Math 222 - Selected Homework Solutions for
Assignment 7
Instructor - Al Boggess
Fall 1998
Section 4.1
12. Let v1 ; : : : vn be a basis for V and let L1 and L2 be two linear transformations mapping V into W . We are to show that if L1 (vi ) = L2 (vi )
for 1 i n, then L1 (v) = L2 (v) for all v 2 V .
Since v1 ; : : : ; vn is a basis, any v 2 V can be expressed as
v=
Xn v
i=1
i i
for some choice of 1 ; : : : ; n . Since L1 and L2 are linear,
L1 (v)
= L1 (1 v1 + : : : + n vn )
= 1 L1 (v1 ) + : : : + n L1 (vn )
L2 (v)
= L2 (1 v1 + : : : + n vn )
= 1 L2 (v1 ) + : : : + n L2 (vn )
and
Since L1 (vi ) = L2 (vi ) for 1 i n, we conclude that L1 (v) = L2 (v),
as desired.
20. We are to show that the kernel of L : V 7! W is zero if and only if L
is one to one.
We know that L(0) = 0. If L is one to one, then 0 is the only vector
which is mapped to 0 (by denition of one to one). Therefore, the
kernel, which by denition is the set of all vectors with L(v) = 0 must
consist of only the zero vector.
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Conversely, suppose the kernel of L = f0g. We must show that L is
one to one. Suppose L(v1 ) = L(v2 ). We must show that v1 = v2 or
equivalently, v1 , v2 = 0. Now if L(v1 ) = L(v2 ), then
L(v1 , v2 ) = L(v1 ) , L(v2 ) since L textrmislinear
=0
and so v1 , v2 belongs to the kernel of L. If the kernel is f0g, then
v1 , v2 = 0 or v1 = v2 as desired.
Section 4.2
14. Let A be the matrix that represents the linear transformation A with
respect to the standard basis. If L(x) = 0 for some x 6= 0, then we are
to show that A is singular. Now, L(x) = A(x), and so if Ax = 0 with
x 6= 0, then the matrix is singular by Theorem 1.4.3.
17. Let E = e1 ; : : : en be a basis for V ; let F = f1 ; : : : ; fm be a basis for
W ; and let G = g1 ; : : : ; gl be a basis for Z . Suppose A is the matrix
representing L1 : V 7! W with respect to bases E and F ; suppose B
is the matrix representing L2 : W 7! Z with respect to bases F and
G. We are to compute the matrix for L2 L1 : V 7! Z .
Since A is the matrix representing L1 : V 7! W with respect to bases
E and F , we have
m
X
L1 (ej ) = aij fi
i=1
Since B is the matrix representing L2 : W 7! Z with respect to bases
F and G, we have
l
X
L2 (fi) = bkigk
k=1
Combining both of these equations, we have
P
L2 (L1 (ej ))
= L2 ( mi=1 aij fi )
P
= mi=1 aij L2 (fi ) since L2 is linear
P P
= mi=1 aij lk=1 bki gk
P P
= lk=1 ( mi=1 bki aij ) gk
P
Let C be the matrix with ckj = mi=1 bki aij ; then
L2 (L1 (ej )) =
2
Xl c
k=1
kj gk
and so ckj is the matrix that represents L2 L1 with respect to E and
G. Note from the formula of ckj that this is just the k; j th entry of
BA, as desired.
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