# How can I find all the matrices that commute with this matrix?

I would like to find all the matrices that commute with the following matrix

$$A = \begin{pmatrix}2&0&0\\ \:0&2&0\\ \:0&0&3\end{pmatrix}$$

I set $AX = XA$, but still can't find the solutions from the equations.

• What goes wrong when you try to find the solutions? Feb 23, 2017 at 19:15
• This question discusses a similar problem.
– Mark
Feb 23, 2017 at 19:17
• create a generic 3x3 matrix. Multiply on the left, multiply on the right. Set the two products equal to each other. What entries give feasible solutions, which do not? Feb 23, 2017 at 19:19

$AX$ doubles the first two rows of $X$, and triples the third row.

$XA$ doubles the first two columns of $X$, and triples the third column. These two must agree.

This gives us a matrix $$X=\left(\begin{matrix}*& * & 0\\*&*&0\\0&0&*\end{matrix}\right)$$

Where $*$ can be anything.

If

$B:= \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)$,

then your matrix becomes $A = 2I + B$. Thus a matrix $C$ will commute with $A$ if and only if $C$ commutes with $B$. But note

$BC = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right) = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ g & h & i \end{matrix} \right)$

and

$CB = \left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 0 & 0 & c \\ 0 & 0 & f \\ 0 & 0 & i \end{matrix} \right)$.

It follows that $BC = CB$ if and only if $c=f = g = h = 0$

>>> from sympy import *
>>> A = diag(2,2,3)
>>> X = MatrixSymbol('X',3,3)
>>> Matrix(A*X - X*A)
Matrix([
[      0,       0, -X[0, 2]],
[      0,       0, -X[1, 2]],
[X[2, 0], X[2, 1],        0]])


If $\rm A X = X A$, then $x_{13} = x_{23} = x_{31} = x_{32} = 0$. The other five entries are unconstrained.

We can also vectorize $\rm A X = X A$, which yields the following homogeneous linear system

$$\left( (\mathrm I_3 \otimes \mathrm A) - (\mathrm A \otimes \mathrm I_3) \right) \mbox{vec} (\mathrm X) = 0_9$$

or,

$$\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix} \mbox{vec} (\mathrm X) = \begin{bmatrix} 0\\0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}$$

Again, we conclude that $x_{13} = x_{23} = x_{31} = x_{32} = 0$.

Block matrices provide an immediate insight. Let

$$A = \left[\begin{array}{l}2&0&0\\0&2&0\\0&0&3\end{array}\right] = \left[\begin{array}{l|l}a&0\\\hline0&3\end{array}\right]$$

The submatrix $a = 2I_2$. Now define the block matrix $$X = \left[\begin{array}{l|l}b&0\\\hline0&c\end{array}\right]$$ with conformal block sizes. That is, $c=constant$ and $$b = \left[\begin{array}{l}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right]$$ has arbitrary complex elements.

The equation to solve is $$[A,X] = AX - XA = \left[\begin{array}{l|l}a&0\\\hline0&3\end{array}\right] \left[\begin{array}{l|l}b&0\\\hline0&c\end{array}\right] - \left[\begin{array}{l|l}b&0\\\hline0&c\end{array}\right] \left[\begin{array}{l|l}a&0\\\hline0&3\end{array}\right] = \left[\begin{array}{l|l}0&0\\\hline0&0\end{array}\right]$$

We have two equations:

$$b a = a b$$ $$3c = 3c$$

The second equation is trivial: $c$ is arbitrary. The first equation is just $$2bI_{2} = 2 I_{2} b$$ Since the identity matrix commutes with every matrix, the $b$ matrix is arbitrary.

To conclude, the solution matrix has five arbitrary complex numbers arranged so: $$X = \left[\begin{array}{ll|l}b_{11}&b_{12}&0\\b_{21}&b_{22}&0\\\hline0&0&c \end{array}\right]$$

One sees the benefit of this form in analyzing matrices of the form $$A = \left[\begin{array}{l}c_{i} I_{i}&0&0 & 0\\0&c_{j}I_{j}&0 & 0\\0&0&c_{k}I_{k} & 0 \\ 0 & 0 & 0 & \ddots\end{array}\right]$$

• If one uses the matrix direct sum $A\oplus B$ (i.e. a block-diagonal matrix with $A,B$ on the diagonal) then the above can be stated as: The matrix $2I_2 \oplus 3$ commutes with any matrix of the form $B \oplus c$, where $B$ is 2-by-2 and $c$ is scalar. (This is similarly convenient for the last matrix, which can be written as $A=\bigoplus_k c_k I_k$.) Feb 24, 2017 at 0:28
• @Semiclassical: An elegant summation and observation. Feb 24, 2017 at 0:47
• This answer shows that matrices of the given form commute with $A$, but not why all matrices that commute with $A$ have the given form. Feb 24, 2017 at 4:01