- #1

steenis

- 312

- 18

- TL;DR Summary
- A one-to-one correspondence between bilinear non-degenerate maps and invertible linear maps

I have the following problem in multilinear algebra:

Let ##W## and ##V## be real finite-dimensional vector spaces, ##V^*## is the dual space of ##V##

Let ##L:W \times V \rightarrow \mathbb{R}## be a non-degenerate bilinear map

Define ##g:W \rightarrow V^*## by ##g(w)(v)=L(w,v)##

To prove: ##g## is an isomorphism

The map ##g## is obviously linear

##g## is injective, because if ##g(w)=0## for ##w \in W##, then for all ##v \in V## we have that ##g(w)(v)=L(w,v)=0##, so, because ##L## is non-degenerate, we have ##w=0##

Remains to prove that ##g## is surjective

If ##dim W=dim V=n##, then we are ready, because in that case ##dim W=dim V^*=n## and ##g## is surjective if and only if ##g## is injective. On the other hand, if ##g## is bijective, then ##g## is an isomorphism and ##dim W=dimV=dim V^*##. However, it is not a-priori given that ##W## and ##V## have the same dimension

Can anybody help me with this problem, can the statement be proven or is the statement not true?

Let ##W## and ##V## be real finite-dimensional vector spaces, ##V^*## is the dual space of ##V##

Let ##L:W \times V \rightarrow \mathbb{R}## be a non-degenerate bilinear map

Define ##g:W \rightarrow V^*## by ##g(w)(v)=L(w,v)##

To prove: ##g## is an isomorphism

The map ##g## is obviously linear

##g## is injective, because if ##g(w)=0## for ##w \in W##, then for all ##v \in V## we have that ##g(w)(v)=L(w,v)=0##, so, because ##L## is non-degenerate, we have ##w=0##

Remains to prove that ##g## is surjective

If ##dim W=dim V=n##, then we are ready, because in that case ##dim W=dim V^*=n## and ##g## is surjective if and only if ##g## is injective. On the other hand, if ##g## is bijective, then ##g## is an isomorphism and ##dim W=dimV=dim V^*##. However, it is not a-priori given that ##W## and ##V## have the same dimension

Can anybody help me with this problem, can the statement be proven or is the statement not true?