2
$\begingroup$

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?

  • 0
    If$I$am not mistaken, what you guys mean is to consider the two cases: First case: where B=\begin{pmatrix} \lambda_{1} &0 \\ 0& \lambda_{2} \end{pmatrix} and in this case A and C have to diagonal matrices, and sure enough diagonal matrices commute with each other. Second case: where B=\begin{pmatrix} \lambda &1 \\ 0& \lambda \end{pmatrix} and in this case too,$I$can find that A and C commute with each other. Is this proof correct?2011-11-20

1 Answers 1

2

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.