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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.

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