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 $$w_1\in W\cap rad(f)=\{0\}\implies w_1=0.$$