Matrix multiplication in general is not commutative. Here is an example:
$A, B \in R^{2 \times 2}$
$$A := \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$$ $$B := \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$$
$$A \cdot B = \begin{pmatrix} 19 & 22 \ 43 & 50 \end{pmatrix} \neq \begin{pmatrix} 23 & 34 \ 31 & 46 \end{pmatrix} = B \cdot A$$
When is 2x2 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: $$\begin{eqnarray} I) & ae + bg &= ae + cf &\Leftrightarrow bg = cf \ II) & af + bh &= be + df\ III) & ce + dg &= ag + ch\ IV) & cf + dh &= bg + dh &\Leftrightarrow cf = bg \end{eqnarray}$$
You might note that (I) is the same as (IV). So you have those equations: $$\begin{eqnarray} I) & bg = cf \ II) & af + bh &= be + df & \Leftrightarrow f (a - d) = b (e - h)\ III) & ce + dg &= ag + ch & \Leftrightarrow g (a - d) = c (e - h) \end{eqnarray}$$
Case #1: a != d and e != h
$$\begin{eqnarray} I) & bg &= cf \ II) & \frac{f}{g} &= \frac{b}{c} \Leftrightarrow cf = bg \end{eqnarray}$$
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{eqnarray} I) & bg &= cf \ II) & 0 &= b (e - h)\ III) & 0 &= c (e - h) \end{eqnarray}
So you end up with: ($e = h$ and $bg = cf$) or ($b = c = 0$)
Case #2.2: e == h
\begin{eqnarray} I) & bg &= cf \ II) & f (a - d) &= 0\ III) & g (a - d) &= 0 \end{eqnarray}
So you end up with: ($a = d$ 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 $A, B \in R^{n \times n}$ are called simultaneous diagonalizable $: \Leftrightarrow$ one matrix $S \in R^{n \times n}$ exists, such that $D_A = S^{-1} \cdot A \cdot S$ and $D_B = S^{-1} \cdot B \cdot S$ with $D_A$ and $D_B$ are diagonal matrices.
Statement: $A, B \in \mathbb{R}^{n \times n}$ are simultaneous diagonalizable $\Rightarrow A \cdot B = B \cdot A$
Proof: As A and B are simultaneous diagonalizable, a matrix $T \in \mathbb{R}^{n \times n}$ exists, such that $D_A = S^{-1} \cdot A \cdot S$ and $D_B = S^{-1} \cdot B \cdot S$ with $D_A$ and $D_B$ are diagonal matrices.
\begin{align} \Rightarrow A \cdot B &= S \cdot D_A 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 \blacksquare \end{align}
Statement: $A \cdot B = B \cdot A \nRightarrow A, B \in \mathbb{R}^{n \times n}$ are simultaneous diagonalizable.
Proof: by Counter-Example $$\begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix} \cdot \begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix} \cdot \begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$$ but \begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix} is not diagonalizable. $\blacksquare$
See also
- When is matrix multiplication commutative? on math.stackexchange.com