Let ${\mathbb K}={\mathbb R}$ or $\mathbb C$. Let $V$ be a vector space over $\mathbb K$ and fix a basis $\cal B$ of $V$. We say that a family of vectors of $V$ is nice (relatively to $\cal B$) if it is made of “disjoint linear combinations”, i.e. if we write out the decomposition of those vectors, no basis vector is used twice. For example, $\lbrace b_1-3b_5+b_7, -4b_2+b_6, b_8 \rbrace$ is nice relatively to $\lbrace b_i \rbrace_{1 \leq i \leq 8}$. Also, say that a subspace is nice if it admits a nice basis. Such a basis is clearly unique, up to rescaling the basis vectors.
We use these notions on $V={\cal M}_{n}(\mathbb K)$, the $n\times n$ square matrices over $\mathbb K$, with the canonical basis ${\cal E}=(E_{ij})_{1 \leq i,j \leq n}$ ($E_{i,j}$ is the matrix of all whose coefficients are zero except the one at $(i,j)$, which is $1$).
Given a subset $X$ of $V$, the commutant ${\bf Com}(X)$ of $X$ is
$ {\bf Com}(X)=\lbrace a \in V | \forall x \in X, ax=xa\rbrace $
Prove or find a counterexample: if $X \subseteq {\cal E}$, then ${\bf Com}(X)$ is a nice subspace (relatively to the canonical basis).
I have checked this combinatorially when $|X| \leq 2$.