Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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
    
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. –  Neal Dec 10 '11 at 20:27

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.