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I know that in general, the only matrices that are multiplicatively commutative are those that are scalar multiples of $I$, the identity matrix.

But what about matrices that are multiplicatively commutative with only invertible matrices? Is it any different? I don't think so, but I'm not certain, and am struggling to prove it.

Simply, with $A$ and $B$ both being $n\times n$ matrices over the reals, what are all $A$ such that $AB = BA$ if $B$ is invertible?

I suppose in group theory this could be phrased as the centre of the general linear group over the reals - $S(GL_n(\mathbb{R}))$.

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    $\begingroup$ A standard proof for the any-kind-of matrices is to consider $AE_{ij} = E_{ij}A$ where $E_{ij}$ has only one nonzero entry, 1, and it is in the $i$th row, $j$th column. Unfortunately $E_{ij}$ is not invertible. However, you can fix this easily by considering $B=I+E_{ij}$. Then $AB = A+AE_{ij}$ and $BA=A+E_{ij}A$ and these are equal iff $AE_{ij}=E_{ij}A$. $\endgroup$ – Jack Schmidt Jun 6 '13 at 18:21
  • $\begingroup$ This isn't true. Matrices that are similar to diagonal matrices are also multiplicatively commutative. $\endgroup$ – Ataraxia Jun 6 '13 at 18:22
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    $\begingroup$ @ZettaSuro: two diagonal matrices commute with each other, but a matrix only commutes with all other matrices if it is a scalar multiple of the identity. Two matrices that are similar to diagonal matrices need not commute with each other (they may not be simultaneously similar). $\endgroup$ – Jack Schmidt Jun 6 '13 at 18:35
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So I solved this using an approach similar to what Jack Schmidt described in the comments to the question.

I used the approach of $E_{i,j}$ being the matrix with 1 at position (i, j) and 0 elsewhere. Then $(I+E_{i,j})A = A(I+E_{i,j})$ iff $E_{i,j}A = AE_{i,j}$. Thus $A$ must be diagonal.

Now I defined the matrix $P_{i,j}$ to be the identity matrix with rows i and j swapped. Now if $A$ is diagonal with $a_{i,i} \neq a_{j,j}$ then $P_{i,j}A \neq AP_{i,j}$ and so $A$ must be $xI$ for some $x \in \mathbb{R}$. If $x = 0$ then $A \neq GL_n(\mathbb{R})$. Also, $AB = xIB = BxI = BA$ so then $Z(GL_n(\mathbb{R})) = \{xI \mid x \in \mathbb{R}, x \neq 0\}$.

Again, big thanks to Jack Schmidt.

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  • $\begingroup$ The $E_{ij}$ suffice to prove that $A$ is scalar. You don't need to add the transpositions. $\endgroup$ – Julien Jun 7 '13 at 2:23
  • $\begingroup$ @julien How does that suffice? Then you only get that A is diagonal but not necessarily scalar. Nothing stops the entries along the diagonal from being different. $\endgroup$ – michek Feb 11 '14 at 16:18
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I think you can use the Schur's lemma in representation theory and then it will be easy.

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  • $\begingroup$ In theory this works. But you need to first show that the module in question is irreducible. That is, of course, kinda trivial. Ok, swatting flies with cannonballs is fun - at least in moderation :-) The bad news is that some users think this is not an answer. I guess one can say that your answer requires quite a bit more than necessary from the reader. Which, in the same vein, is not necessarily a bad thing. $\endgroup$ – Jyrki Lahtonen Dec 11 '15 at 4:42
  • $\begingroup$ @ Jyrki Lahtonen I find this question by chance , and I think we can solve this question by Schur's lemma easily. You say that is swatting flies with cannonballs, I can only say: OK.... I just want to provide another way to solve this question. $\endgroup$ – Xiaosong Peng Dec 11 '15 at 7:35
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If $R$ is a unital ring, the center of $GL_n(R)$ is the set of scalar matrices whose scalar is a central invertible element in the ring. In the case $R=\mathbb{R}$, we see that the center is all nonzero scalar matrices.

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