# Proving existence of basis such that the matrix of $f$ is given by $\begin{pmatrix} A&0\\0&0 \end{pmatrix}$

Consider a sesquilinear form $$f$$ over a finite dimensional vectorspace $$V$$. Show that the is a basis of $$V$$ such that the matrix of $$f$$ is given by $$A_{f} = \begin{pmatrix} A_g &0\\0&0 \end{pmatrix}$$ with $$g=f_{|W}$$ and $$W\le V$$ such that $$V = W \oplus \operatorname{rad}(f)$$. Moreover, $$\operatorname{rad}(g)$$ is trivial.

My attempt:

This seems like a rather trivial theorem. If $$V = W \oplus \operatorname{rad}(f)$$, then $$f = f_{|W} \oplus_{\perp} 0_{|\operatorname{rad}(f)}$$. So it's clear that we only have to consider the matrix of $$f_{|W}=: g$$. This is $$A_g$$, and the matrix of the zero map is of course the zero matrix. This explains the given matrix representation of $$f$$.

How do I prove this rigorously?

So You take an orthonormal basis of $$rad(f)$$ and complete it to an orthonormal basis of $$V$$, the basis vectors not in $$rad(f)$$ forming an orthonormal basis of $$W$$ and rearrange the vectors. Then the matrix representation of $$f$$ with respect to this basis clearly has the desired form. The only thing that deserves a proof is that $$rad(g)=\{0\}$$. For this assume $$w_1\in rad(g)$$ i.e. $$g(w_1,w)=0\text{ for all }w\in W$$ $$\implies f(w_1,w)=0\text{ for all }w\in W$$ $$\implies f(w_1,v)=0\text{ for all }v\in V$$ since any $$v\in V$$ can be written in the form $$v=w+u$$ where $$w\in W,u\in rad (f)$$ and thus $$w_1\in W\cap rad(f)=\{0\}\implies w_1=0.$$