Trace of power of stochastic matrix I would like to know if this statement is true.
Having a stochastic matrix (rows sum up to 1), with a positive (non-negative) diagonal, then it holds that
$$\text{trace}({W^2})\leq \text{trace}({W}),$$
(or more generally, if $p\geq q$, then $\text{trace}({W^p})\leq \text{trace}({W}^q)$.)
In other words, does it hold that $\sum_{i=1}^N\lambda_i^2\leq \sum_{i=1}^N\lambda_i$ ? Note that $\lambda_i$ may be also negative (it is not a positive-definite, or symmetric matrix)!
My intuition says it should hold, since with the powers of a stochastic matrix ${W}$, the spectrum gets smaller and smaller, eventually leading to $\lambda_1=1$ and $\lambda_i=0$, for $i\ne 1$. But is also the SUM of eigenvalues getting smaller with growing power? Thanks.

This is (still) trivially false if you just take a small epsilon
$p_{12}=p_{21}=1-\epsilon$ and $p_{11}=p_{22}=\epsilon$
The trace of $P^2$ is $2((1-\epsilon)^2+\epsilon^2)>2\epsilon$, if take $\epsilon$ to be, say, smaller than 0.2
