This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3.3. VECTOR SPACES 171 Proof. First, let’s check that W is ﬁnite–dimensional, with dimension no
greater than dim.V /. If fv1 ; : : : ; vn g is a basis of V , then fT .v1 /; : : : ; T .vn /g
is a spanning subset of W , so contains a basis of W as a subset.
Now let fw1 ; : : : ; ws g be a basis of W . For each basis element wi ,
let xi be a preimage of wi in V (i.e., choose xi such that T .xi / D wi ).
The map wi 7! xi extends uniquely to a linear map S W W ! V , deﬁned
by S. P ˛i wi / D
i P xi , according to Proposition 3.3.32. We have
T ı S. i ˛i wi / D T . i ˛i xi / D i ˛i T .xi / D i ˛i wi . Thus T ı
S D idW .
In the situation of the previous proposition, let W 0 denote the image of
S . I claim that
V D ker.T / ˚ W 0 Š ker.T / ˚ W:
Suppose v 2 ker.T / \ W 0 . Since v 2 W 0 , there is a w 2 W such that
v D S.w/. But then 0 D T .v/ D T .S.w// D w , and, therefore, v D
S.w/ D S.0/ D 0. This shows that ker.T / \ W 0 D f0g. For any v 2 V ,
we can write v D S ı T .v/ C .v S ı T .v//. The ﬁrst summand is evidently
in W 0 , and the second is in the kernel of T , as T .v/ D T ı S ı T .v/. This
shows that ker.T / C W 0 D V . We have shown that V D ker.T / ˚ W 0 .
Finally, note that S is an isomorphism of W onto W 0 , so we also have
V Š ker.T / ˚ W . We have shown the following:
Proposition 3.3.35. If T W V ! W is a linear map and V is ﬁnite–
dimensional, then V Š ker.T / ˚ range.T /. In particular, dim.V / D
dim.ker.T // C dim.range.T //.
Now let V be a ﬁnite–dimensional vector space and let N be a subspace. The quotient map W V ! V =N is a a surjective linear map with
kernel N . Let S be a right inverse of , as in the proposition, and let M
be the image of S . The preceding discussion shows that V D N ˚ M Š
N ˚ V =N . We have proved the following:
Proposition 3.3.36. Let V be a ﬁnite–dimensional vector space and let N
be a subspace. Then V Š N ˚ V =N . In particular, dim.V / D dim.N / C
dim.V =N /. Corollary 3.3.37. Let V be a ﬁnite–dimensional vector space and let N be
a subspace. Then there exists a subspace M of V such that V D N ˚ M . ...
View Full Document