# Non degenerate bilinear form

Problem:

Let $F:V \times W \to \mathbb{R}$ be a non degenerate bilinear form. The question is: prove that $V$ and $W$ have the same dimension (the vector spaces $V$ and $W$ are finite dimensional)

My answer is: $F$ is non degenerate, then the matrix of $F$ is invertible, which means it's square and this implies that $V$ and $W$ have the same dimension. Is my assumption that the matrix of a non degenerate bilinear form is invertible true? Also let me know if my answer is true?

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Hint: What is the definition of non-degeneracy? If $V$ and $W$ have different dimension, can $F$ be non-degenerate? PS- $F$ does not have a matrix if you have not chosen bases of the vector spaces. – Neal Dec 10 '11 at 17:10
@Dylan Moreland : Yes, I assume $V$ and $W$ to be finite dimensional vector spaces. – M.Krov Dec 10 '11 at 18:47
@Dylan Moreland I have already added that to the problem statement. So, now in finite dimensional case, what do you think about the assumption of invertibility? – M.Krov Dec 10 '11 at 18:56
@Zi2018Alpha Great! What's your definition of non-degenerate? – Dylan Moreland Dec 10 '11 at 18:59
Is the particular definition going to matter too much? Aren't they all more or less "only zero kills everything"? – Neal Dec 10 '11 at 19:34

We can obtain a linear map $V \to W^*$ by sending $x \in V$ to the functional $y \mapsto F(x, y)$ on $W$. That $F$ is non-degenerate implies that this map is injective, so $\dim V \leq \dim W^*$. Since $W$ is finite-dimensional, $W^*$ has the same dimension as $W$ and hence $\dim V \leq \dim W$. Using the analogous map $W \to V^*$, we get the reverse inequality.
You could make a very similar argument using the fact that the row and column rank of a matrix for $F$ are the same. Might write that up later. – Dylan Moreland Dec 10 '11 at 20:23