I know that the converse is true; that is, if A and B are similar matrices, then $A^2$ and $B^2$ are similar . However, I'm not sure about the reverse.
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26Consider $1\times 1$ matrices $A = [1]$ and $B = [-1]$ :) – Antoine Apr 21 '15 at 21:36
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No: consider $$ A=\begin{bmatrix}1&0\\0&1\end{bmatrix},\qquad B=\begin{bmatrix}1&0\\0&-1\end{bmatrix} $$
egreg
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1Well you used one more dimension than you needed to, but I agree that's not good cause for a downvote... – R.. GitHub STOP HELPING ICE Apr 22 '15 at 07:35
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6@R.. Of course dimension one would suffice, but I felt it wouldn't be a clear example to the OP. – egreg Apr 22 '15 at 07:54
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A little more dramatic, should you want another example:
$A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\qquad B=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
Ethan Bolker
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