How to find the set of all matrices that commute with a given matrix?

Question

Is there any smart way to compute the set of all matrices that commute with a given matrix?

Smart way means that involves concept and intutition of linear and abstract algebra and does not involve any calculation. For example, given a matrix $A$ of order $n \times n$ we can solve for $X$ the matrix equation $$AX=XA$$ and get a condition on the entries of $X$. But I am not looking for such a method. I am looking for some smart approach like using linear transformations and visual understanding. Is there any way to do this? I have searched for this question in stack exchange but the answers are out of scope of my present knowledge.

@RodrigodeAzevedo I mean I want to avoid complicated calculations involving matrix multiplication which would have a large complexity for larger values of $n$. — Kishalay Sarkar, Dec 25 '19 at 08:25
@RodrigodeAzevedo As to answer your question we can find a suitable basis by visualizing the situation and accordingly come up with the subspace,that is not very difficult to calculate.What I want to say is that an apporach with linear transformation makes it very much simple as for example we can prove the associativity of matrix multiplication very easily using transformations but if we go with calculation,then we have to handle complicated matrix multiplication. — Kishalay Sarkar, Dec 25 '19 at 08:28
@RodrigodeAzevedo Nothing in that page is useful for this question. What they need to be looking at is at the invariant subspaces of $A$, in particular at the Jordan decomposition of $A$. — MoonLightSyzygy, Dec 25 '19 at 08:33
@MoonLightSyzygy One can mix the Jordan decomposition and the Kronecker product. They are miscible. — Rodrigo de Azevedo, Dec 25 '19 at 08:35
You already have all the polynomials $P(A)$ of $A$ in the commutant. — Gribouillis, Dec 25 '19 at 10:43
Does this answer your question? What commutes with a matrix in Jordan canonical form? — P. Quinton, Dec 25 '19 at 12:36

score 2 · Accepted Answer · answered Dec 25 '19 at 19:31

In this website, the fashion is to explain what is the intuition which makes it possible to solve such or such problem (incidentally, it also earns points...). First of all, if you are just starting out or if you are not working on examples, you are unlikely to have an effective intuition; in other words, intuition is useful (and even, in some difficult cases, essential) only when you have understood a certain number of results and have accumulated experience.

Calculate the commutant $C(A)$ of a matrix $A\in M_n(K)$ is not obvious in the general case.

$\textbf{Proposition 1}$. Let $A=diag(A_1,\cdots,A_k)\in M_n(\mathbb{C})$ s.t., for every $i\not= j$, $spectrum(A_i)\cap spectrum(A_j)=\emptyset$. Then $B\in C(A)$ iff $B$ is in the form $diag(B_1,\cdots,B_k)$ where $A_iB_i=B_iA_i$.

$\textbf{Proposition 2}$. Let $A\in M_n(\mathbb{C})$ be cyclic (that is, its Jordan form has the form $Jordan(A)=diag(\lambda_1I_{i_1}+J_{i_1},\cdots,\lambda_kI_{i_k}+J_{i_k})$, where the $(\lambda_i)$ are distinct and $(J_p)$ is the nilpotent Jordan block of dimension $p$). Then $B\in C(A)$ iff $B$ is a polynomial in $A$ (note for @P. Quinton : in particular, $B\in C(\lambda_kI_{i_k}+J_{i_k})$ iff $B$ is a polynomial in $J_k$). Therefore, $dim(C(A)=n$.

$\textbf{Remarks}$. i) Note that, in general when $n\geq 5$, we don't know how to calculate the Jordan form of a matrix (because we don't know how to calculate its eigenvalues).

ii) According to both previous propositions, $dim(C(A)\geq n$.

Moreover, to study $C(A)$, it suffices to consider the case when $A$ has only one eigenvalue and is not cyclic. Unfortunately (for your intuition), it's the difficult case; I'll just consider the following example

Let $A=diag(J_2,J_2)$. The matrices $B=diag(B_1,B_2)$ where $B_1,B_2\in span(I_2,J_2)$ are in $C(A)$. Yet, $dim(C(A))\not= 4$. Indeed, $dim(C(A))=8$ -because $B\in C(A)$ is in the form

$\begin{pmatrix}a&b&c&d\\0&a&0&c\\e&f&g&h\\0&e&0&g\end{pmatrix}$-.

Conclusion: "Intuition is the anticipated vision of a truth: hypothesis for the scientist, reverie for the poet."

score 1 · Answer 2 · answered Dec 25 '19 at 11:07

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If we take the Jordan form $A=PJP^{-1}$ where $J=\begin{bmatrix}J_1&0&\cdots&0\\0&J_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&J_k\end{bmatrix}$ such that $J_i$ is a Jordan block of size $n_i\times n_i$ ($\sum_{i=1}^k n_i=n$) then all the matrices \begin{align*} X&=P\cdot J'\cdot P^{-1} \end{align*} where $J'=\begin{bmatrix}J_1'&0&\cdots&0\\0&J_2'&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&J_k'\end{bmatrix}$ and $J_i'$ are jordan blocks of size $n_i\times n_i$. Indeed \begin{align*} J_i\cdot J_i' &=\begin{bmatrix} \lambda_i&1&0&\cdots&0\\0&\lambda_i&1&\cdots&0\\0&0&\lambda_i&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda_i \end{bmatrix}\cdot\begin{bmatrix} \lambda_i'&1&0&\cdots&0\\0&\lambda_i'&1&\cdots&0\\0&0&\lambda_i'&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda_i' \end{bmatrix}\\ &=\begin{bmatrix}\lambda_i\lambda_i'&\lambda_i+\lambda_i'&1&\cdots&0\\0&\lambda_i\lambda_i'&\lambda_i+\lambda_i'&\cdots&0\\0&0&\lambda_i\lambda_i'&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda_i\lambda_i' \end{bmatrix}\\&=J_i'\cdot J_i \end{align*} Hence $J$ and $J'$ commute and so does $A$ and $X$.

answered Dec 25 '19 at 11:07

P. Quinton

6,031

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This doesn't include all matrices that commute with $A$ (for example, it won't include multiples of identity matrix if $A$ is not diagonalisable). – Vedran Šego Dec 25 '19 at 11:19
@VedranŠego actually it appears to be the case (as far as I've tried) that it is true for all Jordan matrices $J'$ of size $n\times n$ even if the block size doesn't match. And conversely all comuting matrices have same eigen space so that would be exatcly all comuting matrices. – P. Quinton Dec 25 '19 at 11:22
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In the reference I gave in a comment to the question you will find further counterexample to the claim that your construction represents in general all matrices commuting with the given one. Though the counterexample given in the previous comment already suffices. – user Dec 25 '19 at 13:35
But I guess from your formulation that it is not true that all $n\times n$ Jordan matrices commute. Anyone have a counterexample ? – P. Quinton Dec 25 '19 at 15:27
@P.Quinton, I gave an obvious counterexample, but let me elaborate on it: take a very simple matrix: $A = J_2(1) = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ and explain how can you get from that to $X = \mathrm{I} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$, which obviously commutes with $A$. Which $J'$ would you use? – Vedran Šego Dec 26 '19 at 12:52
@VedranŠego I feel like $J'=I$ works and it is in Jordan normal form and then $X=PJ'P^{-1}=I$ commutes with $A$. I think I'm missing your point. I was asking for two matrices in Jordan form that do not commute. – P. Quinton Dec 26 '19 at 13:29
@P.Quinton, $J'=I$ does not have a Jordan block of order $2 \times 2$, like $J$, so it doesn't fit your construction. I am not saying that your matrices do not commute. They do. I am saying that your construction will not find ALL the matrices that commute with the given $A$, as the OP asked (and I gave multiples of identity when $A$ is non-diagonalisable as an example of that). – Vedran Šego Dec 26 '19 at 23:57

How to find the set of all matrices that commute with a given matrix?

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