Given a square matrix $A$, what are all the possible square matrices $B$ that commutes with $A$? (meaning $AB = BA$)

Question

Let $A \in \mathbb{R}^{n \times n}$, I want to find the set $S$ of all $B \in \mathbb{R}^{n \times n}$ such that $AB = BA$.

My guess is that $S$ is the set of all possible square matrices $B$ that can be written as a polynomial of $A$, i.e. $B = a_0 I + a_1 A + a_2 A^2 + ... + a_n A^n$ for some $a_0, ..., a_n$ and $n$.

However, does anyone know if this set is tight?

In other words, I want to show that $AB = BA$ if and only if $B$ is a polynomial of $A$. Clearly, if $B$ is a polynomial of $A$, then $AB = BA$. However, if $AB = BA$, is $B$ necessarily a polynomial of $A$?

Answer 1

One condition that you might find helpful is that if two $n\times n$ matrices $A$ and $B$ share the same $n$ linearly independent eigenvectors, then $AB=BA$. Although this not an if and only if as $A$ and $B$ might not have any eigenvalues at all and therefore no eigenvectors. For a specific example, consider the matrix $A=\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$ and $B=\begin{bmatrix}2 & 0 \\ 0 & 2 \end{bmatrix}$. We have that $AB=BA$, but $A$ has not eigenvalues over $\Bbb{R}$ and therefore no eigenvectors.

Proof: Let $v_1,\dots,v_n$ be the eigenvectors of $A$. Given that $A$ is an $n\times n$ matrix and has $n$ linearly independent eigenvectors, these eigenvectors form a basis. Thus, to show that $AB=BA$, it suffices to show that $(AB)v_i=(BA)v_i$. Let $\lambda_1,\dots\lambda_n$ be the eigenvalues(possible repeating) corresponding to the eigenvectors $v_1,\dots,v_n$ of $A$ and let $c_1,\dots,c_n$ be the eigenvalues(possibly repeating) corresponding to the eigenvectors $v_1,\dots,v_n$ of $B$. Then we have $$(AB)v_i=A(Bv_i)=A(c_iv_i)$$ $$=c_i(Av_i)=c_i\lambda_iv_i=\lambda_ic_iv_i=(BA)v_i$$

Thus, $AB=BA$.

Answer 2

Let $C_A$ be the space of all real matrices that commute with $A$. Then $C_A$ has the dimension at least $n$. Moreover, the dimension is exactly $n$ if and only if the minimal polynomial of $A$ coincides with the characteristic polynomial of $A$.

Let $J(\lambda, f)$ be the Jordan block of size $f\times f$ with $\lambda$ on the diagonal entries. Suppose that the Jordan form of $A$ is $$ \bigoplus_{i=1}^r \oplus_{j=1}^{e_i} J(\lambda_i, f_{ij}) $$ where $\lambda_i$'s are distinct eigenvalues of $A$ and $\sum\limits_{i=1}^r\sum\limits_{j=1}^{e_i} f_{ij}=n$.

Then we are able to write down the dimension of $C_A$ in terms of $f_{ij}$: $$ \dim C_A=\sum_{i=1}^r\sum_{j=1}^{e_i}\sum_{k=1}^{e_i} \min(f_{ij},f_{ik}). $$ For example, if Jordan form of $A$ is $J(1,2)\oplus J(1,3)\oplus J(3,1)$, then $$ \dim C_A = \min(2,2)+\min(2,3)+\min(3,2)+\min(3,3)+\min(1,1)=10. $$ A computationally useful method is to use Kronecker product. The matrices in $C_A$ can be thought of the solution set of the following linear system: $$ (I_n \otimes A - A^T \otimes I_n)\vec{x}=\vec{0}. $$ Thus, we may use elementary row operations to make the $n^2\times n^2$ matrix $I_n \otimes A - A^T \otimes I_n$ to the reduced row echelon form, to solve the above system.

Answer 3

Start with a simple example: $$ A = \left[\begin{array}{cccc} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}\right] $$ This is composed of two matrices, both of which are in Jordan canonical form: $$ J= \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right] $$ This matrix $J$ has minimal polynomial $(\lambda-2)^2$. And so does $A$. The matrix $A$ commutes with the matrix $B$ given by $$ B= \left[\begin{array}{cc} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Even though $B$ commutes with $A$, the matrix $B$ cannot be written as a polynomial in $A$ because every polynomial acting on $A$ acts in the same way on the upper left $2\times 2$ block as it does on the lower right $2\times 2$ block.