Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|X| \leq 2$.

share|cite|improve this question
Nice bases are not unique: you can always multiply a basis vector by a nonzero constant. But indeed this is all that can be varied, apart from the order of the vectors. – Marc van Leeuwen Oct 22 '12 at 8:01
@MarcvanLeeuwen :Corrected, thanks for the feedback. – Ewan Delanoy Oct 22 '12 at 10:31
up vote 1 down vote accepted

The answer is YES. Let $M=(m_{ij})$ be a matrix. Then $M$ commutes with $E_{i_1j_1}$ iff $m_{j_1j_1}=m_{i_1i_1}$ and $m_{xy}=0$ for any $(x,y) \in I(i_1,j_1)$ where $$I(i_1,j_1)=\lbrace (j_1,t) ; (t,i_1) \rbrace \setminus \lbrace (i_1,i_1);(j_1,j_1)\rbrace.$$

This generalizes immediately : for any $X \subseteq {\cal E}$,

$$ (m \in {\sf Com}(X)) \Leftrightarrow (\forall (i,j)\in X, m_{jj}=m_{ii}) \ \text{and} \ \bigg(\forall (x,y) \in \bigcup_{(i,j)\in X}I(i,j), \ m_{x,y}=0 \bigg) $$

So if we denote by $C_1, \ldots ,C_r$ the connected components of $X$ (viewed as a nondirected graph on $\lbrace 1,2, \ldots ,n \rbrace$), ${\sf Com}(X)$ admits the following nice basis :

$$ \bigg\lbrace \sum_{c\in C_k}E_{cc} \bigg\rbrace_{1 \leq k \leq r} \cup \lbrace E_{xy}\rbrace_{(x,y)\in R}, $$

where $R$ is the “residual” set

$$ R=\lbrace 1,2, \ldots ,n \rbrace ^2 \setminus \Bigg(\bigg(\bigcup_{(i,j)\in X}I(i,j) \bigg) \cup \lbrace E_{cc} \rbrace_{c\in \cup_{k}C_k}\Bigg) $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.