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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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    @Dylan Okay, I see your concern about finite-dimensionality.2011-12-10

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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.

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    Slick! I was thinking of showing that some subspace of higher-dimensional space annihilated everything in the lower-dimensional space and deriving a contradiction, but this amounts to the same thing and is much better.2011-12-10
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The problem is that you have to show that the matrix is invertible - in fact you just have reformulated your problem. Sometimes such a reformulation can be helpful, but it is by no means a solution.