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Verify: $A\in\mathbb C^{2\times 2}$ be such that $A^3=A.$ Then the number of such $A$ is infinite.

My attempt: Choose a nonsingular $P\in\mathbb C^{2\times 2}.$ And let $A=P\begin{pmatrix}1&0\\0&-1\end{pmatrix}P^{-1}.$ Then $A^3=A.$ So infinite such $A$ can occur.

Am I correct?

Sriti Mallick
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  • Can you generalize to other polynomials $p(A)=0$? – lhf May 22 '13 at 12:36
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    You did not prove that $PAP^{-1} \neq QAQ^{-1}$ for infinitely many $(P,Q)$. – Seirios May 22 '13 at 12:37
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    This question has answer here http://math.stackexchange.com/questions/243161/are-there-infinitely-many-a-in-mathbbc2-times-2-satisfying-a3-a?rq=1 – Srijan May 22 '13 at 12:51
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    How about considering $A=\begin{pmatrix}1&0\r&0\end{pmatrix},r\in\mathbb R?$ – Sriti Mallick May 22 '13 at 12:51
  • That would be simpler and clearer. – user1551 May 22 '13 at 12:53
  • The form of this question is problematic. There seems to be "Let" missing at the beginning. Moreover this formulation means that $A$ is given, and asking for their number is meaningless. What you appear to ask is "Show that the number of $A\in\ldots$ such that $A^3=A$ is infinite". – Marc van Leeuwen May 22 '13 at 13:03

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