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If $ A^3=A $ holds for the real valued matrix $A$, what can we conclude about its eigenvalues?

I thought that if I subtracted from $A$ on both sides, I would get $ A^3-A=0 $ which factors as $ A(A-1)(A+1)=0 $.
Can I then correctly conclude by the Cayley-Hamilton theorem that since the characteristic polynomial would be $ p(x)= x(x-1)(x+1)=0 \rightarrow p(A)=0?$
Would that prove that the eigenvalues of the matrix would be $\alpha_k=0,1,-1?$
Thank you for any input!

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    $\begingroup$ Since $A$ annihilates $X^3-1$, its minimal polynomial divides $X^3-1$, hence its roots are among $\{0,1,-1\}$. Since the characteristic polynomial and the minimal polynomial share the same set of roots, this means the eigenvalues are among (a subset of) $\{0,1,-1\}$. $\endgroup$ – Gabriel Romon Oct 30 '17 at 10:24
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    $\begingroup$ @GabrielRomon This looks like heavy artillery to describe the fact if $\lambda$ is an eigenvalue of $A$ and $v$ an eigenvector, then $\lambda v = Av = A^3v= \lambda^3 v$ and so $\lambda = \lambda^3 \in\{-1,0,1\}$. $\endgroup$ – Surb Oct 30 '17 at 10:27
  • $\begingroup$ Is there a general theorem explaining this that I could refer to for greater understanding? $\endgroup$ – Jae Kim Oct 30 '17 at 10:27
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    $\begingroup$ @Surb you're right ! I just wanted to write a proof along the lines of what the OP has (wrongfully) written. $\endgroup$ – Gabriel Romon Oct 30 '17 at 10:33
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    $\begingroup$ @Surb that should be an answer. $\endgroup$ – Randall Oct 30 '17 at 11:37
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Note that for any matrix $B$ with eigenpair $(\alpha, w)$, i.e. $Bw=\alpha w$, it holds $$B^3w = B^2\alpha w =B\alpha^2w = \alpha^3 w.$$ Hence, if $Av=\lambda v$ with $v\neq 0$ and $A^3 =A$, we obtain $$\lambda v = Av = A^3v = \lambda^3v,$$ implying that $\lambda = \lambda^3$ and so $\lambda \in\{-1,0,1\}$.

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    $\begingroup$ Thank you for actually writing out your answer which helped me to better understand. $\endgroup$ – Jae Kim Oct 30 '17 at 20:00
  • $\begingroup$ @JaeKim You're very welcome. By the way, you might be interested to note that if $Bw=\alpha w$ then $B^kw = \alpha^k w$ for any integer $k$ (of course assuming $B$ is invertible and $\alpha\neq 0$ when $k<0$) and the same still hold for any polynomial, i.e. if $f(x)=\sum_{k=-m_1}^{m_2} c_kx^k$, then $f(B)w = f(\alpha)w$. $\endgroup$ – Surb Oct 30 '17 at 20:29

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