Let $H$ be an invertible $n$ by $n$ Hermitian matrix. Use $H$ to define a Hermitian form, $[ \cdot , \cdot ]$ which acts on two column vectors $x,y \in \mathbb{C^n}$ according to its definition $[x,y] = {\bar x}^T H y.$
Let $W$ be a subspace of $\mathbb{C^n}$ such that for all $w_1, w_2 \in W,$ $[w_1,w_2] = 0.$
Prove that $W$ has the property that $dim(W) \leq n/2.$
I have been having some trouble with this proof, I've included a few thoughts below. Any hints/solutions would be appreciated.
My thoughts: Because $H$ is Hermitian, there exists an orthonormal basis $\{e_1, \dots, e_n\}$ such that $H$ is diagonal in this basis, so we may assume we are using that basis. Then each $\lambda_i$ is nonzero and sits on the diagonal. For each $w = (w_1,w_2, \dots, w_n)^T \in W$ and $v = (v_1,v_2, \dots, v_n)^T \in W,$ one has $[w,v] = \sum_{i=1}^n \lambda_i {\bar w_i}v_i = 0.$ Then from here, I have no idea.
Also, I believe that if $n$ is odd, the maximum dimension $W$ could have is $(n-1)/2..$ but I can't prove it.