When Is Matrix Multiplication Commutative?

Matrix multiplication in general is not commutative. Here is an example:

Let $A, B \in \mathbb{R}^{2 \times 2}$:

$$A := \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B := \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$$

Then:

$$A \cdot B = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix} \neq \begin{pmatrix} 23 & 34 \\ 31 & 46 \end{pmatrix} = B \cdot A$$

When Is 2×2 Matrix Multiplication Commutative?

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$$

$$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$

So you get four equations:

$$\begin{align} I) \quad & ae + bg &= ae + cf &\Leftrightarrow bg &= cf \\ II) \quad & af + bh &= be + df \\ III) \quad & ce + dg &= ag + ch \\ IV) \quad & cf + dh &= bg + dh &\Leftrightarrow cf = bg \end{align}$$

You might note that (I) is the same as (IV). So you have these equations:

$$\begin{align} I) \quad & bg &= cf \\ II) \quad & af + bh &= be + df & \Leftrightarrow f (a - d) = b (e - h) \\ III) \quad & ce + dg &= ag + ch & \Leftrightarrow g (a - d) = c (e - h) \end{align}$$

Case #1: a ≠ d and e ≠ h

$$\begin{align} I) \quad & bg &= cf \\ II) \quad & \frac{f}{g} &= \frac{b}{c} \Leftrightarrow cf = bg \end{align}$$

Now (I) and (II) are essentially the same. So we only demand that $bg = cf$ and $a \neq d$ and $e \neq h$ for commutative matrix multiplication of $2 \times 2$ matrices.

Case #2.1: a = d

$$\begin{align} I) \quad & bg &= cf \\ II) \quad & 0 &= b (e - h) \\ III) \quad & 0 &= c (e - h) \end{align}$$

So you end up with: $(e = h \text{ and } bg = cf)$ or $(b = c = 0)$

Case #2.2: e = h

$$\begin{align} I) \quad & bg &= cf \\ II) \quad & f (a - d) &= 0 \\ III) \quad & g (a - d) &= 0 \end{align}$$

So you end up with: $(a = d \text{ and } bg = cf)$ or $(f = g = 0)$

Special Cases

Matrix multiplication is always commutative if:

One matrix is the Identity matrix
One matrix is the Zero matrix
Both matrices are $2 \times 2$ rotation matrices (basically case #2)
Both matrices are Diagonal matrices

Simultaneous Diagonalization

Two matrices $A, B \in \mathbb{R}^{n \times n}$ are called simultaneously diagonalizable if and only if one matrix $S \in \mathbb{R}^{n \times n}$ exists, such that $D_A = S^{-1} \cdot A \cdot S$ and $D_B = S^{-1} \cdot B \cdot S$ where $D_A$ and $D_B$ are diagonal matrices.

Theorem: If $A, B \in \mathbb{R}^{n \times n}$ are simultaneously diagonalizable, then $A \cdot B = B \cdot A$.

Proof: Since A and B are simultaneously diagonalizable, a matrix $S \in \mathbb{R}^{n \times n}$ exists such that $D_A = S^{-1} \cdot A \cdot S$ and $D_B = S^{-1} \cdot B \cdot S$ where $D_A$ and $D_B$ are diagonal matrices.

$$\begin{align} \Rightarrow A \cdot B &= S \cdot D_A \cdot S^{-1} \cdot S \cdot D_B \cdot S^{-1} \\ &= S \cdot D_A \cdot D_B \cdot S^{-1} \\ &= S \cdot D_B \cdot D_A \cdot S^{-1} \\ &= S \cdot D_B \cdot S^{-1} \cdot S \cdot D_A \cdot S^{-1} \\ &= B \cdot A \quad \blacksquare \end{align}$$

Note: The converse is not true: $A \cdot B = B \cdot A \nRightarrow A, B$ are simultaneously diagonalizable.

Proof by counterexample:

$$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \cdot \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \cdot \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$

but $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ is not diagonalizable. $\blacksquare$