Jtdylktuy

I've been able to find matrices B and C that commute with
$$A=pmatrix{1 & 2cr3 & 4}$$
by solving the system AB=BA where
$$B=pmatrix{a & bcr c & d}$$
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.

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.