6

I feel like this is probably a simple proof but I can't quite come up with it in an elegant way nor could I find it here.

Prove that if a matrix $M$ commutes with any matrix then $M$ is of the form $M=\alpha I$.

Proving the contrapositive seems like the natural way to go where we can logically transform $\lnot \forall A(MA = AM)$ into $\exists A (MA \neq AM)$ but assuming that $M \neq \alpha I$ immediately becomes messy. Is there a nice way out of this or is it inevitably going to get messy?

kyp4
  • 720
  • As per the answer below by Dane, this problem is easier if you step away from formal logic and try to get back to concrete things like how matrix multiplication works. Sometimes that happens! – grantfgates May 15 '14 at 01:27
  • I'd be interested in a more interesting solution than the ones below. I feel like one should exist. – abnry May 15 '14 at 01:37

4 Answers4

6

Here is a coordinate-free proof. For any nonzero vector $x$, let $B$ be a linear map whose null space is spanned by $x$. Then $Ax\in\ker B$ because $B(Ax)=ABx=0$. As $\ker B$ is one-dimensional, $Ax=\lambda_xx$ for some scalar $\lambda_x$ that may depend on $x$.

Now, given any two nonzero vectors $x$ and $y$, if they are linearly dependent, then $y=kx$ for some scalar $k$. It follows that $\lambda_xy=\lambda_xkx=k\lambda_xx=kAx=Ay=\lambda_yy$ and hence $\lambda_x=\lambda_y$.

If $x$ and $y$ are linearly independent instead, then $0=Ax+Ay-A(x+y)=\lambda_xx+\lambda_yy-\lambda_{x+y}(x+y)$. By linear independence, we must have $\lambda_x=\lambda_{x+y}=\lambda_y$.

In other words, the factor $\lambda_x$ is identical for every nonzero vector $x$, i.e. there exist some $\lambda$ such that $Ax=\lambda x$ for every $x$, meaning that $A=\lambda I$.

user1551
  • 139,064
  • Very nice argument! I suppose someone might want a lemma establishing existence of such a $B$. – Ted Shifrin May 15 '14 at 04:40
  • I enjoyed this argument as well. – abnry May 15 '14 at 06:35
  • 2
    Alternatively, after the first paragraph, using the basis vectors $e_i$ we can immediately conclude $A$ is diagonal. Then use the fact that $A \bf{1} = \lambda \bf{1}$ to conclude each diagonal element is the same. – abnry May 15 '14 at 06:37
4

We can do this by inspection. Suppose $A$ commutes with every matrix. Let $E_{ij}$ have all entries zero except the $(i,j)$-th entry, which is $1$ . Then $$\delta_{\ell j}a_{ki}=[AE_{ij}]_{k\ell}=[E_{ij}A]_{k\ell}=\delta_{ki}a_{j\ell}$$ so if $\ell\ne j$ setting $k=i$ gives us $a_{j\ell}=0$, hence $A$ is diagonal. If $\ell=j$ and $k=i$ we get $a_{ii}=a_{jj}$, hence $A$ is scalar.

Alex Becker
  • 60,569
3

Here is a hint:

Let $E_{ij}$ be the matrix with a one in the $ij$th position and zeros everywhere else. Write down what it means for $$ M E_{ij} = E_{ij} M $$ to hold, in terms of the entries of $M$.

Cardboard Box
  • 4,997
2

Here's somewhat of an overkill answer for what it is worth.

A normal matrix is a matrix that is unitarily similar to a diagonal matrix. Another characterization is that a matrix $M$ is normal iff $M^* M = M M^*$.

If $M$ commutes with all matrices then it is clear it is normal. From this we have $M = UDU^*$ for some unitary matrix $U$. Write $MU = UDU^* U = UD$ and then use commutativity to write $UM = UD$. Finally cancel the $U$'s to conclude that $M$ must be diagonal.

Then if $P$ is the elementary matrix that swaps rows $i$ and $j$ when applied to $M$ as $PM$, since $MP$ swaps columns and is the same as $MP$, we conclude each diagonal element is the same.

abnry
  • 14,664