About diagonalization of complex quadractic forms

Question

I'm studying diagonalization of quadractic forms by an orthonormal basis in finite-dimensional Euclidean spaces over $\mathbb K=\mathbb R$ or $\mathbb C$.

In this context, quadractic forms are defined to be functions $q: V \to \mathbb K$ given by $q(u)=f(u,u)$, where $f: V\times V\to\mathbb K$ is a sesquilinear form.

When $\mathbb K=\mathbb R$, we can show that any quadractic form is given by a symmetric form (which has a symmetric matrix $A$, that is, $A=A^t$, under the canonical basis), and so, by using the spectral theorem, we are able to find an orthonormal basis that diagonalizes it.

However, when $\mathbb K=\mathbb C$, not any quadractic form is given by a self-adjoint form (just pick $q: \mathbb C \to \mathbb C$ such that $q(a)=i|a|^2$, which is given by the sesquilinear form $f:\mathbb C \times \mathbb C \to \mathbb C$, $f(a,b)=ia\overline b$). Therefore, to use the spectral theorem for self-adjoint operators, we must suppose that the sesquilinear form that generates $q$ is hermitian, that is, $f(x,y)=\overline{f(y,x)}$ (which is equivalent to say that the matrix $A$ of $f$ is self-adjoint, that is, $A=A^\ast$ under the canonical basis).

Then, here is my question. When $\mathbb K=\mathbb C$, the spectral theorem is not only valid for self-adjoint operators, but also for any normal operator, that is, operators satisfying $AA^\ast=A^\ast A$. Then I wonder: is it true, when $\mathbb K=\mathbb C$, that any quadractic form is given by a sesquilinear form with its matrix $A$, under the canonical basis, satisfying $AA^\ast=A^\ast A$?

If this is true, then we would be able to prove that, when $\mathbb K=\mathbb C$, any quadractic form (not only the hermitian ones) is diagonalizable by an orthonormal basis.

Answer 1

I got a negative answer to this question, which is a consequence of the following:

Any quadratic form, $q: V\to \mathbb C$ is represented by a unique sesquilinear form $f: V\times V \to \mathbb C$, given by the following identity: $$ f(u,v)=\frac{q(u+v)-q(u-v)+i(q(u+iv)-q(u-iv))}{4} $$

When we permutate $u$ and $v$, we get that $$ f(v,u)=\frac{q(u+v)-q(u-v)-i(q(u+iv)-q(u-iv))}{4}, $$ however it does not imply that $f(v,u)=\overline{f(u,v)}$, because we are not sure that $q(u+v)-q(u-v)$ and $q(u+iv)-q(u-iv)$ are the real and imaginary parts of $f(u,v)$. This is assured when we suppose that the counterdomain of $q$ is $\mathbb R$.

Putting it all together, we prove that

Any quadratic form $q:V\to \mathbb R$ is represented by a unique hermitian sesquilinear form.