# Find All Matrices $B$ that Commutes With a Given Matrix $A$: $AB=BA$

## Problem 272

Let
$A=\begin{bmatrix} 1 & 3\\ 2& 4 \end{bmatrix}.$ Then

(a) Find all matrices
$B=\begin{bmatrix} x & y\\ z& w \end{bmatrix}$ such that $AB=BA$.

(b) Use the results of part (a) to exhibit $2\times 2$ matrices $B$ and $C$ such that
$AB=BA \text{ and } AC \neq CA.$

## Solution.

### Find all matrices $B$ such that $AB=BA$

Since we want to find $B$ such that $AB=BA$, we need to find $x, y, z, w$ satisfying
$\begin{bmatrix} 1 & 3\\ 2& 4 \end{bmatrix}\begin{bmatrix} x & y\\ z& w \end{bmatrix}=\begin{bmatrix} x & y\\ z& w \end{bmatrix}\begin{bmatrix} 1 & 3\\ 2& 4 \end{bmatrix}.$

Comparing entries, we have the following system of linear equations.
\begin{align*}
x+3z&=x+2y\\
y+3w&=3x+4y\\
2x+4z&=z+2w\\
2y+4w&=3z+4w.
\end{align*}

Simplifying this, we obtain
\begin{align*}
2y-3z &=0\\
3x+3y-3w &=0\\
2x+3z-2w &=0\\
2y-3z &=0.
\end{align*}

Note that the first and the last equations are identical. So we omit the first equation from our consideration.

To solve this system, we use the Gauss-Jordan elimination method. Namely we form the augmented matrix of this system and apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
3 & 3 & 0 & -3 &0 \\
2 & 0 & 3 & -2 & 0 \\
0 & 2 & -3 & 0 & 0
\end{array}\right] \xrightarrow{\frac{1}{3}R_1}
\left[\begin{array}{rrrr|r}
1 & 1 & 0 & -1 &0 \\
2 & 0 & 3 & -2 & 0 \\
0 & 2 & -3 & 0 & 0
\end{array}\right] \xrightarrow{R_2-2R_1}\10pt] \left[\begin{array}{rrrr|r} 1 & 1 & 0 & -1 &0 \\ 0 & -2 & 3 & 0 & 0 \\ 0 & 2 & -3 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_1+\frac{1}{2}R_2 \\ R_2+R_3}} \left[\begin{array}{rrrr|r} 1 & 0 & 3/2 & -1 &0 \\ 0 & -2 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\\[10pt] \xrightarrow{-\frac{1}{2}R_2} \left[\begin{array}{rrrr|r} 1 & 0 & 3/2 & -1 &0 \\ 0 & 1 & -3/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]. \end{align*} The last matrix is in reduced row echelon form. From this, we obtain the general solution \begin{align*} x&=-\frac{3}{2}z+w\\ y&=\frac{3}{2}z, \end{align*} where z and w are free variables. Therefore the matrix B such that AB=BA must be \begin{align*} B&=\begin{bmatrix} -\frac{3}{2}z+w & \frac{3}{2}z\\ z& w \end{bmatrix}\\ &=z\begin{bmatrix} -\frac{3}{2} & \frac{3}{2}\\ 1& 0 \end{bmatrix} +w \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix} \end{align*} for any z and w. ### (b) Find matrices B, C such that AB=BA, AC\neq CA By part (a), if \[B=z\begin{bmatrix} -\frac{3}{2} & \frac{3}{2}\\ 1& 0 \end{bmatrix} +w \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix} for some $z, w$, then we have $AB=BA$.

For example, let $z=2, w=0$. Then the matrix
$B=\begin{bmatrix} -3 & 3\\ 2& 0 \end{bmatrix}$ satisfies $AB=BA$.

To find a matrix $C$ such that $AC \neq CA$, the matrix $C$ must not be of the form of the formula of $B$.

For example, let
$C=\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}.$ You may directly check that $AC\neq CA$.

Or, we can show that $C$ is never be the matrix of the form
$\begin{bmatrix} -\frac{3}{2}z+w & \frac{3}{2}z\\ z& w \end{bmatrix}.$

To see this, compare $(1,2)$ and $(2,1)$ entries. There is no $z$ such that
$\frac{3}{2}z=0 \text{ and } z=1.$ (This is how I found the matrix $C$ as above.)

##### If a Matrix $A$ is Singular, There Exists Nonzero $B$ such that the Product $AB$ is the Zero Matrix
Let $A$ be an $n\times n$ singular matrix. Then prove that there exists a nonzero $n\times n$ matrix $B$ such...