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Problem:

Prove that the commutativity on the set of non-scalar $2\times2$ matrices is an equivalence relation. (That is, for all A; B; and C; if AB = BA and BC = CB then AC = CA:)

For commutativity to be equivalence relation, we have to it is reflexive, symmetric and transitive. The first two properties are obvious. Any help on how to prove the third property?

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What do you mean by non-scalar? Did you want to say nonsingular (=invertible)? en.wikipedia.org/wiki/Invertible_matrix –  Martin Sleziak Nov 20 '11 at 16:42
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I mean by scalar matrix a matrix that is a constant multiple of the identity matrix –  M.Krov Nov 20 '11 at 16:46
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@Martin, it's not true under that interpretation because $I$ is nonsingular and commutes with everything, but not everything commutes. –  Henning Makholm Nov 20 '11 at 16:47
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@Zi2018Alpha, I think the critical property is that a $2\times 2$ matrix has two different eigenvalues iff it is not a multiple of $I$. Is that hint enough? (You can work in $\mathbb C$ establishing your identity even if you're only interested in real matrices at the end of the day). –  Henning Makholm Nov 20 '11 at 16:57
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Henning describes the major case. I think that the other case of commuting 2x2-matrices with repeated eigenvalues needs its own treatment, but is easier in a way. –  Jyrki Lahtonen Nov 20 '11 at 17:14
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1 Answer 1

This is a write-up of the comments adding some thoughts that came to my mind:

  1. The first step is to show that $A$ and $B$ commute iff $S^{-1}AS$ and $S^{-1}BS$ commute. Therefore it follows that for the proof of transitivity, i.e. that we have $A$, $B$ and $C$ with $A$ and $B$ commuting and $B$ and $C$ commuting, we can assume that $B$ is in Jordan normal form. Now because we look at two cases: 2./3. (since $B$ is not a scalar multiple of the identity.
  2. The matrix $B$ is a diagonal matrix with two distinct eigenvalues. In this case, as the OP already remarked it follows that also $A$ and $C$ are diagonal matrices and hence they commute.
  3. The matrix $B$ consists of one Jordan block to a single eigenvalue. In this case, both $A$ and $C$ also have to be Jordan blocks with a single eigenvalue and one checks by direct computation that these commute.
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