# Commutative matrices are multiples of the identity

Assume M is a $2 \times 2$ real matrix such that $MX = XM$ for all real $2\times2$ matrices $X$. Show that $M$ must be some real multiple $q$ of $I$.

I can see that this is logical and have tried a few examples where I have multiplied $qI$ by some random $2\times2$ matrix from both sides and get the same matrix in both cases. How would I go about showing this though? Surely a few examples aren't actually showing anything for the general case, right?

Also, does the property shown in the question hold for all square matrices, or just $2\times2$ ones?

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Here's an answer for the $3\times3$ (and $n\times n$) case, where in an epilogue, you can also see what Didier means by elementary matrices, where I call them $e_ie_j^t$. – bgins May 20 '12 at 13:47
Use the four elementary matrices as X to deduce relations between the coefficients of M. – Did May 20 '12 at 13:47
The same result is true for $n \times n$ matrices. – Mikko Korhonen May 20 '12 at 13:47
This is a near-duplicate of question math.stackexchange.com/questions/142967/… (with $3$ replaced by $2$). – Marc van Leeuwen May 20 '12 at 14:55

Say $\,M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ , and choose $\,X=\left(\begin{matrix}1&0\\0&0\end{matrix}\right)$ , so $MX=XM\Longrightarrow \left(\begin{matrix}a&0\\c&0\end{matrix}\right)=\left(\begin{matrix}a&b\\0 &0\end{matrix}\right)$ and you already got $\,b=c=0\,$ .

Choose now another appropiate matrix $X$ to get more conditions on the other entries of $M$ .

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Let $K$ be a field.

Let $E_{ij}\in M_{n\times n}$ denote the matrix with a $1$ in the $i$-th row and $j$-th column and $0$'s everywhere else. If $i,j,k,l\in\{1,...,n\}$ then you can calculate $E_{ij}E_{kl}$; it's $E_{il}$ if $j=k$ and is the $n\times n$ matrix with $0$'s everywhere if $j\neq k$.

Suppose $M=(m_{ij})\in M_{n\times n}(K)$ commutes with every matrix. Let $k,l\in\{1,...,n\}$ and consider:

$$M E_{kl} = (\sum_{i,j} m_{ij} E_{ij}) E_{kl} = \sum_i m_{ik}E_{il}$$ $$E_{kl}M = E_{kl}(\sum_{i,j} m_{ij} E_{ij}) = \sum_{j} m_{lj}E_{kj}$$

The first is a matrix with $0$'s everywhere but the $l$'th column, the second is a matrix with $0$'s everywhere but the $k$'th row. Letting $k,l$ vary, you can deduce that $M$ must be of the form $m\text{Id}_n$ for some $m\in K$.

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