Suppose that $A$ be a square, invertible matrix such that $A^{4} = A$. Find all real eigenvalues of $A$. Suppose that $A$ be a square, invertible matrix such that $A^4=A$. Find all real eigenvalues of $A$.

Since
$A^4 = A, \tag{1}$
we have
$A(A^3 - I) = A^4 - A = 0, \tag{2}$
and since $A$ is invertible we may write
$A^3 - I = I(A^3 - I) = (A^{-1}A)(A^3 - I)$
$= A^{-1}(A(A^3 - I)) = A^{-1}(A^4 - A) = 0; \tag{3}$
now if $\lambda$ is an eigenvalue of $A$ there is a vector $v \ne 0$ such that
$Av = \lambda v, \tag{4}$
whence
$A^2v = A(Av) = A(\lambda v) = \lambda(Av)$
$= \lambda(\lambda v) = \lambda^2 v; \tag{5}$
likewise
$A^3 v = A(A^2 v) = A(\lambda^2 v)$
$= \lambda^2(Av) = \lambda^2 (\lambda v) =\lambda^3 v;.\tag{6}$
combining (6) with (3) we find
$(\lambda^3 - 1)v = \lambda^3 v - v = A^3 v - Iv$
$=  (A^3 - I)v = 0, \tag{7}$
and since $v \ne 0$ we conclude that
$\lambda^3 - 1 = 0 \tag{8}$
for any eigenvalue $\lambda$ of $A$; the eigenvalues must then lie among the roots of the polynomial $\lambda^3 - 1$, and since
$(\lambda - 1)(\lambda^2 + \lambda + 1) = \lambda^3 - 1 = 0, \tag{9}$
we see that either $\lambda = 1$ or
$\lambda^2 + \lambda + 1 = 0; \tag{10}$
by the quadratic formula, the roots of (10) are
$\lambda = \dfrac{1}{2}(-1 \pm i\sqrt{3}); \tag{11}$
we thus conclude that the only possible real eigenvalue of $A$ is $1$.  
The question remains, is $1$ necessarily an eigenvalue of $A$?  The answer is "no";  to see this, observe that
$(A - I)(A^2 + A + I) = A^3 - I = 0; \tag{12}$
if now there exists a vector $y$ such that
$z = (A^2 + A + I)y \ne 0, \tag{13}$
then
$(A - I)z = (A - I)(A^2 + A + I)y$
$=  (A^3 - I)y = 0, \tag{14}$
that is,
$Az = z; \tag{15}$
$1$ is in fact an eigenvalue in this case; the other option is
$(A^2 + A + I)y = 0 \tag{16}$
for all $y$, or
$A^2 + A + I = 0; \tag{17}$
certainly such matrices exist; an example is
$A = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}. \tag{18}$
The final conclusion is thus
$1$ is an eigenvalue of $A$ if and only if $A^2 + A + I \ne 0$, and no other real eigenvalues are possible.
