All matrices which commute with all $2\times 2$ matrices

Question

I would like to find all matrices which commute with all $2\times2$ matrices. I started solving problem in this way:

1) I have this matrix $A$ with real numbers:
$$A=\left[\begin{array}{cc}a &b\\c &d\end{array}\right]$$
2) Matrix which commute with matrix $A$ is matrix $B$:
$$B = \left[\begin{array}{cc}e& f\\ g &h\end{array}\right]$$
3) When i solve the equation $AB=BA$, I get this:
$$\left[\begin{array}{cc}ae+bg&af+bh\\ce+dg&cf+dh\end{array}\right]=\left[\begin{array}{cc}ea+cf&eb+df\\ga+ch&gb+dh\end{array}\right]$$

How I can get the general look of wanted matrix?

Answer 1

what are the $2 \times 2$ matrices $\pmatrix{a&b\\c&d}$ that commute with $\pmatrix{0&1\\0&0}?$

$$ \pmatrix{0&a\\0&c} = \pmatrix{a&b\\c&d} \pmatrix{0&1\\0&0} = \pmatrix{0&1\\0&0}\pmatrix{a&b\\c&d}=\pmatrix{c&d\\0&0}$$ implies we need $$ a= d, c = 0$$

by considering the matrix $\pmatrix{0&0\\1&0}$ you will find $$ a= d, b = 0$$

the matrix of the form $$\pmatrix{a&0\\0&a}$$ called the scalar matrix is the matrix that commutes with all $2 \times 2$ matrices.

Answer 2

The set of $2\times 2$ matrices with non-zero determinant form a group under the operation of matrix multiplication. For any group $G$ the centre of $G$, denoted $Z(G)$, is defined as $$Z(G)=\{ z \in G : zg = gz \ \ \text{for all} \ \ g \in G \}$$

We use the letter $Z$ because "zentrum" is "centre" in German. This is exactly what you want: you want the $2\times 2$ matrices $z$ for which, for any $2\times 2$ matrix $g$, you have $zg=gz$.

The group of $n\times n$ matrices with real entries is called the general linear group and is denoted by $\mathrm{GL}(n,\mathbb{R})$. In your case, you have $2\times 2$ matrices, and so $n=2$.

It turns out that the centre of $\mathrm{GL}(n,\mathbb{R})$, for all $n$, is given by the matrices which are a multiple of the identity matrix. The identity matrix is the identity element of the group. In the case $n=2$ $$Z(G) = \left\{ \left(\begin{array}{cc} k & 0 \\ 0 & k \end{array}\right) : k \in \mathbb{R}, \ k \neq 0 \right\}$$

If you don't want a group structure then you can include $k=0$.

Edit: Request for proof

The proof can be found by Googling "Centre of the General Linear Group".

Answer 3

we can try four different scenarios with matrices that have 3 zero entires. We can assumed $\mathbf{A}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. We can then try $\mathbf{B}=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. Solving for $a,b,c$ & $d$ we get $\mathbf{AB}=\mathbf{BA}$ when $c=0$ and $b=0$. We can then solve for $\mathbf{B}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ which gives us $\mathbf{AB}=\mathbf{BA}$ when $b=0$ and $c=0$. For $\mathbf{B}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ we get $\mathbf{AB}=\mathbf{BA}$ when $c=0$ and when $a=d$. The last equation, $\mathbf{B}=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, give us $\mathbf{AB}=\mathbf{BA}$ when $b=0$, $c=0$, $a=d$.

These equations imply that $a=d$ and $b=c=0$. this means that $\mathbf{A}=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$ is the form for all possible solutions for $A$ in which every matrix of the form found commute with all matrices $\bold{B}$.