Non-Singular Bilinear Function

[Sudharaka]

Hi everyone, :)

Here's a question that I have now clue on how to solve. I hope you can shed some light on it. :)

Question:

Let $f:V\times V\rightarrow F$ be a nonsingular bilinear function on a vector space $V$ over a field $F$. Prove that for any linear function $\psi\in V^*$ there is unique $v\in V$ such that $\psi(x)=f(x,\,v)$, for any $x\in V$, and that the map $v\rightarrow \psi$ is an isomorphism of $V$ and $V^*$.

 

[johng1]

Hi,
I'm unfamiliar with the term non-singular bilinear function. If this just means an ordinary non-degenerate scalar product, then your question is a standard result for finite dimensional vector spaces. (For infinite dimension, it's false). See the wikipedia page Dual space - Wikipedia, the free encyclopedia for the result.

 

[Sudharaka]

Hi,
I'm unfamiliar with the term non-singular bilinear function. If this just means an ordinary non-degenerate scalar product, then your question is a standard result for finite dimensional vector spaces. (For infinite dimension, it's false). See the wikipedia page Dual space - Wikipedia, the free encyclopedia for the result.

Hi johng, :)

Thanks very much for the reply. Do you know of a link where this result is proved? Assuming non-singular is the same as non-degenerate.

 

[johng1]

I think almost any good linear algebra text should have the result. In particular, Serge Lang's book Linear Algebra has the complete discussion on this topic. I don't know of any on line proofs, but there probably are some.