Example 14: $AB\ne BA$ for two $2\times2$ matrices

Basics you need

Solution

1. Both products are defined and both are $2\times2$

   $A$ and $B$ are square of the same order 2. For $AB$⁠: $(2\times2)(2\times2)$⁠, inner sizes equal, result $2\times2$⁠. For $BA$⁠: the same. So $AB$ and $BA$ exist and have the same order.

   This is the case the textbook raises after Example 13: "perhaps AB and BA could be the same if they were of the same order". We now test that guess.
   If $A$ and $B$ are square matrices of the same order, both $AB$ and $BA$ are defined (Remark after Example 12). Basics: rows, columns and the address

2. Compute $AB$

   Rows of $A$⁠: $[1\ \ 0]$ and $[0\ \ {-1}]$⁠. Columns of $B$⁠: $(0,1)$ and $(1,0)$⁠.

   $$\begin{aligned} c_{11} &= 1\cdot 0 + 0\cdot 1 = 0\\ c_{12} &= 1\cdot 1 + 0\cdot 0 = 1\\ c_{21} &= 0\cdot 0 + (-1)\cdot 1 = -1\\ c_{22} &= 0\cdot 1 + (-1)\cdot 0 = 0 \end{aligned}$$ $$AB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$
   Multiplying $B$ on the left by $A$ keeps row 1 of $B$ and changes the sign of row 2: $A$ multiplies the first row by 1 and the second row by $-1$⁠. Row 2 of $B$ is $[1\ \ 0]$⁠, so it becomes $[-1\ \ 0]$⁠.

3. Compute $BA$

   Rows of $B$⁠: $[0\ \ 1]$ and $[1\ \ 0]$⁠. Columns of $A$⁠: $(1,0)$ and $(0,-1)$⁠. Write $BA=[d_{ik}]$⁠:

   $$\begin{aligned} d_{11} &= 0\cdot 1 + 1\cdot 0 = 0\\ d_{12} &= 0\cdot 0 + 1\cdot(-1) = -1\\ d_{21} &= 1\cdot 1 + 0\cdot 0 = 1\\ d_{22} &= 1\cdot 0 + 0\cdot(-1) = 0 \end{aligned}$$ $$BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$
   Multiplying by $A$ on the right acts on columns instead: it keeps column 1 of $B$ and changes the sign of column 2. Column 2 of $B$ is $(1,0)$⁠, which becomes $(-1,0)$⁠. Acting on rows and acting on columns are different operations, which is the root of $AB\ne BA$⁠.

4. Compare entry by entry

   Both are $2\times2$⁠, so we compare corresponding entries. The $(1,2)$ entries are $1$ in $AB$ and $-1$ in $BA$⁠. They are different, so $AB\ne BA$.

   In fact the $(2,1)$ entries differ too ($-1$ against $1$⁠), and every entry of $BA$ is the negative of the matching entry of $AB$⁠: $BA=-AB$⁠.

   Equality needs all corresponding entries equal, so a single mismatch is a complete proof of inequality. Matching orders did not rescue commutativity: matrix multiplication is not commutative, even for square matrices of the same order.
   Definition 2, Section 3.3.1: equal matrices have the same order and equal corresponding entries. Lesson: equality of matrices

5. The picture behind the numbers (beyond the textbook)

   This step is not in the textbook's solution and is not needed for the answer; it shows what the two matrices do. Multiply each matrix by a column $(x,y)$⁠:

   $$A\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix}, \qquad B\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ x \end{bmatrix}.$$

   So $A$ reflects a point in the $x$-axis and $B$ reflects it in the line $y=x$⁠. $ABP=A(BP)$ means "apply $B$ first, then $A$", and $BAP=B(AP)$ means "apply $A$ first, then $B$". For $P=(2,1)$⁠:

   $B$ first, then $A$⁠:

   $$(2,1)\to(1,2)\to(1,-2)$$

   $A$ first, then $B$⁠:

   $$(2,1)\to(2,-1)\to(-1,2)$$
   The two routes end at different points, so the two products cannot be the same matrix. Check with the matrices of Steps 2 and 3: $AB$ sends $(2,1)$ to $(0\cdot2+1\cdot1,\ -1\cdot2+0\cdot1)=(1,-2)$⁠, and $BA$ sends it to $(0\cdot2-1\cdot1,\ 1\cdot2+0\cdot1)=(-1,2)$⁠. The grouping $A(BP)=(AB)P$ is the associative law of Section 3.4.6.
   Writing $AB$ means $B$ acts first: the factor nearest the column acts on it first. Lesson: properties of the operations

6. Not every pair fails: the textbook's note

   Non-commutativity means $AB=BA$ is not guaranteed, not that it always fails. The textbook's Note gives diagonal matrices:

   $$\begin{aligned} &\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix},\\ &\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix}. \end{aligned}$$

   Entry by entry, the first product is $1\cdot3+0\cdot0=3$⁠, $1\cdot0+0\cdot4=0$⁠, $0\cdot3+2\cdot0=0$⁠, $0\cdot0+2\cdot4=8$⁠; the second is $3\cdot1+0\cdot0=3$⁠, $3\cdot0+0\cdot2=0$⁠, $0\cdot1+4\cdot0=0$⁠, $0\cdot0+4\cdot2=8$⁠.

   For diagonal matrices of the same order, each product just multiplies the diagonal entries in matching places: $1\cdot3$ and $2\cdot4$⁠. Multiplication of numbers is commutative, so the products agree. Our $B$ is not diagonal, so the Note does not apply to the pair $A$⁠, $B$ of this example. (Being non-diagonal does not by itself force $AB\ne BA$⁠: for instance every square matrix commutes with itself. For our pair, Step 4 is what proves it.)

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