Eigenvalues of a sparse 8x8 matrix

Question

I have the following $ 8 \times 8 $ sparse matrix

$ P=\begin{bmatrix} 0.5 & 0.5 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 \\ 0.0 & 0.0 & 0.5 & 0.5 & 0.0 & 0.0 & 0.0 & 0.0\\ 0.0 & 0.0 & 0.0 & 0.0 & 0.5 & 0.5 & 0.0 & 0.0\\ 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.5 & 0.5\\ 0.5 & 0.5 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\ 0.0 & 0.0 & 0.5 & 0.5 & 0.0 & 0.0 & 0.0 & 0.0\\ 0.0 & 0.0 & 0.0 & 0.0 & 0.5 & 0.5 & 0.0 & 0.0\\ 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.5 & 0.5\\ \end{bmatrix} $

I know that $1$ is an eigenvalue because the rows sum to 1 and also that $0$ is an eigenvalue because the matrix is singular( $rank(P)= 4$ so its multiplicity is at least 4).

I'm need to show that there are no other eigenvalues beside $1$ and $0$. Is there any simple way to show that beside using brute force and calculate the eigenvalues directly ?

Thanks!

Answer 1

Using Mathematica \begin{align*} \mathtt{P=\{\{1/2,1/2,0,0,0,0,0,0\},\{0,0,1/2,1/2,0,0,0,0\},\\ \{0,0,0,0,1/2,1/2,0,0\},\{0,0,0,0,0,0,1/2,1/2\},\\ \{1/2,1/2,0,0,0,0,0,0\},\{0,0,1/2,1/2,0,0,0,0\},\\ \{0,0,0,0,1/2,1/2,0,0\},\{0,0,0,0,0,0,1/2,1/2\}\}} \end{align*}

$\mathtt{Eigenvalues[P]}$

$$\mathtt{\{1,0,0,0,0,0,0,0\}}$$

Answer 2

Since $P^3=\frac18ee^T$, $1$ is a simple eigenvalue of $P$ and $0$ is an eigenvalue of algebraic multiplicity $7$.