$A$ is a symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent 
Let $A$ be a real symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent.

I have tried using eigenvalues and only inferred that the eigen values may be $0,1$. But I cannot proceed with this.
 A: Let $A=PDP^{-1}$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}=PDP^{-1}=A$$
A: Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:
We have $A^4 - A = 0$.  Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = \left(A + \frac 12 I\right)^T\left(A + \frac 12 I\right) + \frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible).  Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 \implies A^2 - A = 0 \implies A^2 = A
$$
Thus, $A$ is idempotent.
A: I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:
if
$A^4 = A, \tag 1$
then
$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; \tag 2$
since
$A^T = A, \tag 3$
we have for any vector $x$
$\langle x, Ax \rangle = \langle x, A^4 x \rangle = \langle (A^T)^2 x, A^2 x \rangle = \langle A^2 x, A^2 x \rangle \ge 0; \tag 4$
$\langle x, A^2 x \rangle = \langle A^T x, Ax \rangle = \langle Ax, Ax \rangle \ge 0; \tag 5$
thus
$\langle x, (A^2 + A + I)x \rangle = \langle x, A^2 x \rangle + \langle x, Ax \rangle + \langle x, x \rangle > 0, \tag 6$
by (4), (5) and the fact that $\langle x, x \rangle > 0$; now if any matrix $B$ satisfies
$\langle x, Bx \rangle \ne 0, \forall x, \tag 7$
then $B$ is invertible; if not, 
$\exists y \ne 0, \; By = 0 \Longrightarrow \langle y, By \rangle = 0, \tag 8$
which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that
$A^2 - A = A(A - I) = 0, \tag 9$
or
$A^2 = A. \tag{10}$
A: You are given that
$$
          0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
             \lambda^2+\lambda +1 = (\lambda+1/2)^2+3/4 \ge 3/4.
$$
So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.
A: If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.
