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I've been able to find matrices B and C that commute with $$A=\pmatrix{1 & 2\cr3 & 4}$$ by solving the system AB=BA where $$B=\pmatrix{a & b\cr c & d}$$ However, when I substitute free variables into my solution, I get two different matrices B and C that commute with A, but B and C always seem to commute with one another.

Can I get a little help on this?

Thanks

Update

Matlab code (Uses Symbolic Toolbox) mentioned in my comment below.

clc
P=[1 0 1; 1 1 1;0 1 1];
D=[1 0 0;0 1 0;0 0 2]
A=P*D*inv(sym(P))
syms a b c d e f g h k
B=[a b c;d e f; g h k];
M=[A*B==B*A]
H=[0 1 1 1 0 0 -1 0 0 0;...
    1 2 1 0 -1 0 0 1 0 0;...
    1 1 0 0 0 1 0 0 -1 0;...
    1 0 0 -2 -1 -1 1 0 0 0;...
    0 1 0 1 0 1 0 1 0 0;...
    0 0 1 -1 -1 -2 0 0 1 0;...
    1 0 0 -1 0 0 0 -1 -1 0;...
    0 1 0 0 -1 0 1 2 1 0;...
    0 0 1 0 0 -1 -1 -1 0 0]
rref(sym(H))
B=[3 -3 3;1 1 1;1 1 1]
A*B
B*A
C=[5 -4 3;2 1 1;1 1 2]
C*A
A*C
B*C
C*B

Second Update

Let $A$ be a diagonalizable matrix. Then $A=SDS^{-1}$. Prove: $B$ commutes with $A$ if and only if $S^{-1}BS$ commutes with $D$.

Proof: If we replace $A$ with $SDS^{-1}$, we can write: $$\begin{align*} AB&=BA\\ (SDS^{-1})B&=B(SDS^{-1})\\ SDS^{-1}B&=BSDS^{-1} \end{align*}$$ The last line follows because of the associative property of matrix multiplication. Now we will multiply both sides on the left by $S^{-1}$, then both sides on the right by $S$. $$\begin{align*} DS^{-1}B&=S^{-1}BSDS^{-1}\\ DS^{-1}BS&=S^{-1}BSD\\ D(S^{-1}BS)&=(S^{-1}BS)D \end{align*}$$ And we've shown that $B$ commutes with $A$ if and only if $S^{-1}BS$ commutes with $D$. Now, if $D$ has distinct diagonal elements, and $S^{-1}BS$ commutes with $D$, $S^{-1}BS$ is a diagonal matrix. See also A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal. Then if $C$ also commutes with $A$, $S^{-1}CS$ is also a diagonal matrix. Because all diagonal matrices commute with each other, we can write: $$\begin{align*} (S^{-1}BS)(S^{-1}CS)&=(S^{-1}CS)(S^{-1}BS)\\ S^{-1}BSS^{-1}CS&=S^{-1}CSS^{-1}BS\\ S^{-1}BICS&=S^{-1}CIBS\\ S^{-1}BCS&=S^{-1}CBS \end{align*}$$ Now we can multiply both sides on the left by $S$, then both sides on the right by $S^{-1}$ in our second step. $$\begin{align*} BCS&=CBS\\ BC&=CB \end{align*}$$

Thanks for your help.

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  • $\begingroup$ What is your solution for $B$? (Please do not ask people to work it out for themselves when you already have the answer.) $\endgroup$ – David Dec 17 '18 at 4:38
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If a square matrix $A$ has distinct eigenvalues, the matrices that commute with $A$ are all polynomials in $A$, and these always commute with each other. This is most easily seen by noticing that $A$ is diagonalizable, and all matrices that commute with a diagonal matrix whose diagonal entries are distinct are themselves diagonal.

Thus in order to give an example where two matrices that commute with $A$ don't commute with each other, you'll need $A$ to have a repeated eigenvalue. The easiest example is $A=I$. If you want a less trivial example (not a multiple of $I$) you'll need at least $3 \times 3$ matrices.

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  • $\begingroup$ Thanks for the so quick response. This worked. I used A=[2 -1 1;1 0 1;1 -1 2] which has eigenvalues 1, 1, and 2 and whose eigenvectors are independent. I let B=[a b c; d e f; g h k] and solved the system AB=BA using Matlab (posted above). I was able to determine to matrices [3 -3 3;1 1 1;1 1 1] and C=[5 -4 3;2 1 1;1 1 2], both of which commuted with A, but did not commute with each other. However, I would really love to see a proof of your first paragraph or be given a good reference online to find a detailed proof. Thanks for your help. :-) $\endgroup$ – David Dec 17 '18 at 6:06
  • $\begingroup$ I outlined the proof in that paragraph. If $A$ has distinct eigenvalues, it is diagonalizable, say $A = S D S^{-1}$ where $D$ is diagonal with the same eigenvalues as $A$. $B$ commutes with $A$ iff $S^{-1} B S$ commutes with $D$. If $C$ commutes with $D$, then $C_{ij} D_{jj} = (CD)_{ij} = (DC)_{ij} = D_{ii} C_{ij}$, so if $D_{ii} \ne D_{jj}$ we must have $C_{ij} = 0$. Thus the only matrices that commute with a diagonal matrix whose diagonal entries are all distinct is a diagonal matrix. And all diagonal matrices commute with each other. $\endgroup$ – Robert Israel Dec 17 '18 at 13:17
  • $\begingroup$ Due to your help, I was finally able to figure it out. See above. Thanks. $\endgroup$ – David Dec 17 '18 at 19:42

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