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I'm tasked with showing that matrix $A$ commutes with every $2\times2$ matrix if and only if $A = \begin{bmatrix}a & 0\\0 & a\end{bmatrix}$ for some a.

I was able to prove in the first direction, assuming that $A = \begin{bmatrix}a & 0\\0 & a\end{bmatrix}$, by just multiplying the matrix by $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ in AB = BA format, and showing how AB does equal BA.

$$ (=>)$$

$$\begin{bmatrix}a & 0\\0 & a\end{bmatrix}\begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}a & 0\\0 & a\end{bmatrix}$$

$$\begin{bmatrix}a^2 & ab\\ac & ad\end{bmatrix} = \begin{bmatrix}a^2 & ab\\ac & ad\end{bmatrix}$$

But now I have to prove in the other direction, assuming that A already does commute with every $2\times 2$ matrix, and I'm not particularly sure how I can do that. Somehow, I have to show that if A commutes with $\begin{bmatrix}a & b\\c & d\end{bmatrix}$, then A will be $\begin{bmatrix}a & 0\\0 & a\end{bmatrix}$, and I don't know how I can carry that out.

Where can I start? Ideas are appreciated.

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Your proof of the first direction is flawed, since choosing the same variable name $a$ in both matrices causes Loss Of Generality. With different names the argument is not really harder. –  Marc van Leeuwen Sep 19 '12 at 16:42
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2 Answers

up vote 2 down vote accepted

Consider the elementary matrices:

$$ \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix},\quad \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix},\quad \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix},\quad \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix} $$ What do you get when you force $A$ to commute with each of them?

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In fact, a matrix commutes with every matrix iff it commutes with the elementary matrices, because the elementary matrices form a basis for the space of matrices. –  lhf Sep 19 '12 at 16:52
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For the opposite direction, it helps to write down the fact that you matrix commutes with particular matrices, which you can choose as simple as you like, so that the computations won't be too hard. With a few good choices you may get to your goal, as you need to establish only $3$ equations in the $4$ unknowns $a,b,c,d$ (namely $a=d$, $b=0$, and $c=0$). You might find a very small set of matrices such that any matrix commuting with them automatically commutes with everyone (because the former condition forces them to be of the form $$ \begin{pmatrix}a&0\\0&a\end{pmatrix}$$). Your particluar matrix should of course not be so simple as to be the null matrix, or more generally a multiple of the identity matrix, since you know that those commute with everyone, so the won't give you any useful equations. But there are some very simple choices left.

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