Proving that a symmetric and idempotent matrix has all eigenvalues equal to 1

Question

Let $A^{n \times n}$ be a symmetric and idempotent matrix. $A^t=A$ and $A^2=A$. Prove that $A$ has $n$ eigenvalues that are all equal to 1.

I'm having some difficulty proving that. Here's what I've done

$$Au=λu\Rightarrow A^2=λAu=A^2 λ^2u$$ and thus

$$λ^2-λ=0 \Rightarrow λ=0, λ=1$$

I dont know how to eliminate the possibility of $λ$ being $0$. My guess is that I have to find a way using the symmetry of $A$ to prove either that $dimKer(A)=0$ which implies that there is no eigenvalues equal to zero, or that $A$ has positive eigenvalues (by proving that $A$ is positive-definite ? ).

Any ideas on that?

Answer 1

The difficulty in proving this stems from the fact that it is false. A simple counter-example is provided by taking $A$ to be a diagonal matrix

$A = [d_{ij}], \tag 1$

where

$d_{ij} = 0, \; i \ne j, \tag 2$

but

$d_{ii} \in \{0, 1\}; \tag 3$

as long as

$\exists i, \; d_{ii} = 1, \tag 4$

we have

$A \ne 0; \tag 5$

nevertheless

$A^T = A \tag 6$

and

$A^2 = A \tag 7$

both continue to bind.

Answer 2

The proposition is incorrect.

Counter example #1: $A = 0$.

Counter example #2: $A = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$.

As shown in your statement, a symmetric idempotent matrix's eigenvalues must be either $0$ or $1$.