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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 $|X| \leq 2$.

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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
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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

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

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