Let $A=(a_{ij})$ be an $n \times n$ matrix.
We say that $A=(a_{ij})$ is a right stochastic matrix if each entry $a_{ij}$ is nonnegative and the sum of the entries of each row is $1$. That is, we have
\[a_{ij}\geq 0 \quad \text{ and } \quad a_{i1}+a_{i2}+\cdots+a_{in}=1\] for $1 \leq i, j \leq n$.

Let $A=(a_{ij})$ be an $n\times n$ right stochastic matrix. Then show the following statements.

(a)The stochastic matrix $A$ has an eigenvalue $1$.

(b) The absolute value of any eigenvalue of the stochastic matrix $A$ is less than or equal to $1$.

Proof.

(a) The stochastic matrix $A$ has an eigenvalue $1$.

We compute that
\[\begin{bmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} &a_{22} & \dots & a_{2n} \\
\vdots & \vdots & \dots & \vdots \\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix}
=
\begin{bmatrix}
a_{11}+a_{12}+\cdots+a_{1n} \\
a_{21}+a_{22}+\cdots+a_{2n} \\
\vdots \\
a_{n1}+a_{n2}\cdots+a_{nn}
\end{bmatrix}
=1\cdot \begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix}.\] Here the second equality follows from the definition of a right stochastic matrix.
(Each row sums up to $1$.)
This computation shows that $1$ is an eigenvector of $A$ and $\begin{bmatrix}
1 \\
\vdots \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $1$.

(b) The absolute value of any eigenvalue of the stochastic matrix $A$ is less than or equal to $1$.

Let $\lambda$ be an eigenvalue of the stochastic matrix $A$ and let $\mathbf{v}$ be a corresponding eigenvector.
That is, we have
\[A\mathbf{v}=\lambda \mathbf{v}.\]

Comparing the $i$-th row of the both sides, we obtain
\[a_{i1}v_1+a_{i2}v_2+\cdots+a_{in}v_n=\lambda v_i \tag{*}\] for $i=1, \dots, n$.
Let
\[|v_k|=\max\{|v_1|, |v_2|, \dots, |v_n|\},\] namely $v_k$ is the entry of $\mathbf{v}$ that has the maximal absolute value.

Note that $|v_k|>0$ since otherwise we have $\mathbf{v}=\mathbf{0}$ and this contradicts that an eigenvector is a nonzero vector.
Then from (*) with $i=k$, we have
\begin{align*}
|\lambda|\cdot |v_k| &= |a_{k1}v_1+a_{k2}v_2+\cdots+a_{kn}v_n|\\
& \leq a_{k1}|v_1|+a_2|v_2|+\cdots+ a_{kn}|v_{n}| &&(\text{by the triangle inequality and } a_{ij} \geq 0)\\
&\leq a_{k1}|v_k|+a_2|v_k|+\cdots+ a_{kn}|v_{k}| && (\text{since } |v_k| \text{ is maximal})\\
&=(a_{k1}+a_{k2}+\cdots+a_{kn})|v_k|=|v_k|.
\end{align*}

Since $|v_k|>0$, it follows that
\[\lambda \leq 1\] as required.

Remark.

A stochastic matrix is also called probability matrix, transition matrix, substitution matrix, or Markov matrix.

The post Eigenvalues of a Stochastic Matrix is Always Less than or Equal to 1 first appeared on Problems in Mathematics.

(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).

The post Stochastic Matrix (Markov Matrix) and its Eigenvalues and Eigenvectors first appeared on Problems in Mathematics.