0, .12, .24, .36, .46, .54, .55, .43, 0. The MATLAB code is E = diag(ones(6, 1), 1);
K = 64 (2 eye(7) E E ); D = 80 (E eye(7)); (K + D)\ones(7, 1), (K D )\ones(7, 1).
Problem Set 8.2, page 366
1
1
1 A = 1
0
0 1
c
1 ; nullspace contains c ;
1
c
0
1
0 is not orthogonal to that nullspace.
0
2 AT y = 0 for y = (1, 1, 1); current = 1 along edge 1, edge 3, back on edge 2 (full loop).
67
1
1
3 U = 0 1
0
0
0
1 ; tree from edges 1 and 2.
0
4 Ax = b is solvable for b = (1, 1, 0) and not solvable for b = (1, 0, 0); b must be orthogonal to
y = (1, 1, 1); b1 b2 + b3 = 0 is the third equation after elimination.
5 Kirchhos Current Law AT y = f is solvable for f = (1, 1, 0) and not solvable for f = (1, 0, 0);
f must be orthogonal to (1, 1, 1) in the nullspace.
c
2 1 1
3
1
T
6 A Ax = 1
2 1 x = 3 = f produces x = 1 + c ; potentials 1, 1, 0 and
1 1
2
0
0
c
currents Ax = 2, 1, 1; f sends 3 units into node 1 and out from node 2.
1
3 1 2
1
5/4
c
T
5
7
7A
A = 1
2
3 2 ; f = 0 yields x = 1 + c ; potentials 4 , 1, 8
2
2 2
4
1
7/8
c
and currents CAx = 1 , 3 , 1 .
444
1
1
0
0
1
1
1
1
0
1
0
1
and y = 1 ,
8 A = 0 1
1
0 leads to x =
1
0
0 1
0
1
1
0
0 1
1
0
0
0
1 .
1
1
9 Elimination on Ax = b always leads to y T b = 0 which is b1 + b2 b3 = 0 and b3 b4 + b5 = 0
(y s from Problem 8 in the left nullspace). This is Kirchhos Voltage Law around the loops.
1
1
0
0
is the matrix that keeps
0 1
1
0
edges 1, 2, 4; other trees
10 U = 0
0 1
1
from 1, 2, 5; 1, 3, 4; 1, 3, 5;
0
0
0
0
1, 4, 5; 2, 3, 4; 2, 3, 5; 2, 4, 5.
0
0
0
0
2 1 1
0
diagonal entry = number
1
3 1 1 of edges into the node
11 AT A =
1 1
3 1 o-diagonal entry = 1
0 1 1
2
if nodes are connected.
12 (1) The nullspace and rank of AT A and A are always the same
semidenite because x T AT Ax = Ax
2
(2) AT A is always positive
0. Not positive denite because rank is only 3 and
(1, 1, 1, 1) is in the nullspace
(3) Real eigenvalues all 0 because positive semidenite.
4 2 2
0
1
2
0
8 3 3
x = gives potentials x = ( 5 , 1 , 1 , 0) (grounded x4 = 0
13 AT CAx =
12 6 6
2 3
0
8 3
0 3 3
6
1
2
and solved 3 equations); y = CAx = ( 3 , 2 , 0, 1 , 1 ).
3
22
14 AT CAx = 0 for x = (c, c, c, c); then f must be orthogonal to x .
68
15 n m + 1 = 7 7 + 1 = 1 loop.
16 5 7 + 3 = 1; 5 8 + 4 = 1.
17 (a) 8 independent columns
(b) f must be orthogonal to the nullspace so f1 + + f9 = 0
(c) Each edge goes into 2 nodes, 12 edges make diagonal entries sum to 24.
18 Complete graph has 5 + 4 + 3 + 2 + 1 = 15 edges; tree has 5 edges.
Problem Set 8.3, page 373
1 = 1 and .75; (A I )x = 0 gives x = (.6, .4).
.6 1
1
1
1
;
2 A=
.4
1
.75
.4
.6
.6 1
10
1
1
.6
=
Ak approaches
.4 1
00
.4
.6
.4
.6
.4
.
3 = 1 and .8, x = (1, 0); = 1 and .8, x = ( 5 , 4 ); = 1,
99
1
,
4
and
1
,
4
x = ( 1 , 1 , 1 ).
333
4 AT always has the eigenvector (1, 1, . . . , 1) for = 1.
5 The steady state is (0, 0, 1) = all dead.
6 If Ax = x , add components on both sides to nd s = s. If = 1 the sum must be s = 0.
.8 .3
.6 1
1
11
=
; A16 has the same factors except now (.5)16 .
7
.2 .7
.4
1
.5
.4 .6
.6 + .4a .6 .6a
k
k
with 2 a 1.
8 (.5) 0 gives A A ; any A =
3
.4 .4a .4 + .6a
9 u 1 = (0, 0, 1, 0); u 2 = (0, 1, 0, 0); u 3 = (1, 0, 0, 0); u 4 = u 0 . The eigenvalues 1, i, 1, i are
all on the unit circle. This Markov matrix contains zeros; a positive matrix has one largest
eigenvalue.
10 M 2 is still nonnegative; [ 1 1 ]M = [ 1 1 ] so multiply by M to nd
[ 1 1 ]M 2 = [ 1 1 ] columns of M 2 add to 1.
11 = 1 and a + d 1 from the trace; steady state is a multiple of x 1 = (b, 1 a).
12 Last row .2, .3, .5 makes A = AT ; rows also add to 1 so (1, . . . , 1) is also an eigenvector of A.
13 B has = 0 and .5 with x 1 = (.3, .2) and x 2 = (1, 1); e.5t approaches zero and the
solution approaches c1 e0t x 1 = c1 x 1 .
14 Each column of B = A I adds to zero. Then 1 = 0 and e0t = 1.
15 The eigenvector is x = (1, 1, 1) and Ax = (.9, .9, .9).
16 (I A)(I + A + A2 + . . .) = I + A + A2 + . . . A + A2 + A3 + . . .) = I . This says that
(
0 .5
, A2 = 1 I , A3 = 1 A, A4 = 1 I and the
I + A + A2 + . . . is (I A)1 . When A =
2
2
4
10
1
1+ 2 +. . . 1 + 1 +. . .
21
2
4
=
= (I A)1 .
series adds to
1
1
1+ 2 +. . . 1+ 2 +. . .
22
69
17
0
1
0
4
and
have max < 1.
0
.2 0
8
130
.5 1
; I
has no inverse.
18 p = and
6
32
.5 0
0
19 = 1 (Markov), 0 (singular), .2 (from trace). Steady state (.3, .3, .4) and (30, 30, 40).
20 No, A has an eigenvalue = 1 and (I A)1 does not exist.
Problem Set 8.4, page 382
1 Feasible set = line segment from (6, 0) to (0, 3); minimum cost at (6, 0), maximum at (0, 3).
2 Feasible set is 4-sided with corners (0, 0), (6, 0), (2, 2), (0, 6). Minimize 2x y at (6, 0).
3 Only two corners (4, 0, 0) and (0, 2, 0); choose x1 very negative, x2 = 0, and x3 = x1 4.
4 From (0, 0, 2) move to x = (0, 1, 1.5) with the constraint x1 + x2 + 2x3 = 4. The new cost
is 3(1) + 8(1.5) = $15 so r = 1 is the reduced cost. The simplex method also checks
x = (1, 0, 1.5) with cost 5(1) + 8(1.5) = $17 so r = 1 (more expensive).
5 Cost = 20 at start (4, 0, 0); keeping x1 + x2 + 2x3 = 4 move to (3, 1, 0) with cost 18 and
r = 2; or move to (2, 0, 1) with cost 17 and r = 3. Choose x3 as entering variable and move
to (0, 0, 2) with cost 14. Another step to reach (0, 4, 0) with minimum cost 12.
6 c = [ 3 5 7 ] has minimum cost 12 by the Ph.D. since x = (4, 0, 0) is minimizing. The dual
problem maximizes 4y subject to y 3, y 5, y 7. Maximum = 12.
Problem Set 8.5, page 387
1
2
0
sin(j +k)x
j +k
cos(j + k)x dx =
2
= 0 and similarly
0
2
0
notice j k = 0). If j = k then
2
1
(1)(x) dx
1
= 0,
1
(1)(x2
1
2
0
cos(j k)x dx = 0 (in the denominator
2
cos jx dx = .
1 ) dx = 0,
3
1
(x)(x2
1
1 ) dx = 0. Then 2x2 = 2(x2 1 ) +
3
3
2
0(x) + 3 (1).
3 w = (2, 1, 0, 0, . . .) has w =
4
1
(1)(x3
1
cx) dx = 0 and
Choose c so that
1
1
5.
1
(x2
1
1 )(x3 cx) dx = 0 for all c (integral of an odd function).
3
1
c
x(x3 cx) dx = [ 5 x5 3 x3 ]1 1 =
2
5
c 2 = 0. Then c = 3 .
3
5
5 The integrals lead to a1 = 0, b1 = 4/ , b2 = 0.
6 From equation (3) the ak are zero and bk = 4/k. The square wave has f
equation (6) is 2 = (16/
8
2
)( 11
2
+
1
32
+
1
52
2
= 2 . Then
2
+ ) so this innite series equals /8.
= 1 + 1 + 1 + 1 + = 2 so v = 2; v 2 = 1 + a2 + a4 + = 1/(1 a2 ) so
2
4
8
2
v = 1/ 1 a2 ; 0 (1 + 2 sin x + sin2 x) dx = 2 + 0 + so f = 3 .
v
2
70
1
+ 1 (square wave) so as are 1 ,
2
2
2
2
x dx/2 = , other ak = 0, bk
0
9 (a) f (x) =
(b) a0 =
0, 0, . . ., and bs are 2/ , 0, 2/3 , 0, 2/5 , . . .
= 2/k.
10 The integral from to or from 0 to 2 or from any a to a + 2 is over one complete period
2
0
of the function. If f (x) is odd (and periodic) then
0
f (x) dx =
f (x) dx +
0
f (x) dx and
those integrals cancel.
11 cos2 x = 1 + 1 cos 2x; cos(x +
2
2
1
0
0
0
cos x 0
0 1
d
12 dx sin x = 0
1
0
cos 2x 0
0
0
sin 2x
0
0
0
)
3
0
0
0
0
2
= cos x cos sin x sin
3
0
1
0 cos x
0 sin x .
2 cos 2x
0
sin 2x
3
=
1
2
cos x
3
2
sin x.
13 dy/dx = cos x has y = yp + yn = sin x + C .
Problem Set 8.6, page 392
1 (x, y, z ) has homogeneous coordinates (x, y, z, 1) and also (cx, cy, cz, c) for any nonzero c.
2 For an ane transformation we need T (origin). Then (x, y, z, 1) xT (i ) + yT (j ) + zT (k ) +
T (0).
3 T T1 =
1
1
1
4
6
c
c
1
1
0
1/11
1
1
1
2
2
, ST =
1/8.5
1
1
5
1
=
c
c
1
4
1
1
1
c
3
1
1
6
8
1
, T S =
is translation along (1, 6, 8).
c
c
c
c
4c
3c
1
1
1
1
5 S=
3
c
4 S=
1
1
for a 1 by 1 square.
2
2
2
1
=
2
2
2
2
2
4
1
.
9 n = ( 2 , 2 , 1 ) has n = 1 and P = I nn T =
333
1
9
5 4 2
4
5 2 .
2 2
8
1
, use v T S .
71
10 Choose (0, 0, 3) on the plane and multiply T P T+ =
11 (3, 3, 3) projects to
1
(1, 1, 4)
3
1
9
5 4 2
4
5 2
2 2
8
6
6
3
0
0
.
0
9
and (3, 3, 3, 1) projects to ( 1 , 1 , 5 , 1).
333
12 A parallelogram (or a line segment).
13 The projection of a cube is a hexagon.
1 8 4
1
11
11
1
14 (3, 3, 3)(I 2nn T ) = ( 1 , 1 , 3 ) 8
1 4 = ( 3 , 3 , 3 ).
33
4 4
7
7
15 (3, 3, 3, 1) (3, 3, 0, 1) ( 3 , 7 , 8 , 1) ( 7 , 7 , 1 , 1).
3
3
3
33
16 v = (x, y, z, 0) ending in 0; add a vector to a point.
17 Rescaled by 1/c because (x, y, z, c) is the same point as (x/c, y/c, z/c, 1).
Problem Set 9.1, page 402
1 Without exchange, pivots .001 and 1000; with exchange, pivots 1 and 1.
1
larger than the entries below it, lij = entry/pivot has |lij | 1. A = 0
1
9
36
30
When the pivot is
1
1
1 1 .
1
1
2 A1 = 36
192 180 .
30 180
180
1
11/16
1.833
0
1.80
3 A = 1 = 13/12 = 1.083 compared with A 6 = 1.10 .
1
47/60
0.783
3.6
0.78
x > 6.
b
< .04 but
4 The largest x = A1 b is 1/min ; the largest error is 1016 /min .
5 Each row of U has at most w entries. Then w multiplications to substitute components of x
(already known from below) and divide by the pivot. Total for n rows is less than wn.
6 L, U , and R need
12
n
2
multiplications to solve a linear system. Q needs n2 to multiply the
right side by Q1 = QT . So QR takes 1.5 times longer than LU to reach x .
7 On column j of I , back substitution needs 1 j 2 multiplications (only the j by j upper left block
2
is involved). Then
1
(12
2
+ 22 + + n2 ) 1 ( 1 n3 ).
23
72
8
1
0
2
2
2
2
0 1
0
2
0
2
2
1
0
2
2
2
0 1
2
1 0
2
0
0 1
0
= U with P =
2
0 0
1
0
2
2
0
0
0
1
1
0
and L =
0
0 = U with P = 0
1
1
1
.5
1
0
0
0
2
2
0
; A 1 0 1
1
020
0
1
0
0
1 and L = 0
1
0 .
0
.5 .5
1
9 The cofactors are C13 = C31 = C24 = C42 = 1 and C14 = C41 = 1.
10 With 16-digit oating point arithmetic the errors x y computed for = 103 , 106 , 109 ,
1012 , 1015 are of order 1016 , 1011 , 107 , 104 , 103 .
1
3
1 1
1
=
11 cos = 1/ 10, sin = 3/ 10, R = 10
3
1
3
5
1
10
10
14
0
8
.
12 Eigenvalues 4 and 2. Put one of the unit eigenvectors in row 1 of Q: either
1 1
2 4
1
and QAQ1 =
or
Q = 2
1
1
0
4
1 3
4 4
1
1
and QAQ =
.
Q = 10
3
1
0
2
13 Changes in rows i and j ; changes also in columns i and j .
14 Qij A uses 4n multiplications (2 for each entry in rows i and j ). By factoring out cos , the
entries 1 and tan need only 2n multiplications, which leads to
23
n
3
for QR.
Problem Set 9.2, page 408
A = 3, c = 3/1 = 3; A = 2 + 2, c = (2 + 2)/(2 2) = 5.83.
A = 2, c = innite (singular matrix); A = 2, c = 1.
1
A = 2, c = 2/.5 = 4;
2
A = 2, c = 1;
3 For the rst inequality replace x by B x in Ax
Bx B
A
x . Then AB = max( AB x / x ) A
4 Choose B = A1 and compute I = 1. Then 1 A
x ; the second inequality is just
B.
A1 = c(A).
5 If max = min = 1 then all i = 1 and A = SIS 1 = I . The only matrices with A =
A1 = 1 are orthogonal matrices.
6
AQ
R = R and in reverse R Q1
7 The triangle inequality gives
Ax + B x
A = A.
Ax + B x . Divide by
x
and take the
maximum over all nonzero vectors to nd A + B A + B .
8 If Ax = x then Ax / x = || for that particular vector x . When we maximize the ratio
over
0
9
0
1
0
all vectors we get A ||.
1
00
01
+
=
has (A) = 0 and (B ) = 0 but (A + B ) = 1; also AB =
0
10
10
0
has (AB ) = 1; thus (A) is not a norm.
0
73
10 The condition number of A1 is A1
(A1 )1 = c(A). Since AT A and AAT have the same
nonzero eigenvalues, A and AT have the same norm.
11 c(A) = (1.00005 + (1.00005)2 .0001)/(1.00005
).
12 det(2A) is not 2 det A; det(A + B ) is not always less than det A + det B ; taking |det A| does
not help. The only reasonable property is det AB = (det A)(det B ). The condition number
should not change when A is multiplied by 10.
13 The residual b Ay = (107 , 0) is much smaller than b Az = (.0013, .0016). But z is much
closer to the solution than y .
659,000 563,000
6
1
. Then A > 1, A1 > 106 , c > 106 .
14 det A = 10 so A =
913,000
780,000
15 x = 5, x 1 = 5, x = 1; x = 1, x 1 = 2, x = .7.
16 x2 + +x2 is not smaller than max(x2 ) and not larger than x2 + +x2 +2|x1 ||x2 |+ = x 2 .
1
n
1
n
1
i
Certainly x2 + +x2 n max(x2 ) so x n x . Choose y = (sign x1 , sign x2 , . . . , sign xn )
1
n
i
to get x y = x 1 . By Schwarz this is at most x y = n x . Choose x = (1, 1, . . . , 1)
for maximum ratios n.
17 The largest component |(x + y )i | = x + y
is not larger than |xi | + |yi | x
+y
.
The sum of absolute values |(x + y )i | is not larger than the sum of |xi | + |yi |. Therefore
x +y
1
x
1
+y
1.
18 |x1 |+2|x2 | is a norm; min |xi | is not a norm; x + x
is a norm; Ax is a norm provided A is
invertible (otherwise a nonzero vector has norm zero; for rectangular A we require independent
columns).
Problem Set 9.3, page 417
1 S = I and T = I A and S 1 T = I A.
2 If Ax = x then (I A)x = (1 )x . Real eigenvalues of B = I A have |1 | < 1 provided
is between 0 and 2.
1
1
which has || = 2.
3 This matrix A has I A =
1 1
4 Always AB
A
nd B 3 B 2
B . Choose A = B to nd B 2
BB
3
B
2
. Then choose A = B 2 to
. Continue (or use induction). Since B max |(B )| it is no
surprise that B < 1 gives convergence.
5 Ax = 0 gives (S T )x = 0. Then S x = T x and S 1 T x = x . Then = 1 means that the
errors do not approach zero.
01
with ||max = 1 .
6 Jacobi has S 1 T = 1
3
3
10
0 1
3
7 Gauss-Seidel has S 1 T =
with ||max =
01
9
1
9
= (||max for Jacobi)2 .
74
8 Jacobi has S 1 T =
a
1
0 b
0 b/a
=
with || = |bc/ad|1/2 . Gaussd
c
0
c/d
0
1
a0
0 b
0
b/a
=
with || = |bc/ad|.
Seidel has S 1 T =
cd
0
0
0 bc/ad
9 Set the trace 2 2 + 1 2 equal to ( 1) + ( 1) to nd opt = 4(2 3) 1.07. The
4
eigenvalues 1 are about .07.
11 If the iteration gives all xnew = xold then the quantity in parentheses is zero, which means
i
i
Ax = b . For Jacobi change the whole right side to xold .
13 u k /k = c1 x 1 + c2 x 2 (2 /1 )k + + cn x n (n /)k c1 x 1 if all ratios |i /1 | < 1. The
1
1
largest ratio controls, when k is large. A =
0
1
1
0
has |2 | = |1 | and no convergence.
14 The eigenvectors of A and also A1 are x 1 = (.75, .25) and x 2 = (1, 1). The inverse power
method converges to a multiple of x 2 .
15 The j th component of Ax 1 is 2 sin
j
n+1
sin
(j 1)
n+1
sin
sin(a + b) = sin a cos b + cos a sin b, combine into 2 sin
(j +1)
.
n+1
j
n+1
cos
The last two terms, using
.
n+1
The eigenvalue is 1 =
.
n+1
2 2 cos
1
2
5
14
is converging to the eigenvector direction
16 u 0 = , u 1 = , u 2 = , u 3 =
0
1
4
13
1
with max = 3.
1
1
1
1 2 1
1 2
1 5
1 14
17 A1 =
gives u 0 = , u 1 =
, u2 =
, u3 =
.
312
31
94
27 13
0
1
1 cos sin
cos (1 + sin2 )
sin3
and A1 = RQ =
.
18 R = QT A =
0 sin2
sin3
cos sin2
19 If A is orthogonal then Q = A and R = I . Therefore A1 = RQ = A again.
20 If A cI = QR then A1 = RQ + cI = Q1 (QR + cI )Q = Q1 AQ. No change in eigenvalues
from A to A1 .
21 Multiply Aq j = bj 1 q j 1 + aj q j + bj q j +1 by q T to nd q T Aq j = aj (because the q s are
j
j
orthonormal). The matrix form (multiplying by columns) is AQ = QT where T is tridiagonal.
Its entries are the as and bs.
22 Theoretically the q s are orthonormal. In reality this algorithm is not very stable. We must
stop every few steps to reorthogonalize.
23 If A is symmetric then A1 = Q1 AQ = QT AQ is also symmetric. A1 = RQ = R(QR)R1
= RAR1 has R and R1 upper triangular, so A1 cannot have nonzeros on a lower diagonal
than A. If A is tridiagonal and symmetric then (by using symmetry for the upper part of A1 )
the matrix A1 = RAR1 is also tridiagonal.
24 The proof of || < 1 when every absolute row sum < 1 uses |
aij xj |
|aij ||xi | < |xi |.
(Note |xi | |xj |.) The Gershgorin circle theorem (very useful) is proved after its statement.
75
25 The maximum row sums give all || .9 and || 3. The circles around diagonal entries give
tighter bounds. The circle | .2| .7 contains the other circles | .3| .5 and | .1| .6
and all three eigenvalues. The circle | 2| 2 contains the circle | 2| 1 and all three
eigenvalues 2 + 2, 2, and 2 2.
26 The circles | aii | ri dont include = 0 (so A is invertible!) when aii > ri .
27 From the last line of code, q 2 is in the direction of v = Aq 1 h11 q 1 = Aq 1 (q T Aq 1 )q 1 . The
1
dot product with q 1 is zero. This is Gram-Schmidt with Aq 1 as the second input vector.
28 r 1 = b 1 Ab = b (b T b /b T Ab )Ab is orthogonal to r 0 = b : the residuals r = b Ax
are orthogonal at each step. To show that p 1 is orthogonal to Ap 0 = Ab , simplify p 1 to cP 1 :
P 1 = Ab
2
b (b T Ab )Ab and c = b T b /(b T Ab )2 . Certainly (Ab )T P 1 = 0 because AT = A.
2
(That simplication put 1 into p 1 = b 1 Ab + (b T b 21 b T Ab + 1 Ab
2
)b /b T b . For a
good discussion see Numerical Linear Algebra by Trefethen and Bau.)
Problem Set 10.1, page 427
1 Sums 4, 2 + 2i, 2 cos ; products 5, 2i, 1.
1
2 In polar form these are 5ei , 5e2i , 5 ei , 5.
3 Absolute values r = 10, 100,
1
,
10
100; angles , 2, , 2.
2
4 |z w| = 6, |z + w| 5, |z/w| = 3 , |z w| 5.
5 a + ib =
3
2
+ 1 i,
2
1
2
+
3
i,
2
i, 1 +
2
3
i;
2
w12 = 1.
1
6 1/z has absolute value 1/r and angle ; r ei times rei = 1.
a b
c
ac bd
real part
=
7
b
a
d
bc + ad
imaginary part
A1 A2
x
b
1 = 1 .
8
A2
A1
x2
b2
9 2 + i; (2 + i)(1 + i) = 1 + 3i; ei/2 = i; ei = 1;
1i
1+i
= i; (i)103 = (i)3 = i.
10 z + z is real; z z is pure imaginary; z z is positive; z/z has absolute value 1.
11 If aij = i j then det(A I ) = 3 6 = 0 gives = 0, 6i, 6i (the conjugate of 6i).
12 (a) When a = b = d = 1 the square root becomes 4c; is complex if c < 0
(b) = 0
and = a + d when ad = bc
(c) the s can be real and dierent.
13 Complex s when (a + d)2 < 4(ad bc); write (a + d)2 4(ad bc) as (a d)2 + 4bc which is
positive when bc > 0.
14 det(P I ) = 4 1 = 0 has = 1, 1, i, i with eigenvectors (1, 1, 1, 1) and (1, 1, 1, 1)
and (1, i, 1, i) and (1, i, 1, i) = columns of Fourier matrix.
15 det(P6 I ) = 6 1 = 0 when = 1, w, w2 , w3 , w4 , w5 with w = e2i/6 as in Figure 10.3.
16 The block matrix has real eigenvalues; so i is real and is pure imaginary.
76
17 (a) 2ei/3 , 4e2i/3
(b) e2i , e4i
(c) 733i/2 , 49e3i (= 49),
50ei/4 , 50ei/2 .
2
18 r = 1, angle
; multiply by ei to get ei/2 = i.
1
19 a + ib = 1, i, 1, i, 2
i
.
2
20 1, e2i/3 , e4i/3 ; 1, ei/3 , ei/3 ; 1.
21 cos 3 = Re(cos +i sin )3 = cos3 3 cos sin2 ; sin 3 = Im(cos +i sin )3 = 3 cos2 sin
sin3 .
22 If z = 1/z then |z |2 = 1 and z is any point ei on the unit circle.
23 (a) ei is at angle = 1 on the unit circle; |ie | = 1e = 1
e
i(/2+2n)e
candidates i = e
.
(b) Spiral in to e2
24 (a) Unit circle
(c) There are innitely many
(c) Circle continuing around to angle = 2 2 .
Problem Set 10.2, page 436
9 = 3, v = 3, u H v = 3i + 2, v H u = 3i + 2 (conjugate of u H v ).
2
0
1+i
31
H
H
are Hermitian matrices.
2 A A= 0
2
1 + i and AA =
13
1i 1i
2
1
u=
3 z = multiple of (1 + i, 1 + i, 2); Az = 0 gives z H AH = 0H so z (not z !) is orthogonal to all
columns of AH (using complex inner product z H times column).
4 The four fundamental subspaces are C(A), N(A), C(AH ), N(AH ).
5 (a) (AH A)H = AH AHH = AH A again
Az
2
(b) If AH Az = 0 then (z H AH )(Az ) = 0. This is
= 0 so Az = 0. The nullspaces of A and AH A are the same. AH A is invertible when
N(A) = {0}.
0
6 (a) False: A =
1
1
0
(b) True: i is not an eigenvalue if A = AH
(c) False.
7 cA is still Hermitian for real c; (iA)H = iAH = iA is skew-Hermitian.
8 Orthogonal, invertible, unitary, factorizable into QR.
001
9 P 2 = 1 0 0 , P 3 = I , P 100 = P 99 P = P ; = cube roots of 1 = 1, e2i/3 , e4i/3 .
010
10 (1, 1, 1), (1, e2i/3 , e4i/3 ), (1, e4i/3 , e2i/3 ) are orthogonal (complex inner product!) because
P is an orthogonal matrixand therefore unitary.
254
11 C = 4 2 5 = 2+5P +4P 2 has = 2+5+4 = 11, 2+5e2i/3 +4e4i/3 , 2+5e4i/3 +4e8i/3 .
542
77
12 If U H U = I then U 1 (U H )1 = U 1 (U 1 )H = I so U 1 is also unitary.
Also (U V )H (U V ) = V H U H U V = V H V = I so U V is unitary.
13 The determinant is the product of the eigenvalues (all real).
14 (z H AH )(Az ) = Az
2
is positive unless Az = 0; with independent columns this means z = 0;
H
so A A is positive denite.
1
1 + i
20
1
15 A = 3
1+i
1
0 1
16 K = (iAT in Problem 15) =
1
3
1
3
1
1i
.
1
1
1 i
2i
0
1i
1
0 i
1 i
1
3
1
1+i
1 + i
1
;
s are imaginary.
1 i
cos + i sin
0
1i
1
1
has || = 1.
17 Q = 2
2
i
1
0
cos i sin
i1
1 + 3 1 + i
1
0 1 1+ 3
1i
2
1
18 V = L
with L = 6 + 2 3 has || = 1.
L
1+i 1+ 3
0 1
1 i
1+ 3
V = V H gives real , trace zero gives = 1, 1.
19 The v s are columns of a unitary matrix U . Then z = U U H z = (multiply by columns)
= v 1 (v H z ) + + v n (v H z ).
1
n
20 Dont multiply eix times eix ; conjugate the rst, then
2
0
e2ix dx = [e2ix /2i]2 = 0.
0
21 z = (1, i, 2) completes an orthogonal basis for C3 .
22 R + iS = (R + iS )H = RT iS T ; R is symmetric but S is skew-symmetric.
23 Cn has dimension n; the columns of any unitary matrix are a basis: (i, 0, . . . , 0), . . .,
(0, . . . , 0, i)
24 [ 1 ] and [ 1 ]; any [ ei ];
a
b + ic
b ic
d
w
;
z
ei z
ei w
with |w|2 + |z |2 = 1.
25 Eigenvalues of AH are complex conjugates of eigenvalues of A: det(A I ) = 0 gives det(AH
I ) = 0.
26 (I 2uu H )H = I 2uu H ; (I 2uu H )2 = I 4uu H + 4u (u H u )u H = I ; the matrix uu H
projects onto the line through u .
27 Unitary means U H U = I or (AT iB T )(A + iB ) = (AT A + B T B ) + i(AT B B T A) = I . Then
AT A + B T B = I and AT B B T A = 0 which makes the block matrix orthogonal.
28 We are given A + iB = (A + iB )H = AT iB T . Then A = AT and B = B T .
29 AA1 = I gives (A1 )H AH = I . Therefore (A1 )H = (AH )1 = A1 and A1 is Hermitian.
1i 1i
1 0 1 2 + 2i 2
= S S 1 .
30 A =
6
1
2
04
1+i
2
78
Problem Set 10.3, page 444
1 Equation (3) is correct using i2 = 1 in the last two rows and three columns.
1
11
1
1
1 1 i2
1
1
1
1
1
= 1 F H.
2F
=
4
2
2
1
1 1 1
1
1
1 i2
i
i
1
1
1
11
2
1 i
1
1
1
.
3 F =
1
1 1 1
1
2
i
i
1
1i
1
1
1
1
4 D=
and F3 = 1 e2i/3 e4i/3 .
e2i/6
e4i/6
1 e4i/3 e2i/3
5 F 1 w = v and F 1 v = 1 w .
4
4000
0 0 0 4
2
and (F4 )4 = 16I .
6 (F4 ) =
0 0 4 0
0400
2
2
0
0
0
2
1
1
1
0
0
1
0
0
0
0
7 c = = F c ; .
1
2
0
1
2
2
0
0
0
1
1
0
0
0
0
0
8 c (1, 1, 1, 1, 0, 0, 0, 0) (4, 0, 0, 0, 0, 0, 0, 0) (4, 0, 0, 0, 4, 0, 0, 0) which is F8 c . The second
vector becomes (0, 0, 0, 0, 1, 1, 1, 1) (0, 0, 0, 0, 4, 0, 0, 0)
(4, 0, 0, 0, 4, 0, 0, 0).
9 If w64 = 1 then w2 is a 32nd root of 1 and
w is a 128th root of 1.
10 For every integer n, the nth roots of 1 add to zero.
11 The eigenvalues of P are 1, i,
0
23
12 = diag(1, i, i , i ); P = 0
1
i2 = 1, and i3 = i.
10
T
3
0 1 and P lead to 1 = 0.
00
13 e1 = c0 + c1 + c2 + c3 and e2 = c0 + c1 i + c2 i2 + c3 i3 ; E contains the four eigenvalues of C .
14 Eigenvalues e1 = 2 1 1 = 0, e2 = 2 i i3 = 2, e3 = 2 (1) (1) = 4, e4 = 2 i3 i9 = 2.
Check trace 0 + 2 + 4 + 2 = 8.
1
15 Diagonal E needs n multiplications, Fourier matrix F and F 1 need 2 n log2 n multiplications
each by the FFT. Total much less than the ordinary n2 .
16 (c0 + c2 )+(c1 + c3 ); then (c0 c2 )+ i(c1 c3 ); then (c0 + c2 ) (c1 + c3 ); then (c0 c2 ) i(c1 c3 ).
These steps are the FFT!