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Usually, matrices are non-commutative objects. However, I want to know under what conditions matrices do commute, i.e. for a given (invertible) matrix $A$, find the set of all possible $B$ such that

$$AB=BA$$

This is connected to the center of the group. In fact, if we preserve the group structure, the center of $A$ should be the entire group. From the Wikipedia, such case does exists, i.e. for the Heisenberg group representation

$$\begin{pmatrix}1&a&c\\0&1&b\\0&0&1\end{pmatrix}$$, the subset(in this case, also the subgroup) was $$\begin{pmatrix}1&0&c\\0&1&0\\0&0&1\end{pmatrix}$$

Is there a general condition for the set(or group of) matrix $B$ for an arbitrary matrix $A$? Consider the $2\times 2$, $3\times 3$, $4\times 4$, and $5\times 5$. The $1\times 1$ the scalar case was trivally satisfied. For the interests of the physics, also consider $7\times 7$ matrix because ".. if the (cross) product is limited to non-trivial binary products with vector results, it exists only in three and seven dimensions."

What's the condition such that the matrix $AB=BA$?

ShoutOutAndCalculate
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    If the eigenvectors of A are the same as those of B the matrices will commute. – user317176 Nov 16 '24 at 23:24
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    The connection to "center of the group" is rather tenuous unless you are working in $\mathsf{GL}_n(k)$ so that you have a group. Matrices need not form a group under multiplication. Also, "the center of $A$" doesn't make sense: $A$ is a matrix, not a group. Perhaps you mean the centralizer in $\mathsf{GL}_n(k)$? And by "be the entire group" do you mean "should form a group"? – Arturo Magidin Nov 16 '24 at 23:29
  • @user317176 that's a good point, but I read that even for symmetric matrices it's not quite sufficient. – ShoutOutAndCalculate Nov 16 '24 at 23:32
  • @ArturoMagidin I'm actually intend to formulate the conditions in terms of a field where the additive and the multiplicative identities were included, that's why I consider the group structure as a sub condition. But your are right, it doesn't have to be a group(in fact, I encountered the non group structure in the related topic myself, the field is a very naive far shot), it could be a set equipped with certain algebra. – ShoutOutAndCalculate Nov 16 '24 at 23:35
  • @Nic I'm asking for specific case of $2\times 2$, $3\times 3$, and $4\times 4$ and $5\times 5$ cases, not the arbitrary dimension, but more generally in the restricted dimension. That question and Kevin Carlson's answer https://math.stackexchange.com/a/1018128/603316 doesn't address it. – ShoutOutAndCalculate Nov 16 '24 at 23:41
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    see http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy Nov 17 '24 at 00:17

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