Do two matrices commute only if one matrix is the inverse of the other?

Question

Is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? The only way I could believe this was the 2 commutative type properties I know of: $$AA^{-1}=A^{-1}A=I$$ and $$adj A*A=A*adjA = |A|I$$ where $adj A$ is the adjoint of matrix A.

The first equation when expanded with $A^{-1}=\frac{adjA}{|A|}$ you end up getting the second equation. So both are the same equation. These are the only two properties that explains commutativity as far as I'm aware of(are there any more?) and my conclusion was based on these only. Also multiplication with identity matrix is also commutative but any other counter cases? So is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? Thanks!!

Answer 1

Any two diagonal matrices commute, because:

$$\begin{bmatrix}a_1&0&\cdots&0\\0&a_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&a_n\end{bmatrix}\cdot\begin{bmatrix}b_1&0&\cdots&0\\0&b_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&b_n\end{bmatrix}=\begin{bmatrix}a_1b_1&0&\cdots&0\\0&a_2b_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&a_nb_n\end{bmatrix}$$

and "ordinary" multiplication commutes.

Now, if a matrix $P$ is an arbitrary invertible matrix, and $D_1$ and $D_2$ are arbitrary diagonal matrices, this means that the matrix $P^{-1}D_1P$ will also commute with $P^{-1}D_2P$ because:

$$(P^{-1}D_1P)(P^{-1}D_2P)=P^{-1}D_1D_2P=P^{-1}D_2D_1P=(P^{-1}D_2P)(P^{-1}D_1P)$$

Answer 2

Not at all. For instance, the identity matrix commutes with every other square matrix of the same size. As another example, the matrices $\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right]$ commute, but none of them is a multiple of the other one.

Answer 3

For any matrix $A$, the matrix $A^n$ whatever be $n$ satisfies $A A^n = A^n A$.

Also for any two real numbers $a, b$ the matrices $1\quad a\choose 0\quad 1$ and $1\quad b\choose 0\quad 1$ can be multiplied in any order to get the same product.

Answer 4

Another commuting matrix examples come from:

Lemma: Any square matrix $A$ conmutes with a linear combination of itself and the identity, that is:

$A\cdot (\alpha A-\beta I)= A \cdot(\alpha A-\beta I)$

Proof:

$$A\cdot (\alpha A-\beta I) = \alpha A^2 -\beta A = A\cdot(\alpha A - \beta I)$$