Cases where matrix multiplication follows the commutative law

Question

I am aware that matrix multiplication does not generally follow the commutative law. However, I have identified some cases in which matrix multiplication is commutative, and I would like to inquire if there are any additional cases that I may have missed.

The cases I have considered so far are:

$1.$A and B are both diagonal matrices, including scalar, identity, and null matrices.(both are in the same order)

$2.$A and B are both equal matrices.

I am interested in knowing if there are any other scenarios where matrix multiplication follows the commutative law. Specifically, I would like to explore cases beyond diagonal matrices and matrices that are equal to each other.

I would appreciate any insights or examples you can provide regarding cases where matrix multiplication is commutative. Thank you!

Answer 1

If $A$ is a polynomial in $B$, that is, if $A=\sum_0^nc_jB^j$ for some numbers $c_j$ and $n$, then $AB=BA$.

If $A$ and $B$ are simultaneously diagonalizable, that is, if there is an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal, then $AB=BA$.

Answer 2

If you denote $\mathcal{C}(S)$ the set $\{M | \forall A \in S, \, AM = MA\}$ for $S$ a subset of $\mathcal{M}_n(k)$, which is called in French the "commutant". You have multiple results: The "bicommutant" theorem which gives you $$\mathcal{CC} (A) = \mathcal{C}(\mathcal{C}(A)) = k [A].$$

If $A$ has $n$ different eigenvalues, $\mathcal{C}(A) = k [A]$ and in general if $A$ id diagonalizable, we have an eigenspaces decomposition $E = E_1 \oplus \cdots \oplus E_k$, then $\mathcal{C} (A)$ is exactly the vector space of endomorphism that stabilize each $E_i$, so we have $$\dim \mathcal{C}(A) = \sum_{i=1}^k \dim E_i^2$$

In general we always have $\dim \mathcal{C}(A) \geqslant n$.

Here are some references in French (sorry)

http://mpstar.lamartin.free.fr/fichiers/matieres-640-1413608066.pdf

https://agreg.org/data/uploads/sujets/MG/MG17.pdf