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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.

  • 2
    What goes wrong when you try to find the solutions? – 2017-02-23
  • 0
    [This question](https://math.stackexchange.com/questions/296857/finding-all-matrices-b-such-that-ab-ba-for-a-fixed-matrix-a?rq=1) discusses a similar problem. – 2017-02-23
  • 6
    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? – 2017-02-23

4 Answers 4

21

$AX$ 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.

20

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 $c=f = g = h = 0$

4

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 $$A = \left[\begin{array}{l}c_{i} I_{i}&0&0 & 0\\0&c_{j}I_{j}&0 & 0\\0&0&c_{k}I_{k} & 0 \\ 0 & 0 & 0 & \ddots\end{array}\right]$$

  • 2
    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 $A=\bigoplus_k c_k I_k$.) – 2017-02-24
  • 0
    @Semiclassical: An elegant summation and observation. – 2017-02-24
  • 6
    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. – 2017-02-24
3
>>> 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 $x_{13} = x_{23} = x_{31} = x_{32} = 0$.