Unique steady state vector in relation to regular transition matrix

Question

Define a steady-state vector for a transition matrix $T$ as a probability vector $v$ such that $Tv = v$ ($1$ is the eigenvalue for $v$).

Define a transition matrix $T$ as regular if there exist a $k \ge 1$ such that each entry of $T^k$ is non-zero.

Consider the following theorem:

Suppose $T$ is a regular transition matrix such that $x_k = T^kx_0$ (a discrete dynamic system), then there is a unique steady-state vector $v$ such that $Tv = v$. Furthermore, if $x_0$ is any probability vector: $$\lim_{k\to\infty} x_k = \lim_{k\to\infty} T^kx_0 = v$$

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Here's the confusion:

Suppose $\lambda$ is an eigenvalue for the $x_0$ mentioned above, then $$T^kx_0 = \lambda^kx_0 \Rightarrow \lim_{k\to\infty} T^kx_0 = \lim_{k\to\infty} \lambda^kx_0$$ Suppose we have real entries and real eigenvalues, then we have few cases:

1) $\lim_{k\to\infty} \lambda^k = \pm\infty$ when $\lambda > 1$ or $\lambda < -1$

2) $\lim_{k\to\infty} \lambda^k = 0$ when $ -1 < \lambda < 1$

3) $\lim_{k\to\infty} \lambda^k = 1$ when $\lambda = 1$

4) $\lim_{k\to\infty}$ oscillate between $1$ and $-1$ when $\lambda = -1$

It seems to me that, besides the trivial case of 3), none of these cases can produce $\lim_{k\to\infty} \lambda^kx_0 = v$. Since $v$ is a unique steady-state vector, in the other $3$ cases, $v$ becomes $0$, $\pm\infty$, or oscillate between $x_0$ and $-x_0$.

I think I'm missing something but I don't know what it is. Can someone care to explain what I'm missing?

Answer 1

Suppose $\lambda$ is an eigenvalue for the $x_0$ mentioned above

If the matrix $T$ is a transition matrix and your assuming $x_0$ is a probability vector and an eingenvector associated to the eingenvalue $\lambda$, $x_0$ must be the steady-state vector $v$ associated with the aingenvalue $\lambda=1$.

Answer 2

The Perron Frobenius theorem for non negative regular matrices states that there is a real positive eigenvalue $\lambda$ of multiplicity one that has positive left & right eigenvectors and for any other eigenvalue $\mu$ we have $|\mu| < \lambda$.

In this case, since $T$ is stochastic, we have $\|T\|_1 = 1$ and $e^T T = e^T$, hence we see that $\lambda = 1$.

By considering the above, if $V J V^{-1}$ is the Jordan form of $T$, we see that $J^k \to \begin{bmatrix} 1 & \\ & 0\end{bmatrix} $, where the $0$ is a block diagonal of sufficient size. In particular, $T^k \to T_\infty$, where $T_\infty$ is a rank one matrix and hence of the form $T_\infty = a b^T$ for some $a,b$. Since $e,v$ are left and right eigenvectors and $v$ is positive, we see that $T_\infty = v e^T$.

In particular, $\lim_k T_k x_0 = T_\infty x_0 = v e^T x_0 = v$.