Why are the eigenvalues of a Markov matrix bounded within the unit circle? Define a Markov matrix as $P \in M_n([0, 1])$ such that for every row $i \in \{1, \dots, n\}$, $\sum_{j=1}^n P_{ij} = 1$.
It's commonly noted the eigenvalues of $P$ lie within the unit circle, i.e., $\sigma(P) \subset \{ z \in \mathbb C : |z| \leq 1 \}$.
This fact is important because it is necessary to show that a diagonalizable Markov chain will converge to a steady state distribution, but I could not find a proof for this bound.
I eventually proved this using the Gershgorin circle theorem, but I'm curious what other approaches exist for proving this?
 A: Let $G(P) = \cup_{i=1}^n \{ z \in \mathbb C : |z - P_{ii}| \leq \sum_{j \neq i} P_{ij} \}$ be the union of the Gershgorin circles of $P$.
By the definition of the Markov matrix, the radius of every Gershgorin circle is bounded such that $\sum_{j \neq i} P_{ij} \leq 1$.
Additionally, every Gershgorin circle is centered at $1 - \sum_{j \neq i} P_{ij}$.
Therefore, every circle is constrained within the unit circle, implying that $G(P) \subset \{ z \in \mathbb C : |z| \leq 1 \}$ as well.
Finally, by application of the Gershgorin circle theorem, $\sigma(P) \subset G(P) \subset \{ z \in \mathbb C : |z| \leq 1 \}$.
A: using Schur Triangularization $P= UTU^*$.
$P\mathbf 1 = \mathbf 1$ so $P$ has an eigenvalue of 1. Let $\lambda_m$ be the maximum modulus eigenvalue of $P$, with modulus of $1+\delta$, for some $\delta \geq 0$.
$1+k\delta \leq \vert \lambda_m \vert^k \leq \big \Vert T^k\big \Vert_F = \big \Vert P^k\big \Vert_F = \sqrt{ \sum_{i=1}^n\sum_{j=1}^n (p_{i,j}^{(k)}})^2\leq  \sum_{i=1}^n\sum_{j=1}^n p_{i,j}^{(k)}=\sum_{i=1}^n(1) = n$
justification:  Bernouli Inequality, unitary invariance of Frobenius norm, triangle inequality, $P^k\mathbf 1 = \mathbf 1$
This holds for all $k\in \big\{1,2,3,...\big\}\implies \delta =0\implies \vert \lambda_m \vert= 1$
