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I am stuck on a seemingly simple problem: if $\mathbf{M},\mathbf{N}$ are $n\times n$ and have all eigenvectors in common, then $\mathbf{MN}=\mathbf{NM}$. I can prove this if they are diagonalisable, but I am stuck on the general case. I am probably being daft, but I run into a problem when the eigenbasis does not have dimension $n$. Any help?

uuuh
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  • Basis of eigenvectors means diagonalisable. – Daniel Fischer Sep 04 '14 at 23:34
  • @DanielFischer Sorry, mistake on my part; I meant that they both have the same set of eigenvectors – uuuh Sep 04 '14 at 23:37
  • Well, if they are not diagonalisable, then they need not commute. – Daniel Fischer Sep 04 '14 at 23:38
  • @DanielFischer Ah that would explain my frustration. Are there simple examples of, say $2\times 2$ non-diagonalisable matrices that share eigenvectors but fail to commute? – uuuh Sep 04 '14 at 23:40
  • $2\times 2$ doesn't work (at least for real matrices), I think. But $4\times 4$: $$\begin{pmatrix} a & -b & 0 & 0\ b & a & 0 & 0\ 0 & 0 & c & -d\ 0 & 0 & d & c \end{pmatrix}; \qquad \begin{pmatrix} a & 0 & -b & 0 \ 0 & c & 0 & -d \ b & 0 & a & 0 \ 0 & d & 0 & c\end{pmatrix}$$ with generic nonzero $a,b,c,d$. – Daniel Fischer Sep 04 '14 at 23:45
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    @DanielFischer Wait sorry, I don't get the same eigenvectors. Compare here and here. Is there a mistake? – uuuh Sep 04 '14 at 23:56
  • Related. – Git Gud Sep 05 '14 at 00:39
  • I intended the matrices to be real. Generically, neither of the two matrices has any (real) eigenvectors (for unfortunate choices of $a,b,c,d$, they might have). – Daniel Fischer Sep 05 '14 at 10:59

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As mentioned, this is not true in the general (non-diagonalizable) case. To see why, note that for any $A$ and invertible $S$, the eigenvectors of $SAS^{-1}$ are $\{Sv \mid v \text{ eigenvector of } A\}$. If $A$ does not have a basis of eigenvectors, then $S$ may permute eigenvectors of $A$, but act freely on the remaining vectors in a basis, so that $SAS^{-1}$ does not commute with $A$. For an explicit counterexample, take

$$A = \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}, \; S = \begin{bmatrix} 0 & 2 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}, \; B = SAS^{-1} = \begin{bmatrix} 2 & -2 & 0\\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$$

Then $A, B$ have the same set of eigenvectors, namely (multiples of) $\begin{bmatrix}1 \\ 0 \\0\end{bmatrix}$ and $\begin{bmatrix}0 \\ 0 \\1\end{bmatrix}$, but $AB = \!\begin{bmatrix} 2 & -1 & 0\\ 0 & 1 & 0 \\ 0 & 2 & 2 \end{bmatrix} \ne \!\begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 1 & 2 \end{bmatrix} = BA$.

A. J. Pan-Collantes
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zcn
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