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I am trying to prove the following standard result:

Let $V$ be a finite dimensional vector space over a field $F$ and $f:V\times V\to F$ be a symmetric bilinear form on $V$. Let $W$ be a subspace of $V$ such that $f$ is non-degenerate on $W$. Then $$V=W\oplus W^\perp$$

(Here $W^\perp=\{v\in V:f(w,v)=0\text{ for all } w\in W\}$).

Here is what I tried:

The bilinear form gives us a map $L_f:V\to V^*$ defined as $$(L_fu)v=f(u,v),\quad\forall u,v\in V$$ Let $W^0$ denote the annihilator of $W$. We show that

$$v\in W^\perp \text{ if and only if } L_fv\in W^0$$

Let $v\in W^\perp$. Then $(L_fv)w=f(v,w)=0$ for all $w\in W$. Therefore $L_fv\in W^0$. Now say $L_fv\in W^0$ for some $v\in V$. Then $(L_fv)w=0$ for all $w\in W$, giving $f(v,w)=0$ for all $w\in W$. Therefore $v\in W^\perp$.

Also, it is clear that $W\cap W^\perp=0$.

Since $\dim W^0=\dim V-\dim W$, we would be done if we could show that $\dim W^\perp \geq \dim W^0$.

From the observation done above, it is natural to consider the map $T:W^\perp\to W^0$ defined as $Tv=L_fv$ for all $v\in W^\perp$. We just need to show that $T$ is surjective. But here I am stuck.

Can somebody help.

Thanks.

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  • $\begingroup$ What's $W^{\perp}$? Do you have an inner product in $V$? If not, it's just the definition of what you're trying to prove $\endgroup$ Commented Nov 30, 2014 at 7:55
  • $\begingroup$ I have added the definition of $W^\perp$. $V$ is not an inner product space. There is a bilinear form given on $V$ which defines $W^\perp$. $\endgroup$ Commented Nov 30, 2014 at 8:16
  • $\begingroup$ Ok, I understand now. This is just the proof of Riesz theorem for finite dimensional vector spaces. $\endgroup$ Commented Nov 30, 2014 at 8:21
  • $\begingroup$ So do you see how we can prove that $T:W^\perp\to W^0$ as defined in the last paragraph is a surjection, and hence an isomorphism? $\endgroup$ Commented Nov 30, 2014 at 8:23
  • $\begingroup$ Yes, do you know that every functional is of the form $f(-, v)$ for some $v \in V$? $\endgroup$ Commented Nov 30, 2014 at 8:24

1 Answer 1

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You can do the following: define $\flat\colon V \to W^\ast$ putting $v \mapsto v_\flat\big|_W$, where $v_\flat(w) = f(v,w)$. Clearly $\ker \flat = W^\perp$, and $\flat$ is surjective. It follows that $$\dim V = \dim W + \dim W^\perp,$$even if $W$ is degenerate. However, we have $$\dim V = \dim W + \dim W^\perp = \dim(W+W^\perp)-\dim(W \cap W^\perp).$$But $W$ is non-degenerate if and only if $W \cap W^\perp = \{0\}$, so $W\oplus W^\perp = V$ if and only if $W$is non-degenerate.

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