Find two non-zero matrices B such that AB = BA

Question

Consider a $2\times 2$ matrix $$A:=\begin{pmatrix}2 & -2 \\-2 & -2\end{pmatrix}.$$

How would I work out two different non-zero matrices $B$ so that $AB = BA$?

At the moment I only have $$B=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}.$$

Answer 1

The most trivial answer is $B=\lambda I, \lambda A$. But I want to solve it.

Let $B$ as a matrix with arbitary components. So let $B=\begin{pmatrix}a & b \\c & d\end{pmatrix}$. Then $AB, BA$ will be like...

$$ AB=\begin{pmatrix}2a-2c & 2b-2d \\-2a-2c & -2b-2d\end{pmatrix}, BA=\begin{pmatrix}2a-2b & -2a-2b \\2c-2d & -2c-2d\end{pmatrix} $$

When comparing componentwise, we get $b=c=-\frac12(a-d)$.

So $B$ will be this form... $B=\begin{pmatrix}a & b \\b & a+2b\end{pmatrix}$, and it is, this form:

$$ B=\lambda I+\mu A $$

Answer 2

Let $B = \begin{bmatrix} a && b \\ c && d \end{bmatrix} $

Then

$\frac{1}{2} AB = \begin{bmatrix} a - c && b - d \\ - a - c && - b - d \end{bmatrix}$

and

$\frac{1}{2} BA = \begin{bmatrix} a - b && - a - b \\ c - d && -c - d \end{bmatrix}$

Hence, we want

$a - c = a - b \Rightarrow c = b $

$b - d = -a - b \Rightarrow 2 b = d - a $

$-a - c = c - d \Rightarrow 2 c = d - a $

$ -b - d = -c - d \Rightarrow c = b $

This means that we can choose $a$ and $d$ arbitrarily, then we will have $b = c = \frac{1}{2} (d - a) $

For example if $a = 1 , d = 1 $ then $b = c = 0$ which is the identity matrix.

If $a = 1 , d = 5 $ then $ b = c = 2 $

If $a = 10 , d = 2 $ then $b = c = -4 $

The general form of matrix $B$ is

$B = \begin{bmatrix} t && \frac{1}{2}(s - t) \\ \frac{1}{2} (s - t) && s \end{bmatrix} $

where $t, s \in \mathbb{R} $

If we let $\lambda = \frac{1}{2} (s + t) $ and $\mu = -\frac{1}{4} ( s - t ) $, then

$ B = \begin{bmatrix} \lambda + 2\mu && -2\mu \\ -2\mu && \lambda - 2\mu \end{bmatrix} = \lambda \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} + \mu \begin{bmatrix} 2 && - 2 \\ -2 && -2 \end{bmatrix} = \lambda I + \mu A $

Answer 3

To reduce the degrees of freedom, let's start by symmetric matrices. If you further assume that the diagonal entries in each matrix are equal, you can check that $$ A=\begin{pmatrix}a & b\\b & a \end{pmatrix}, \quad B=\begin{pmatrix}c & d\\d & c \end{pmatrix} $$

do the trick.