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. $\blacksquare$
See Also
- When is matrix multiplication commutative? on math.stackexchange.com