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

Question

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.

Answer 1

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 $$A^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}=PDP^{-1}=A$$

Answer 2

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 $A^2-A=0$.

Answer 3

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

$A^2 = A. \tag{10}$

Answer 4

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.

Answer 5

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 $A^2=A$.