U B Department of Mathematics
Math 201 Matrix Theory
Date: May 06, 2008
Time: 17:00-18:40
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:
Math 201 Number :
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Spring 2008 Second Midterm Solutions
IMPORTANT
1. Write your name, surname on top of each page. 2. The exam consists of 4 questions some of which have more than
one part. 3. Read the questions carefully and write your answers neatly under the corresponding questions. 4. Show all
your work. Correct answers without sucient explanation might not get full credit. 5. Calculators are not allowed.
Q1
Q2
Q3
Q4
total
10 pts
14 pts
21 pts
15 pts
60 pts
1.) Let A be a 3 3 invertible matrix.
a) [4] Write bases for the 4 fundamental subspaces of A. (Justify your answer.)
Solution:
Note that since A is invertible rank (A) = 3 and therefore Row(A)=Column(A)=R3 . Null space
and left nullspace have dimension zero and therefore their bases are empty.
0
0
1
0 , 1 , 0 .
Row space basis=column space basis =
0
0
1
b) [6] Write bases for the 4 fundamental subspaces of the 3 6 matrix B = [ A A ]. (Justify your
answer.)
Solution:
A is row equivalent to I.
Row space basis of B =
1001
1
Column space basis of B = 0 ,
0
T
00
0
1 ,
0
,010010
0
0
1
T
,001001
T
To nd the nullspace we need to solve Bx = 0 [I I ]x = 0 which gives
Nullspace basis of B =
1 0 0 1 0 0
T
, 0 1 0 0 1 0
Left nullspace basis is empty since dim(left nullspace)=3-3=0.
T
, 0 0 1 0 0 1
T
2.) Let Pn be the vector space of polynomials of degree less than or equal to n. Let T : P2 P4 be
the transformation T (q (x)) = x2 q (x) for all q (x) P2 .
a) [4] Show that T is a linear transformation
b) [4] Find the matrix of T with respect to the standard basis of P2 .
c) [4] Find the matrix of T with respect to the basis B = {q1 , q2 , q3 } of P2 where q1 = 1 + x2 ,
q2 = 1 + 2x + 3x2 and q3 = 4 + 5x + x2 .
d) [2] Use the matrix you found in part c) to compute T (2 + 2x + 4x2 ).
Solution:
a-) Let p(x) = a0 + a1 x + a2 x2 , q (x) = b0 + b1 x + b2 x2 , ai , bi R
T (p(x) + q (x)) = (a0 + b0 )x2 + (a1 + b1 )x3 + (a2 + b2 )x4 = T (p(x)) + T (q (x))
T (kp(x)) = (ka0 )x2 + (ka1 )x3 + (ka2 )x4 = kT (p(x)), k R.
0
0
1
2
0 , 1 , 0
b-) Standard basis of P2 : {1, x, x } denoted as
1
0
0
Observe that T (1) = x , T (x) = x , T (x ) = x . Hence, AT =
2
3
2
4
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
c-) Observe: T (q1 ) = x2 + x4 , T (q2 ) = x2 + 2x3 + 3x4 , T (q3 ) = 4x2 + 5x3 + x4
000
0 0 0
B
Hence, in basis B : AT = 1 1 4
0 2 5
131
d-) Note that 2 + 2x + 4x2 = q1 + q2 . Therefore, coordinate vector respect to basis B is
000
0
0 0 0 1
0
1 1 4 1 = 2 2x2 + 2x3 + 4x4 , as expected.
0 2 5 0
2
131
4
110
T
.
3.) Let W be the span of vectors {x1 , x2 , x3 } where xT = [1 1 1 1], xT = [0 1 1 1], xT = [0 0 1 1].
1
2
3
a) [6] Apply the Gram-Schmidt process to construct an orthogonal basis for W .
b) [5] Find the orthogonal complement of W .
c) [2] Using parts a) and b) nd an orthonormal basis for R4 which is not the standard [4] basis.
d) Find coordinates of the vector v T [1 2 3 4] with respect to the basis you found in part c).
=
100
1 1 0
e) [4] Find a QR decomposition of A =
1 1 1 .
111
Solution:
1
1
a-) u1 =
1
1
3
uT x 2
1 1
1
u2 = x2 proju1 x2 = x2 T u1 =
4 1
u1 u1
1
3
1
Choose u2 =
1 .
1
0
uT x 3
uT x 3
1 2
1
2
u3 = x 3 T u1 T u2 =
3 1
u1 u1
u2 u2
1
0
2
Choose u3 =
1 .
1
b-) We want u4 W . If we use vectors {u1 , u2 , u3 } as rows of a matrix then u4 should be in the
a
1
111
b
3
1 1 1 = 0
null space of that matrix, that is,
c
0 2 1 1
d
0
0
Solving this system we get u4 =
1
1
3
0
0
1
1
1
1 1
1 1
, v3 = 2 , v4 = 0
c-) Normalizing u1 , ..., u4 : v1 = , v2 =
2 1
12 1
6 1
2 1
1
1
1
1
Now {v1 , v2 , v3 , v4 } is an orthonormal basis for R4 .
1
a
b 2
d-) Let V = [v1 : v2 : v3 : v4 ]. We want to nd a, b, c, d satisfying V = . Since V is an
c 3
4
d
orthogonal matrix,
1
1
1
1
5
1
a
1
2
2
2
2
1
1
3 1
b
3
12
12 2
= V T 2 = 12 12
=
0
2
1
1
3
c
3 3/6
6
6
6
1
1
4
d
4
0
0
1/ 2
2
2
T
T
T
v1 x1 v1 x2 v1 x3
T
T
v2 x2 v2 x3
e-) A = x1 : x2 : x3 = v1 : v2 : v3 0
T
0
0
v3 x3
A=
1
2
1
2
1
2
1
2
3
12
1
12
1
12
1
12
0
2
2
6
0
1
6
0
1
3
2
3
12
0
1
2
12
2
6
6
4.) Decide whether the following are TRUE or FALSE. If true prove, if false give a counter example.
a) [5] Let A be an n n matrix. Suppose B is obtained by adding the same number x to each entry
in the ith row of A (i n) and C is obtained by subtracting x from each entry in the ith row of
1
A. Then det A = (det B + det C ).
2
Solution:
R1
R1
.
.
R1
.
.
.
. Then, B = Ri + (x, x) , C = Ri (x, x)
Let A =
.
.
.
Rn
.
.
Rn
Rn
Then, by linearity property of the determinant with respect to ith row:
R1
R1
R1
R1
.
.
.
.
.
.
.
.
det(B ) = det . + det (x, .., x) , det(C ) = det . det (x, .., x)
.
.
.
.
.
.
.
.
Rn
Rn
Rn
Rn
Hence, det(B ) + det(C ) = 2det(A) and the proposition is TRUE!
b) [5] If the entries of each row of an n n matrix A add up to zero then det A = 0.
Solution:
TRUE!!
||
Let A = c1 c2
||
cT
1
.. |
. . .
. . cn , AT =
. . .
.. |
cT
n
Now doing the row operation AT
cT (cT +...+cT )cT
n
1
2
1
0 .. 0
cT
2
cT
3
. . .
. . .
cT
n
Since row operations do not change the determinant and we have det(A) = det(AT ) = 0.
c) [5] Let X, Y, Z be subspaces of a vector space V . If X is orthogonal to Y and Y is orthogonal to
Z , then X is orthogonal to Z .
Solution:
FALSE!!
Counter example:
X = span {(1, 0, 0)} = {k (1, 0, 0) | k R}, a subspace of R3
Y = span {(0, 0, 1)} = {k (0, 0, 1) | k R}, a subspace of R3
Z = span {(1, 1, 0)} = {k (1, 1, 0) | k R}, a subspace of R3
As you see all conditions are satised but there are (innitely many) vectors in X which are not
orthogonal to Y . Hence, X and Y are not orthogonal to each other.