and the description of the null space gives that on transposing the m rows of H
hm,1
h1,1
hm,2
h1,2
h1 = . , . . . hm = .
.
.
.
.
hm,n
h1,n
we have N (f) = [cfw_ h1 , . . . , hm ]. ([Strang 93] describes this space as the transpose of the row space o
170
Linear Algebra, by Hefferon
Three.VI.3.24
(a) The representation of
v1
f
v2 1v1 + 2v2 + 3v3
v3
is this.
RepE3 ,E1 (f) = 1
2
3
By the denition of f
v1
v1
1
v1
N (f) = cfw_ v2 1v1 + 2v2 + 3v3 = 0 = cfw_ v2 2 v2 = 0
v3
v3
3
v3
and this second descr
Answers to Exercises
169
followed by matrix-vector multiplication
0
1/2 0
projM (1)
00
2
1/2 0
1/2
0
1
0 1 = 0
1/2
2
1
gives the answer.
Three.VI.3.14 No, a decomposition of vectors v = m + n into m M and n N does not depend on the
bases chosen for t
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Linear Algebra, by Hefferon
and representing the vector with respect to the concatenation gives this.
1
1
1
= 2
1
3
1
1
Keeping the M part yields the answer.
1
2
projM,M (
)=
3
2
The third part is also a simple calculation (there is a 1 1 matrix in t
Answers to Exercises
167
(b) Because R3 is three-dimensional and P is two-dimensional, the complement P must be a line. Anyway,
the calculation as in Example 3.5
x
x
1 0 3
0
P = cfw_ y
y =
012
0
z
z
gives this basis for P .
3
BP = 2
1
1
1
0
3
(c) 1 = (5
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Linear Algebra, by Hefferon
(b) As in the answer to the prior part, we can describe M as a span
3/2
1
M = cfw_c
c R
3/2
1
BM =
and then M is the set of vectors perpendicular to the one vector in this basis.
M = cfw_
u
v
(3/2) u + 1 v = 0 = cfw_ k
2
Answers to Exercises
165
Three.VI.3: Projection Into a Subspace
Three.VI.3.10
(a) When bases for the subspaces
1
1
BM =
2
1
BN =
are concatenated
B = BM
1
2
,
1
1
BN =
and the given vector is represented
3
2
1
1
=1
2
1
+1
then the answer comes from retain
164
Linear Algebra, by Hefferon
(the three conditions in the lower left are redundant but nonetheless
only if
adg
abc
10
b e h d e f = 0 1
cfi
ghi
00
correct). Those, in turn, hold if and
0
0
1
as required.
This is an example, the inverse of this matrix
Answers to Exercises
163
and the two projections are easy.
proj[1 ] (
2
)=
3
2
3
1
1
1
1
1
1
1
1
=
51
21
proj[2 ] (
2
)=
3
2
3
1
1
1
1
1
1
1
1
=
1
1
2
1
Note the recurrence of the 5/2 and the 1/2.
(c) Represent v with respect to the basis
r1
.
RepK (v ) =
162
Linear Algebra, by Hefferon
(the second vector is not dependent on the third because it has a nonzero second component, and the rst
is not dependent on the second and third because of its nonzero third component), and then apply the
Gram-Schmidt proce
Answers to Exercises
161
or we just could modify the Gram-Schmidt procedure to throw out any zero vectors. The second thing
to worry about if we drop the phrase linearly independent from the question is that the set might be
innite. Of course, any subspac
160
Linear Algebra, by Hefferon
and then normalize.
1/ 2
1/ 6
1/ 2 , 1/6
0
2/ 6
Three.VI.2.13 Reducing the linear system
xyz+w=0
x
+z
=0
1 +2
xy z+w=0
y + 2z w = 0
and parametrizing gives this description of the subspace.
0
1
1
2
cfw_ z + w z, w R
0
Preface
These are answers to the exercises in Linear Algebra by J Hefferon.
An answer labeled here as, for instance, One.II.3.4, matches the question numbered 4 from the rst
chapter, second section, and third subsection. The Topics are numbered separately
Answers to Exercises
Three.VI.1.20
157
(a) Fixing
1
1
s=
as the vector whose span is the line, the formula gives this action,
x
y
x
y
1
1
1
1
1
1
1
1
=
x+y
2
1
1
=
(x + y)/2
(x + y)/2
which is the eect of this matrix.
1/2
1/2
1/2
1/2
(b) Rotating the enti
4
Linear Algebra, by Hefferon
Finally, for the third case, if both a and c are 0 then the system
0x + by = j
0x + dy = k
might have no solutions (if the second equation is not a multiple of the rst) or it might have innitely many
solutions (if the second
156
Linear Algebra, by Hefferon
Three.VI.1.14 Any vector in Rn is the projection of some other into a line, provided that the dimension n is
greater than one. (Clearly, any vector is the projection of itself into a line containing itself; the question is
Chapter One
Chapter One: Linear Systems
Solving Linear Systems
One.I.1: Gausss Method
One.I.1.17 We can perform Gausss Method in dierent ways so these exhibit one possible way to get the
answer.
(a) Gausss Method
3y =
13
(1/2)1 +2 2x +
(5/2)y = 15/2
give
154
Linear Algebra, by Hefferon
(d) Assume that T = P1 T P. For the squares: T 2 = (P1 T P)(P1 T P) = P1 T (PP1 )T P = P1 T 2 P.
Higher powers follow by induction.
(e) These two are matrix equivalent but their squares are not matrix equivalent.
1
0
0
0
0
Answers to Exercises
Three.V.2.25
153
(a) The adapted form of the arrow diagram is this.
h
Vwrt B1 Wwrt D
H
Q
id P
id
h
Vwrt B2 Wwrt D
H
Since there is no need to change bases in W (or we can say that the change of basis matrix P is the
identity), we have
152
Linear Algebra, by Hefferon
Three.V.2.19 For reexivity, to show that any matrix is matrix equivalent to itself, take P and Q to be
identity matrices. For symmetry, if H1 = PH2 Q then H2 = P1 H1 Q1 (inverses exist because P and
Q are nonsingular). Fina
Answers to Exercises
7
One.I.1.36 Solving the system
(1/3)(a + b + c) + d = 29
(1/3)(b + c + d) + a = 23
(1/3)(c + d + a) + b = 21
(1/3)(d + a + b) + c = 17
we obtain a = 12, b = 9, c = 3, d = 21. Thus the second item, 21, is the correct answer.
One.I.1.3