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Let $A$ and $B$ be $2\times 2$ matrices such that $AB=BA$. Is there a short proof that they are simultaneously diagonalizable? I'm aware to the usual proof of simultaneously diagonalizability in general, but can this proof be shortened if we assume that the matrices are $2\times2$?

boaz
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1 Answers1

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Here is a counterexample. Let $$ A=B=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}. $$ Then $AB=BA=0$, but they are not simultaneously diagonalisable.

Dietrich Burde
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  • This seems to answer a different question than was intended. The theorem the OP was thinking about says that if two diagonalizable matrices commute, then they are simultaneously diagonalizable. Now it is good to point out that the word 'diagonalizable' is important, but the question if for the correct version of the theorem there is a quicker proof in the 2 by 2 case than in the general case is still interesting IMO – Vincent Mar 09 '24 at 21:47
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    @Vincent Yes, you are right. The intended question was probably what you say. On the other hand, the proof of simultaneous disgonalization is so easy in general, that it hardly can't be easier in this special case. You would have to "define" what you mean by "easier". Perhaps avoiding the term "eigenvalue" or "eigenvector"? But this would be unreasonable, in my opinion. For the general proof (with distinct eigenvalues) see here, for example. – Dietrich Burde Mar 10 '24 at 10:20
  • Yes I agree. One advantage you have in the 2 by 2 case is that either eigenvalues are distinct or the matrix is a scalar multiple of the identity. The distinct eigenvalue case is easier that the general case, I believe, because eigenvectors are a familiar concept you have probably heard about before when you learn this theorem, while the general case only becomes as easy (or identical) once someone introduces the concept of 'eigenspace' (or you think of it yourself). Coming up with the idea eigenspaces out of nowhere is not so easy, even if they look very natural in retrospect. – Vincent Mar 11 '24 at 14:31
  • In the 2x2 case you can just apply the reasoning from the linked question without worrying about multiplicities. But that is the only simplification I can see, so your comment is very much justified – Vincent Mar 11 '24 at 14:34