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Let $A, B, C \in M_n(\mathbb{C}) $. Is it true that if $AB=BA$ and $BC=CB$ then $AC=CA$ in general? I could find an example in $M_2(\mathbb{C} )$,but I don't know how to approach the general case.

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    $\begingroup$ If $B$ is the null matrix… $\endgroup$
    – egreg
    Apr 4, 2019 at 6:42

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No it's not true in general. Just set $B$ to be the identity matrix and $A$ and $C$ to be not equal and not the identity matrix.

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    $\begingroup$ $A$ and $C$ not equal and not the identity matrix does not guarantee that $AC \neq CA$. $\endgroup$ Apr 4, 2019 at 7:37
  • $\begingroup$ I know, but given that $A \neq I$, $C \neq I$ and $A \neq C$ it is pretty simple to come up with a counter example $\endgroup$
    – Cettt
    Apr 4, 2019 at 7:39
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If $B=I$ then $AB=BA$ and $BC=CB$ for any two matrices $A$ and $C$. There is no reason why $AC=CA$. For a specific counterexample let $A$ have $(0,1)$ in both the rows and let $C$ have the rows $(1,0), (0,0)$.

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Need not be true. For example, take $B $ to be the identity matrix and $A $ and $C $ to be two non-commutative matrices.

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ALready this question has been answered mostly by way of counterexamples. Let me add some remark that might also be helpful.

As matrix $B$ appears in both the conditions you can formulate your question in a $B$-centric way (pun intended!). That is, if two matrices $A,C$ commute with $B$, do they commute with each other? (One can ask this question in any non-abelian group)

So it boils down to checking if the centralizer of a subgroup is an abelian group. (Or centralizer in an algebra is a commutative subalgebra).

Zero matrix and identity matrix commute with everything else; but in general two martices don't commute giving the answer you are looking for. Any scalar matrix taken as $B$ will provide counter-exmaples.

Let $A$ be an abelian group and $G$ be a non-abelian group. Pick two elements $g,h\in G$ such that $gh\ne hg$.

In $A\times G$, the element $(e,g)$ commutes with $(a,e)$. Same way $(e, h)$ commutes with $(a,e)$. However $(e,g),\ (e,h)$ don't commute with each other.

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In general, this is not true. However, if all eigenvalues of A are distinct with distinct eigenvectors (s.t they form a basis), B will share the same eigenvectors (and therefore the same eigenbasis, spanning all of space). Now, if we apply the same to B, with all it's eigenvalues being distinct, then B and C share the same eigenbasis. Thus, A and C share the same eigenbasis. This is equivalent to A and C commuting.

$AB = BA \iff A = PD_AP^{-1},\;B = PD_BP^{-1}$.

$BC = CB \iff B = PD_BP^{-1},\;C = PD_CP^{-1}$.

If $A\vec{v_i}=\lambda_i\vec{v}_i$, $A(B\vec{v}_i) = BA\vec{v_i} = \lambda_i(B\vec{v}_i)$

and thus, $B\vec{v_i}$ lies in the eigenspace of A corresponding to $\lambda_i$. If all $\lambda_i$ are distinct, every eigenspace corresponding to one eigenvalue $\lambda_i$ of A has dimension 1. This means that $B\vec{v}_i = \mu_i\vec{v}_i$. Thus is every eigenvector of A also an eigenvector of B. Doing this once more, assuming every $\mu_i$ are distinct, we get that every eigenvector of B is an eigenvector of C. Thus, every eigenvector of A is an eigenvector of C. This means that we can write

$C = PD_CP^{-1}$, and since $A = PD_AP^{-1}$ we would have that $AC = CA$.

This is probably not the only solution, but is used in quantum mechanics to deduce commutation relations of hermitian operators.

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