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Let $A \in\mathrm{SL}(2,\mathbb{C})$ and $p > 1$ be a natural number. Under which conditions the following statement is true?

$A^p$ is a diagonal matrix implies $A$ is a diagonal matrix

For the case $p=2$ and $A^2 \neq \pm \mathrm{Id}$ it is a simple calculation to show that it is true.

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  • $\begingroup$ @alex.jordan, that's what I also thought. Deleting my comment. $\endgroup$ – Andreas Caranti Jun 1 '13 at 19:12
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Even when $p=2$ and $A^2\neq \pm I_2$, this is false. So I'm not sure what your computations were. Take $$ A=\pmatrix{0&\lambda\\1&0}\qquad \qquad A^2=\pmatrix{\lambda&0\\0&\lambda}. $$ As soon as $A^p$ has a repeated eigenvalue, we can build a counterexample. In other words, over $\mathbb{R}$ or $\mathbb{C}$, $\lambda I_m$ has a nondiagonal $p$th root for every $n\geq 2$ and $p\geq 2$. As mentioned by Georges Elencwajg and alex.jordan, it suffices to scale the appropriate rotation ($2\pi/p$, unless $p$ is even, $\lambda<0$, and you really want to do this in $\mathbb{R}$, in which case you should scale the rotation of angle $\pi/p$).

But if $A^p$ has no repeated eigenvalue, in $M_n(K)$ and for any $p\geq 2$, we have a sufficient condition.

If $A^p$ is diagonal with pairwise distinct diagonal coefficients, then $A$ is diagonal.

Proof: since $A$ commutes with $A^p$, it leaves the eigenspaces of $A^p$ invariant. By assumption, these are one-dimensional. So they must be eigenspaces for $A$ as well. QED.

More generally, if for some polynomial $p(X)\in K[X]$, $p(A)$ is diagonalizable with pairwise distinct coefficients, then $A$ is diagonalizable in the same basis.

Of course, all this falls apart as soon as $p(A)$ has an eigenvalue of multiplicity at least two.

Remark: it is good to think about this in terms of commutant. Given a diagonalizable matrix $B\in M_n(K)$, denote $C(B):=\{A\in M_n(K)\,;\,AB=BA\}$. This is a unital subalgebra of $M_n(K)$. When $B$ is in diagonalized form, $C(B)$ is block diagonal with a block $M_{n_j}(K)$ corresponding to each eigenvalue $\lambda_j$ of multiplicity $n_j$. So when all the eigenvalues of $B$ have multiplicity $1$, all these blocks reduce to scalars, and the commutant is the algebra all diagonal matrices in this basis. In your case, $A$ belongs to the commutant of $B=A^p$.

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    $\begingroup$ And conversely it can be shown that if $A^p$ is diagonal with two equal diagonal entries, that it is possible that $A$ is not diagonal using a dilated rotation. $\endgroup$ – alex.jordan Jun 1 '13 at 19:15
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The displayed statement is false for all $p\gt 2$, as proved by the rotation matrix of angle $2\pi/p$.

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  • $\begingroup$ I think you meant $p\geq 2$. $\endgroup$ – Julien Jun 1 '13 at 19:20
  • $\begingroup$ @julien: no, I meant what I wrote. In the case $p=2$ the rotation matrix of angle $2\pi/2$ is diagonal, so it doesn't prove anything about the displayed statement. $\endgroup$ – Georges Elencwajg Jun 1 '13 at 19:30
  • $\begingroup$ Right. I was thinking, in the case $p=2$, about the rotation of angle $\pi/2$, whose square is indeed diagonal as it is $-I_2$. So you get a generic couterexample for all $p\geq 2$ with the rotation of angle $\pi/p$. $\endgroup$ – Julien Jun 1 '13 at 19:35
  • $\begingroup$ @julien. Yes, your rotations would include the case $p=2$ and are thus more elegant. What happened is that since I thought the OP had settled the case $p=2$, I didn't really bother about that value. $\endgroup$ – Georges Elencwajg Jun 1 '13 at 19:47

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