Matrix representation proof

For each sesquilinear form $$f$$ on a finite $$K$$-vector space $$V$$ there can be found a basis for $$V$$ such that the matrix of $$f$$ is given by $$A_f = \begin{pmatrix} A_g & 0 \newline 0 & 0 \end{pmatrix}$$ with $$g = f_{|W}$$ and $$W \le V$$ such that $$V = W \oplus rad(f)$$.

I'm using this theorem (which I don't know how to proof, so if someone could take a look at it, I would be very thankful!): Suppose $f: V \times V \to K$ is a sesquilinear form and $W \le V$ such that $V = W \oplus rad(f)$. Proof that $rad(f_{|W})$ is trivial.. (This theorem should also state that $$f = f_{|W} \oplus_{\perp} 0_{|rad(f)}$$, I will be using this).

My attempt:

Let $$\{e_1,\dots,e_k\}$$ be a basis for $$W$$ and $$\{e_{k+1},\dots,e_n\}$$ a basis for $$rad(f)$$. Then $$\{e_1,\dots,e_n\}$$ is a basis for $$V$$. Choose $$w_1,w_2 \in W$$ and $$u_1,u_2\in rad(f)$$. Then $$w_1 = \sum_{i=1}^k \lambda_i e_i$$ and $$w_2 = \sum_{i=1}^k \mu_i e_i$$, and

$$f(w_1+u_1,w_2+u_2) = f(w_1,w_2)+f(w_1,u_2)+f(u_1,w_2)+f(u_1,u_2) \\= f_{|W}(w_1,w_2)+f(u_1,u_2)$$

We know that (matrix representation) $$f(w_1,w_2) = (\lambda_1 \,\, \dots \,\,\lambda_k)A_g \begin{pmatrix} \mu_1^{\sigma} \\ \vdots \\\mu_k^{\sigma}\end{pmatrix}$$, with $$A_g = \begin{pmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \ddots & \vdots \\ a_{k1} & \cdots & a_{kk}\end{pmatrix}.$$

It is clear that the only function on the $$rad(f)$$ is the zero form. Therefore the matrix representation of $$0_{|rad(f)}$$ is just the zero $$n-k$$-matrix $$O_{n-k}$$.

Since $$f = f_{|W} \oplus_{\perp} 0_{|rad(f)}$$, we know that each element in $$V$$ (which can be written as a linear combination of elements $$e_1,\dots,e_n$$) will be mapped by either $$f_{|W}$$ or the zero form. ($$V = W \oplus rad(f)$$).

So with the given order of the basis of $$V$$, we get the asked form. On the diagonals of $$A_f$$ we have the matrix respresentations of both $$f_{|W}$$ and then $$0_{|rad(f)}$$. All other elements should be 0, because $$V$$ is the DIRECT sum of $$W$$ and $$rad(f)$$, meaning there will be no element of $$W$$ (and thus the basis of $$W$$) mapped by the zero form $$0_{|rad(f)}$$. The same counts for elements in $$rad(f)$$: no element in the radical of $$f$$ will be mapped to another element by the function $$f_{|W}$$.

Is this correct? Thanks a lot.