If $A^2$ is diagonalizable, is it necessary true that $A$ is diagonalizable?
Also, the opposite: If $A$ is diagonalizable, is it necessary true that $A^2$ is diagonalizable?
I'm not sure yet, tried to prove with $A^2 = P^{-1}DP$ , but no success.
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5See this for the first question. For the second, if $A=P^{-1}DP$, what is $(P^{-1}DP)(P^{-1}DP)$? – David Mitra Jul 04 '14 at 22:04
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The first proposition is false, as for example $A^2 = 0$ for nonzero (nilpotent) $A$.
For example: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
Of course if $A$ is diagonalizable, then $A^2$ (and indeed any polynomial in $A$) is also diagonalizable: $D = P^{-1} A P$ diagonal implies $D^2 = P^{-1} A^2 P$.
hardmath
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Why? I thought about saying: if $A^2$ is diagonalizable, that it might has a characteristic polynomial which won't be the same characteristic polynomial for $A$. – Ilan Aizelman WS Jul 04 '14 at 22:03
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Still didn't get the answer, feeling stupid. if $A^2$ is diagonalizable and it equals to zero, how can $A$ be the matrix u just posted? – Ilan Aizelman WS Jul 04 '14 at 22:11
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It is actually true iif $A^2$ has only positive eigenvalues.
If $\pi$ is a polynomial that factors completely into distinct factors (hope I'm getting this right in english, please correct otherwise) and for which $\pi(A^2)=0$ (such a polynomial exists because $A^2$ is diagonalizable), ie : $$\pi(X)=\prod_i (X-\lambda_i)$$ with distincts $\lambda_i>0$, and $\pi(A^2)=\prod_i (A^2-\lambda_i I)=0$
Then $P(X)=\pi(X^2)=\prod_i (X^2-\lambda_i)=\prod_i (X-\sqrt{\lambda_i})(X+\sqrt{\lambda_i})$ is also completely factored into distinct factors and verify $P(A)=0$ which imply with kernel lemma that $A$ is also diagonalizable.
The other side is trivial as mentionned by hardmath
Bertrand R
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@enzotib Yes, that's a bad habit I have. I'll correct it for the sake of compleness). Anyway, the other answer was accepted – Bertrand R Jul 04 '14 at 22:17
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I upvoted this, not only because the circumstances in which the proposition holds are worth articulating, but because the method of proof points to another class of "counterexamples" than the one I gave. The trick here is that the real numbers give square roots of all the positive eigenvalues. If the field we work over lacks square roots for an eigenvalue, say $\mathbb{Q}$ lacking $\sqrt{2}$, we can construct say $A=\begin{pmatrix}0&2\1&0\end{pmatrix}$, which cannot be diagonalized simply because the diagonal entries don't exist in the field. – hardmath Jul 05 '14 at 23:07
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Why positive? It seems that as long as there is no zero, we are good. – athanos lee Oct 26 '22 at 04:14
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@athanoslee: You are right over the reals, see this related Question. The typo "iif" makes it look like an "iff" was intended, but I don't think that's what Bertrand R meant. – hardmath Oct 26 '22 at 11:45