Can any complex $n\times n$ matrix be in a maximal commuting set of $\lfloor n^2/4\rfloor + 1$ matrices?

Question

The maximum number of mutually commuting linearly independent complex matrices of order $n$ is equal to $\lfloor n^2/4\rfloor + 1$ ($\lfloor \cdot \rfloor$ denotes the integer part of $\cdot$). This result is attributed to Schur and a short proof can be found here. In an answer to this question, an explicit set $\mathcal{M}$ of linearly independent and mutually commuting matrices is written for $2n\times 2n$ complex matrices: $$ \begin{pmatrix}a\cdot\text{Id}_n & M_{n\times n} \\ 0_{n\times n} &a\cdot \text{Id}_n\end{pmatrix}, $$ where $a$ is a complex number, $\text{Id}_n$ is the order $n$ identity matrix and $M$ is an arbitrary complex matrix.

Now my question: can any nonzero $n\times n$ complex matrix be part of a subspace of ($\lfloor n^2/4\rfloor +1$) linearly independent mutually commuting matrices? If the answer is negative, what would be necessary?

To avoid misunderstanding, let me add some redundancy and rephrase my question as: given an arbitrary nonzero $n\times n$ complex matrix $A_0$, is it possible to find $\lfloor n^2/4\rfloor$ linearly independent $A_i$, $i=1,\ldots,\lfloor n^2/4\rfloor$, such that $A_iA_j-A_jA_i=0$ for $i,j=0,\ldots,\lfloor n^2/4\rfloor$?

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Update

The answer to my main question - if any nonzero matrix can be in the maximal subspace of mutually commuting matrices - is negative in general as Ewan Delanoy pointed out with the example of diagonalizable matrices with distinct eigenvalues.

The secondary question - what is necessary to be in the maximal commuting set - seems more complicated. A set of mutually commuting matrices can be simultaneously triangularized, then only upper (lower) triangular matrices need be analyzed. For $2n\times 2n$ matrices an explicit set $\mathcal{M}$ of solutions is provided above and for $(2n+1)\times(2n+1)$ matrices the following set, also called $\mathcal{M}$, is a maximal commuting subspace: $$ \begin{pmatrix}a\cdot\text{Id}_n & M_{n\times n+1} \\ 0_{n+1\times n} &a\cdot \text{Id}_{n+1}\end{pmatrix}. $$

This set of matrices is also suggested in the paper by M. Mirzakhani, "A simple proof of a theorem of Schur", already cited above.

Now I would like to change my secondary question for a more specific one: must any maximal commuting subspace have the form of $\mathcal{M}$?

I think the answer is affirmative and, if that is the case, it will answer this question in MO: "How many commuting nilpotent matrices are there?" because the set $\mathcal{M}$ would be the maximal abelian nilpotent subalgebra plus (a multiple of) the identity matrix.

Answer 1

The answer is easily negative when $n\geq 4$ (so that $\lfloor \frac{n^2}{4}\rfloor+1 \gt n$), because then if you take $A_0$ to be diagonalizable with distinct eigenvalues, the commutant of $A_0$ (i.e. the space of matrices $B$ satisfying $A_0B=BA_0$) already has dimension $n$.

This also indicates a partial answer to your second question : a necessary condition is that the commutant of $A$ must have dimension at least $\lfloor \frac{n^2}{4}\rfloor+1$. This is probably not sufficient however, because there is no reason why the matrices in that commutant should commute with each other.

Answer 2

In case $A$ is nilpotent or unitriangular this is true, because there is an invertible matrix $S$ such that $SAS^{-1}$ is contained in the above set $\mathcal{M}$. Then $A\in S^{-1}\mathcal{M}S$, and hence is part of such a set of pairwise commuting matrices.

For matrices $A$ where its Jordan form cannot be chosen in $\mathcal{M}$, it need not be true.

Remark: The result has been generalized later by Jacobson for arbitrary fields:

N. Jacobson: Schur's theorem on commutative matrices.
Bull. Amer. Math. Soc. 50 (1944), 431-436.

See also our paper Minimal faithful representations of reductive Lie algebras, where it is Proposition $2.3$:

Proposition 2.3: Let $M$ be a commutative subalgebra of $M_d(K)$ over an arbitrary field $K$. Then $\dim M\le \lfloor d^2/4 \rfloor +1$, and the bound is attained.