I have this question:
In general, the matrix multiplication is not commutative. Suppose that the matrix $A = (a_{ij})_{2\times2}$ is in the form:
$$ A=\begin{bmatrix}1&0\\-1&0\\\end{bmatrix} $$
Find the general matrix $X = (x_{ij})_{2\times2}$ such that
$$ AX=XA $$
So, in this case, is the vector $X$ simply the same as the vector $A$? or is vector $X$ the same as vector $A$ multiplied by vector $A$ (which comes out to be just vector $A$)? Or is it not that simple and it requires me to work it out?
$\endgroup$ 11 Answer
$\begingroup$$$\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} \begin{bmatrix} 1& 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1& 0 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}$$
$$\begin{bmatrix} x_{11}-x_{12} & 0 \\ x_{21}-x_{22} & 0 \end{bmatrix} = \begin{bmatrix} x_{11} & x_{12} \\ -x_{11} & -x_{12} \end{bmatrix}$$
$$x_{11}-x_{12}=x_{11} \implies x_{12} =0$$$$x_{21}-x_{22}=-x_{11} \implies x_{21}=x_{22}-x_{11}$$
This tells us that knowing $x_{11}$ and $x_{22}$ is sufficient for determining $X$. Suppose that $x_{11} = a$ and $x_{22}=b$:
$$X = \begin{bmatrix} a & 0 \\ b-a & b \end{bmatrix}$$
is the general solution for any $a,b \in \mathbb{R}$. If the matrix $X$ is not of this form, $A$ and $X$ won't commute with each other.
$\endgroup$ 3
