Notes on Homework 8
1. Before starting, it’s worth recalling that if W and V are vector spaces, and if (w1 , · · · , wn ) is a basis,
and if v1 , · · · , vn are arbitrarily chosen vectors, then there is a unique linear transformation f ∈
L(W, V ) such that for each i, f (wi ) = vi . To see how this is defined, if w ∈ W, then there is a unique
choice of numbers a1 , · · · , an ∈ F such that w = ∑ ai wi ; if we set f (w) = ∑ ai f (wi ) = ∑ ai vi , this
turns out to be a linear transformation. (Nothing can go wrong, in some sense, since there’s only one
way to represent a given w as a sum of the wi .)
In this example, note that z = x + 2y.
(a) Suppose there were such a linear transformation f ; then
f ( z) = f ( x + 2y)
= f ( x) + 2 f ( y)
2
0
=
+2
3
1
2
=
;
5
since this doesn’t hold, there is no such linear transformation.
(b) The restrictions are compatible, in that f ( x) + 2 f ( y) = f ( z); there is such a linear transformation.
(c) Check:
0
1
+2
f ( x) + 2 f ( y) =
1
1
1
=
3
3
,
6=
3
so no such linear transformation is possible.
2. We need to show that for each v ∈ V there are numbers a1 , · · · , an ∈ F such that v =
a1 f (w1 ) + · · · + an f (wn ). So, suppose v ∈ V. Since f is surjective, there is some w ∈ W
such that f (w) = v. Since w1 , · · · , wn spans W, there are numbers a1 , · · · , an ∈ F such that
w = a1 w1 + · · · + an wn . Then
v = f (w)
= f ( a1 w 1 + · · · + an w n )
= f ( a1 w1 ) + · · · + f ( an wn ) additivity
= a1 f (w1 ) + · · · + an f (wn ) homogeneity
∈ span ( f (w1 ), · · · , f (wn )).
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
3. What follows gives a quick indication of why each statement is true. Since in lecture we gave short
shrift to row spaces, those questions won’t be graded. In each of these questions, I’ll identify the
matrix A with a linear transformation f ∈ L(Fn , Fm ). Recall the dictionary from class:
f
A
im ( f )
col( A)
dim im ( f )
rank ( A)
ker( f )
nullspace( A)
dim ker( f ) nullity ( A)
Recall the basic result
rank( A) + nullity ( A) = n.
(1)
(a) If rank ( A) = m then the system Ax = b has at least one solution for every b ∈ Fm . True.
Since rank ( A) = m = dim im ( f ), the image of f is an m-dimensional subspace of Fm ,
i.e., the whole thing; so each b is in the image of f , and each b admits a solution to
Ax = b.
(b) If rank ( A) = n then the system Ax = b has at least one solution for every b ∈ Fm . False, in
the sense of “not necessarily”; consider the case where n = 1 and m = 2 (or anything
larger), so that A looks something like, say, A = ( a11 a12 ). Suppose that some a1 j 6= 0.
Then A has rank n = 1. But since the target space has dimension m > 1, this means
that there are things in Fm which aren’t in the image of f .
(c) If rank ( A) = m then the system Ax = b has at most one (i.e., no solution or a unique solution)
for every b ∈ Fm . False, in the sense of “not necessarily”. The condition that solutions
are unique is equivalent to the condition that the kernel is trivial, so that the nullity is
zero; but especially if n > m, it is easy to cook up examples, compatible with (1), in
which the nullity is positive.
(d) If rank( A) = n then the system Ax = b has at most one (i.e., no solution or a unique solution)
for every b ∈ Fm . True; by (1), if rank( A) = n then nullity ( A) = 0, which is equivalent
to f being injective (one-to-one).
(e) If Ax = b, then x ∈ row( A). False, in general; if x is any element of Fn , then it’s the
solution to Ax = b for some b ∈ Fm .
(f) If Ax = b, then b ∈ col( A). True; the column space is the same as the image.
(g) If B ∈ Mat (m, n, F) is row equivalent to A then col( A) = col( B). False; in solving equations, manipulating rows should change the right-hand side of the equation, too.
(h) If B ∈ Mat (m, n, F) is row equivalent to A then row( A) = row( B). True; the operations
on rows preserve the span of the vectors.
(i) If the columns of A are linearly independent then the rows of A are linearly independent. False;
consider the case n ≥ 2, m = 1, in which A consists of a single, nonzero column.
(j) If col( A) = Fm and m ≤ n then row( A) = Fn . False; if m < n, then the row space is
spanned by at most m vectors, and thus can’t be all of Fn .
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
4.
(a) By definition of the map, to compute [ f (v1 )]E we simply multiply;
[ f (v1 )]E = [ f ]E ←E [v1 ]E
−3 −8
5
=
10 27
3
−39
=
.
131
(b) Let z1 =
1
1
, z2 =
2
3
. Then
[id ]E ←C = ([id( z1 )]E [id ( z2 )])
1 2
=
1 3
so that
[id ]C←E = ([id]E ←C )−1
3 −2
=
.
−1 1
Then we can compute:
[ f ]C←C = [id]C←E [ f ]E ←E [id ]E ←C
3 −2
−3 −8
1 2
=
−1 1
10 27
1 3
−107 −292
=
.
48
131
(c)
[ f (v2 )]C = [ f ]C←C [v2 ]C
−107 −292
3
=
.
48
131
4
−1489
=
668
5.
1
0
(a) Let e1 =
, e2 =
. In each case, to write down the matrix of the rele0
1
vant function, we need to calculate [ f 21 (e1 )]E and [ f 21 (e2 )]E . Unfortunately, the picture
makes it a little hard to figure out which points on the house correspond to e1 and e2 .
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
3
0
However, let v1 =
and v2 =
. On one hand, these points on the house are
0
4
distinctive. On the other hand, by homogeneity, we have
f 21 (v1 ) = f 21 (3e1 )
= 3 f 21 (e1 )
1
f 21 (e1 ) = f 21 (v1 )
3
and similarly
1
f 21 (v2 ).
4
3
8
Since f 21 (v1 ) =
and f 21 (v2 ) =
, we have
0
4
1 2
[ f ]E ←E =
.
0 1
f 21 (e2 ) =
(b) Same strategy, different numbers. Without any annotation, the results are:
1 0
.
iii. [ f 31 ]E ←E =
1 5/4
−1 0
iv. [ f 41 ]E ←E =
.
0 2
1 9/4
v. [ f 51 ]E ←E =
.
2 6/4
0 1
vi. [ f 61 ]E ←E =
.
−1 0
6.
(a) We did this in class; the results follow from calculus. Specifically, if p( z), q( z) ∈ P3 (R)[ z],
then
d
D ( p( z) + q( z)) =
( p( z) + q( z))
dz
d
d
p( z) + q( z)
=
dz
dz
= D ( p( z)) + D (q( z)),
and if λ ∈ R then
d
(λ p( z))
dz
d
= λ p( z)
dz
= λ D ( z) .
D (λ p( z)) =
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
(b) Let v1 = 1, v2 = z, v3 = z2 , v4 = z3 . Then
[ D ]B←B = ([ D (v1 )]B [ D (v2 )]B [ D (v3 )]B [ D (v4 )]B )
= ([0]B [1]B [2z]B [3z2 ]B )


0 1 0 0
0 0 2 0 

=
0 0 0 3  .
0 0 0 0
(c) We have
[ D ]C←B = ([ D (v1 )]C [ D (v2 )]C [ D (v3 )]C [ D (v4 )]C )
= ([0]C [1]C [2z]C [3z2 ]C )


0 1 0 0
0 0 1 0

=
0 0 0 1
0 0 0 0
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008