Need help for a Problem about Hermitian Form

Question

Let X be a linear space over C, q is a nondegenerate Hermitian form on it. V is a subspace of X and codim V is finite. $V^q=\{x∈X\; | \;q(x,y)=0,y∈V\}$.

Pleae prove that $\dim⁡ V^q≤\dim ⁡X/V$ and $\dim⁡ V^q=\dim⁡ X/V$ if and only if $V=(V^q)^q$.

I really have no idea about how to prove it, please help me. Thank you very much.

Answer 1

Let $\pi:V^q\to X/V$ be the linear mapping $\pi(x):=x+V$.

It is injective (1-1). Proof: For $x_1,x_2\in V^q$, $\pi(x_1)=\pi(x_2)\iff (x_1-x_2)\in V$; then $q(x_1-x_2,x_1-x_2)=0$ since $x_1-x_2\in V^q$ (which is a subspace). So $x_1=x_2$ since $q$ is non-degenerate.

It follows that $\dim V^q\le \dim X/V$. ($\pi$ preserves linear independence since it is 1-1.)

The dimension is preserved when $\pi V^q=X/V$. It should be clear that $V\subseteq V^{qq}$. Let $w\in V^{qq}$, then $w+V=\pi(x)$ for some $x\in V^q$, that is, $w-x\in V$, i.e. $w=x+y$ with $y\in V$. Then $$0=q(w,x)=q(x+y,x)=q(x,x)+q(y,x)=q(x,x)$$ It follows that $x=0$ and $w=y\in V$. Thus $V^{qq}=V$.