# Stochastic Matrix (Markov Matrix) and its Eigenvalues and Eigenvectors

## Problem 34

(a) Let

$A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{bmatrix}$ be a matrix such that $a_{11}+a_{12}=1$ and $a_{21}+a_{22}=1$. Namely, the sum of the entries in each row is $1$.

(Such a matrix is called (right) stochastic matrix (also termed probability matrix, transition matrix, substitution matrix, or Markov matrix).)

Then prove that the matrix $A$ has an eigenvalue $1$.

(b) Find all the eigenvalues of the matrix
$B=\begin{bmatrix} 0.3 & 0.7\\ 0.6& 0.4 \end{bmatrix}.$

(c) For each eigenvalue of $B$, find the corresponding eigenvectors.

## Hint.

1. For (a), consider the vector $\mathbf{x}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
2. For (b), use (a) and consider the trace of $B$ and its relation to eigenvalues.
For this relation, see the problem Determinant/trace and eigenvalues of a matrix.

## Solution.

### (a) Prove that the matrix $A$ has an eigenvalue $1$.

Let $\mathbf{x}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and we compute
\begin{align*}
A \mathbf{x}=\begin{bmatrix}
a_{11} & a_{12}\\
a_{21}& a_{22}
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=\begin{bmatrix}
a_{11}+a_{12} \\
a_{21}+a_{22}
\end{bmatrix}
=\begin{bmatrix}
1 \\
1
\end{bmatrix}
=1\cdot \mathbf{x}.
\end{align*}
This shows that $A$ has the eigenvalue $1$.

### (b) Find all the eigenvalues of the matrix $B$.

Note that the matrix $B$ is of the type of the matrix in (a).
Thus the matrix $B$ has the eigenvalue $1$. Since $B$ is $2$ by $2$ matrix, it has two eigenvalues counting multiplicities. To find the other eigenvalue, we note that the trace is the sum of the eigenvalues.

Thus we have
$\tr(B)=0.3+0.4=1+\lambda,$ where $\lambda$ is the second eigenvalue. Hence another eigenvalue is $\lambda=-0.3$.

### (c) For each eigenvalue of $B$, find the corresponding eigenvectors.

By solving $(B-I)\mathbf{x}=\mathbf{0}$ and $(B+0.3I)\mathbf{x}=\mathbf{0}$, we find that
$\begin{bmatrix} 1 \\ 1 \end{bmatrix}t \quad \text{ and } \quad \begin{bmatrix} -7 \\ 6 \end{bmatrix}t$ for any nonzero scalar $t$ are eigenvectors corresponding to eigenvalues $1$ and $-0.3$, respectively.

## Comment.

For some specific matrices, we can find eigenvalues without solving the characteristic polynomials like we did in part (b).

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