Markov chains: showing $P$ has unique eigenvalue $1$

Question

I have a $4\times 4$ matrix and I tried solving for the determinant of $P-\lambda I$. This came out really messy and when I put the matrix into a matrix calculator my solution was $1,0$ and $-1$. Does this still mean $1$ is a unique eigenvalue solution? Is there a quicker method of proving $1$ is a unique eigenvalue?

Answer 1

You correctly computed the eigenvalues: they are (with multiplicity), $\{-1,0,0,1\}$. So $1$ is not a unique eigenvalue.

Looking at the matrix, you can see that several rows are equal. Each time this happens, you get $0$ as an eigenvalue.