I'd like to prove that this matrix is idempotent using a more algebraic proof for matrices with a similar definition to A, rather than deriving its eigenvalues or calculating $A^2$ .
$A=$ $\begin{bmatrix}0.5&0&-0.5\\0&1&0\\ -0.5&0&0.5\end{bmatrix}$
I know that A is symmetric (so $A'=A$) and positive semi-definite, so how could I use these properties to prove idempotency?
On
One way to show an object is idempotent is to double it, subtract the identity and check whether the result is (multiplicatively) self-inverse. For instance in $\mathbb Z/10\mathbb Z$ we have $2×5-1\equiv-1$ which is self-inverse, so $5$ is idempotent in $\mathbb Z/10\mathbb Z$. (Note that this works in domains not of characteristic $4n$ for positive $n$.)
In this problem doubling and subtracting the identity leads to
$\begin{bmatrix}0&0&-1\\0&1&0\\-1&0&0\end{bmatrix},$
in which an interchange of the first and third rows and the diagonal elements of the resulting diagonal matrix are trivially self-inverse. So the self-inverse condition holds for the derived matrix assuring the original one will be idempotent.
Since it is symmetric, it is diagonalisable. It's clear from $A$'s second column that $1$ is an eigenvalue and, since the third column is minus the first one, $\det A=0$, and therefore $0$ is an eigenvalue too. And $\operatorname{tr}A=2$. So, $1$ is, in fact, a double eigenvalue. Therefore, $A$ is similar to$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix},$$and so it is idempotent.