# Center of matrices

I encountered a neat problem in a problem in particle physics

So given $n$ skew symmetric matrices $A_1,...,A_n$ in $\mathbb{C}^{d \times d}.$

I would like to call this the commutator property: $A_1,...,A_n$ have the $\textbf{commutator property}$, if the only skew-symmetric matrix that has zero trace and commutes with all the $A_i$, zero trace because $i \mathbb{1}$ will of course always commute, is the zero matrix.

Now, if a matrix $B$ commutes with some $A_1,...,A_n$ then $A_i$ and $B$ are simultaneously diagonalizable, because they are normal. Unfortunately, this does not give us any good condition on all of the $A_1,...,A_n.$

So I was thinking: Assuming you have a bunch (let's say 50) of those matrices and a computer. How do you decide whether the matrices have the commutator property?

• The space of skew-symmetric matrices with trace 0 is called $\mathfrak{so}(d)$. Define $V_0=\mathfrak{so}(d)$, and inductively $V_{i+1}$ as the centralizer of $A_{i+1}$ in $V_{i}$, finally check if $V_n=0$. Each $V_i$ can be effectively computed. Indeed $V_{i+1}$ is the kernel of the map $V_{i}\to M_d(\mathbf{C})$, $M\mapsto MA_{i+1}-A_{i+1}M$. – YCor Sep 18 '16 at 17:51
• $i1$ is not skew-symmetric, it is skew-Hermitian. Do you mean skew-Hermitian instead of skew-symmetric? – Robert Israel Sep 18 '16 at 18:25
• A more explicit version of what YCor wrote earlier: both the commutation relations $A_iB-BA_i=0$ and the trace condition $\operatorname{tr} B=0$ are homogeneous linear equations in the entries of the unknown matrix $B$. With a computer at hand, it is easy to check if a system of homogeneous linear equations has a nonzero solution. – Federico Poloni Sep 18 '16 at 18:32
• If the question is meant for skew-hermitian, then it's simpler to compute the centralizer within all matrices. Indeed this centralizer is stable under taking the projection to skew-hermitian matrices modulo hermitian matrices, so the centralizer in skew-hermitian matrices with trace zero is zero if and only if the centralizer in all matrices is reduced to multiples of identity. (By the way, this is the same as asking whether this bunch of matrices acts irreducibly.) – YCor Sep 18 '16 at 19:41
• @YCor oh sorry, yes I indeed meant skew-hermitian. I am not sure I understand your comment though. If I interpret "if the centralizer in all matrices is reduced to multiples of identity" correctly, then you say that every centralizer has to reduce to multiples of the identity?-This does not seem correct, so I apparently misinterpret your comment. – Theophile1987 Sep 18 '16 at 21:06