Theorem 5.12

Let $A$ be a square matrix with complex entries. Then $\lim\limits_{m \rightarrow \infty} A^m$ exists if and only if both of the following conditions hold:

Every eigenvalue of $A$ is contained in $S$ where $S=\{\lambda \in \mathbf{C}: |\lambda| < 1 \text{ or }\lambda = 1\}$
If 1 is an eigenvalue of $A$, then the dimension of the eigenspace corresponding to 1 equals the multiplicity of 1 as an eigenvalue of 1.

Proof: [TODO]

Theorem 5.13

Let $A \in M_{n \times n}(\mathbf{C})$ satisfy the following two conditions

Every eigenvalue of $A$ is contained in $S$ where $S=\{\lambda \in \mathbf{C}: |\lambda| < 1 \text{ or }\lambda = 1\}$
$A$ is diagonalizable.

Then $\lim\limits_{m \rightarrow \infty} A^m$ exists

Proof:

Since $A$ is diagonalizable, then we know that there exists an invertible matrix $Q$ such that $D = Q^{-1}AQ$. Suppose that

$$ \begin{align*} D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix} \end{align*} $$

$\lambda_1,...,\lambda_n$ are the eigenvalues of $A$. But condition (i) requires that for each $i$, $\lambda_i = 1$ or $|\lambda_i| < 1$. Thus

$$ \begin{align*} \lim\limits_{m \rightarrow \infty} {\lambda_i}^m &= \begin{cases} 1 \quad \text{if } \lambda_i = 1 \\ 0 \quad \text{otherwise } \end{cases} \end{align*} $$

$$ \begin{align*} D^m = \begin{pmatrix} \lambda_1^m & 0 & \cdots & 0 \\ 0 & \lambda_2^m & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n^m \end{pmatrix} \end{align*} $$

and so the sequence $D, D^2 ...$ converges to a limit $L$. Therefore,

$$ \begin{align*} \lim\limits_{m \rightarrow \infty} A^m = \lim\limits_{m \rightarrow \infty} (QDQ^{-1}) = QLQ^{-1} \end{align*} $$

Theorem 5.14

Let $M$ be an $n \times n$ matrix having real nonnegative entries, let $v$ be a column vector in $\mathbf{R}^n$ having nonnegative coordinates, and let $u \in \mathbf{R}^n$ be the column vector in which each coordinate equals 1. Then

$M$ is a transition matrix if and only if $u^tM = u^t$;
$v$ is a probability vector if and only if $u^tv = (1)$.

Proof (a):

$\Rightarrow$: Suppose $M$ is a transition matrix. Then by the definition of matrix-vector multiplication,

$$ \begin{align*} u^tM &= \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{m2} \\ \end{pmatrix} = \begin{pmatrix} w_1 & w_2 & \cdots & w_n \end{pmatrix} \end{align*} $$

where $w_i$ is the sum

$$ \begin{align*} w_i &= 1a_{1i} + 1a_{2i} + ... + 1a_{ni} \\ &= a_{1i} + a_{2i} + ... + a_{ni} \\ &= 1 \quad \text{(since $M$ is a transition matrix)} \end{align*} $$

$\Leftarrow$: [TODO: same argument as $\Rightarrow$]

Proof (b): [TODO: also very similar to $(a)$]

Theorem 5.14 (Corollary)

The product of two $n \times n $ transition matrices is an $n \times n$ transition matrix. In particular, any power of a transition matrix is a transition matrix.
The product of a transition matrix a probability vector is a probability vector.

Proof:

[TODO]

Definition

A transition matrix is called regular if some power of the matrix contains only nonzero (i.e., positive) entries.

Definition

Let $A \in M_{n \times n}(\mathbf{C})$. For $1 \leq i,j \leq n$, define $\rho_i(A)$ to be the sum of the absolute values of the entries of row $i$ of $A$, and define $\nu_j(A)$ to be equal to the sum of the absolute values of the entries of column $j$ of $A$. Thus $$ \begin{align*} \rho_i(A) = \sum_{j=1}^n |A_{ij}| \quad \text{ for $i = 1,2,...,n$} \end{align*} $$ and $$ \begin{align*} \nu_j(A) = \sum_{i=1}^n |A_{ij}| \quad \text{ for $j = 1,2,...,n$} \end{align*} $$ The row sum of $A$ denoted $\rho(A)$, and the column sum of $A$, denoted $\nu(A)$, are defined as $$ \begin{align*} \rho(A) = \max\{\rho_i(A): 1 \leq i \leq n\} \text{ and } \nu(A) = \max\{\nu_j(A): 1 \leq j \leq n\} \end{align*} $$

TODO: definition of Gershgorin’s circle

Theorem 5.15 (Gershgorin's Circle Theorem)

Let $A \in M_{n \times n}(\mathbf{C})$. Then every eigenvalue of $A$ is contained in a Greshgorin disk.

Theorem 5.15 (Corollary 1)

Let $\lambda$ be any eigenvalue of $A \in M_{n \times n}(\mathbf{C})$. Then $|\lambda| \leq \rho(A)$.

Theorem 5.15 (Corollary 2)

Let $\lambda$ be any eigenvalue of $A \in M_{n \times n}(\mathbf{C})$. Then $|\lambda| \leq min\{\rho(A),\nu(A)\}$.

Theorem 5.15 (Corollary 3)

If $\lambda$ is an eigenvalue of a transition matrix, then $|\lambda| \leq 1$.

Theorem 5.16

Every transition matrix has 1 as an eigenvalue.

Theorem 5.17

Let $A \in M_{n \times n}(\mathbf{C})$ be a matrix in which each entry is a positive real number, and let $\lambda$ be a complex eigenvalue of $A$ such that $|\lambda| = \rho(A)$. Then $\lambda = \rho(A)$ and $\{u\}$ is a basis for $E_{\lambda}$, where $u \in \mathbf{C}^n$ is the column vector in which each coordinate equals 1.

Proof [TODO: we did some version of this in class]

Theorem 5.17 (Corollary 1)

Let $A \in M_{n \times n}(\mathbf{C})$ be a matrix in which each entry is a positive, and let $\lambda$ be an eigenvalue of $A$ such that $|\lambda| = \nu(A)$. Then $\lambda = \nu(A)$ and the dimension of $E_{\lambda}$ equals 1.

Theorem 5.17 (Corollary 2)

Let $A \in M_{n \times n}(\mathbf{C})$ be a transition matrix in which each entry is positive, and let $\lambda$ be an eigenvalue of $A$ other than 1. Then $|\lambda| < 1$. Moreover, the eigenspace corresponding to the eigenvalue 1 has dimension 1.

Okay, so only if the transition matrix itself has positive entries (regular is not enough), then $\lambda < 1$ for eigenvalues that are not 1 AND the dimension of the eigenspace corresponding to eigenvalue 1 is 1.

Theorem 5.18

Let $A$ be a regular transition matrix, and let $\lambda$ be an eigenvalue of $A$.

$|\lambda| \leq 1$.
If $|\lambda| = 1$, then $\lambda = 1$, and $\dim(E_{\lambda}) = 1.$

This is the other result I was looking for. If $A$ is regular, then we’ll have the normal restriction of $|\lambda| \leq 1$.

Theorem 5.18 (Corollary)

Let $A$ be a regular transition matrix that is diagonalizable. Then $\lim\limits_{m \rightarrow \infty} A^m$ exists.

Theorem 5.19

Let $A$ be a regular transition matrix, and let $\lambda$ be an eigenvalue of $A$.

The multiplicity of 1 as an eigenvalue of $A$ is 1.
$\lim\limits_{m \rightarrow \infty} A^m$ exists.
$L = \lim\limits_{m \rightarrow \infty} A^m$ is a transition matrix.
$AL = LA = L$
The columns of $L$ are identical. In fact, each column of $L$ is equal to the unique probability vector $v$ that is also an eigenvector of $A$ corresponding to the eigenvalue 1.
For any probability vector $w$, $\lim\limits_{m \rightarrow \infty} (A^m w) = v.$

References:

Linear Algebra 5th Edition