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?

  • $\begingroup$ 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 $e_ie_j^t$. $\endgroup$ – bgins May 20 '12 at 13:47
  • 1
    $\begingroup$ Use the four elementary matrices as X to deduce relations between the coefficients of M. $\endgroup$ – Did May 20 '12 at 13:47
  • $\begingroup$ The same result is true for $n \times n$ matrices. $\endgroup$ – Mikko Korhonen May 20 '12 at 13:47
  • $\begingroup$ This is a near-duplicate of question math.stackexchange.com/questions/142967/… (with $3$ replaced by $2$). $\endgroup$ – 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 $M$ .

| cite | improve this answer | |

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 $m\in K$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.