The unitary reduction of normal matrices is a well-known fact: if $A\in M\_n(\mathbb C)$ commutes with its Hermitian adjoint $A^\*$, then there exists a unitary $U\in\mathbb U\_n$ and a diagonal matrix $D$ such that $A=U^\*DU$. And conversely, such a $U^\*DU$ is normal.

Besides, the theorem of Amitsur & Levitski tells us that the standard polynomial $\mathcal S\_{2n}$ in non-commutative variables vanishes identically over $M\_n(k)$, where
$$\mathcal S\_p(A\_1,\ldots,A\_p):=\sum\_\sigma \epsilon(\sigma)A\_{\sigma(1)}\cdots A\_{\sigma(p)},$$
where $\sigma$ runs over the symmetric group $\mathfrak S\_p$.

Now, let me say that a matrix $A$ is $q$-normal if $\mathcal S\_{2q}$ vanishes identically over the sub-algebra spanned by $A$ and $A^\*$. For instance, a $1$-normal matrix is normal, whereas every $n\times n$ matrix is $2n$-normal.

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> Given a $q$-normal matrix $A$. Does there exist a unitary matrix $U$ and a block-diagonal matrix $D$ with diagonal blocks of sizes $\le q$, such that $A=U^\*DU$ ?
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