# Eigenvalues of a real matrix

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!

• 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) $\{0,1,-1\}$. – Gabriel Romon Oct 30 '17 at 10:24
• @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 $\lambda = \lambda^3 \in\{-1,0,1\}$. – Surb Oct 30 '17 at 10:27
• Is there a general theorem explaining this that I could refer to for greater understanding? – Jae Kim Oct 30 '17 at 10:27
• @Surb you're right ! I just wanted to write a proof along the lines of what the OP has (wrongfully) written. – Gabriel Romon Oct 30 '17 at 10:33
• @Surb that should be an answer. – Randall Oct 30 '17 at 11:37

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 $\lambda \in\{-1,0,1\}$.
• @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 $f(B)w = f(\alpha)w$. – Surb Oct 30 '17 at 20:29