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.

 

[blue_raver22]

Hello,

First, let's define some terms. A bilinear function $f:V\times V\rightarrow F$ is a function that takes two inputs from a vector space $V$ and outputs a scalar from a field $F$. In this case, we are told that $f$ is nonsingular, meaning that it is not equal to zero for any input.

Now, let's consider the linear function $\psi\in V^*$, where $V^*$ is the dual space of $V$. The dual space of a vector space is the set of all linear functions from that vector space to its underlying field. In other words, $\psi$ takes in an element from $V$ and outputs a scalar from $F$.

We want to prove that there exists a unique $v\in V$ such that $\psi(x)=f(x,v)$ for any $x\in V$. This means that $\psi$ and $f(x,v)$ are essentially the same function, but with different inputs. We can rewrite this as $f(x,v)=\psi(x)=\psi(v)$.

To prove uniqueness, we can assume that there exists another $v'\in V$ such that $\psi(x)=f(x,v')$. This would mean that $\psi(v)=f(v,v')=\psi(v')$. But since $f$ is nonsingular, $f(v,v')=0$ if and only if $v=v'$. Therefore, $v=v'$, proving uniqueness.

Next, we want to show that the map $v\rightarrow \psi$ is an isomorphism of $V$ and $V^*$. An isomorphism is a bijective function that preserves structure. In this case, we want to show that the map is both injective (one-to-one) and surjective (onto).

To show injectivity, we can use the uniqueness we just proved. If $v$ and $v'$ both map to the same $\psi\in V^*$, then $f(x,v)=\psi(x)=f(x,v')$ for all $x\in V$. But this means that $v=v'$, again using the nonsingularity of $f$.

To show surjectivity, we need to show that for any $\psi\in V^*$, there exists a corresponding $v\in V$. We already know that $\psi(v)=f(v,v')