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.

Let
\[A=\begin{bmatrix}
\frac{1}{7} & \frac{3}{7} & \frac{3}{7} \\
\frac{3}{7} &\frac{1}{7} &\frac{3}{7} \\
\frac{3}{7} & \frac{3}{7} & \frac{1}{7}
\end{bmatrix}\] be $3 \times 3$ matrix. Find

\[\lim_{n \to \infty} A^n.\]

(Nagoya University Linear Algebra Exam)

Hint.

1. The matrix $A$ is symmetric, hence diagonalizable.
2. Diagonalize $A$.
3. You may want to find eigenvalues
   without computing the characteristic polynomial for $A$.

Solution.

Note that the matrix $A$ is symmetric, hence it is diagonalizable.
We observe several things to simplify the computation.

First, note that the sum of the entries in each row is $1$.
(Such a matrix is call a stochastic matrix. See the comment below.)
Thus the matrix $A$ has eigenvalue $1$ and $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ is an eigenvector.

Also note that if we add $2/7(=-\lambda)$ to diagonal entries, then every entry becomes $3/7$.
That is,
\[A+\frac{2}{7}I=\begin{bmatrix}
\frac{3}{7} & \frac{3}{7} & \frac{3}{7} \\
\frac{3}{7} &\frac{3}{7} &\frac{3}{7} \\
\frac{3}{7} & \frac{3}{7} & \frac{3}{7}
\end{bmatrix},\] which is clearly singular.

From this observation, we notice that $-2/7$ is an eigenvalue of $A$ and eigenvectors are nonzero solution of $x_1+x_2+x_3=0$. Therefore $\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
0 \\
1 \\
-1
\end{bmatrix}$ are linearly independent eigenvectors for the eigenvalue $-2/7$.

Hence the geometric (and hence algebraic) multiplicity of the eigenvalue $-2/7$ is $2$.
(Note that we found all eigenvalues of $A$ without actually computing the characteristic polynomial.)

Now the invertible matix $P=\begin{bmatrix}
1 & 1 & 0 \\
1 &-1 &1 \\
1 & 0 & -1
\end{bmatrix}$ diagonalize $A$. Namely we have
\[P^{-1}AP=\begin{bmatrix}
1 & 0 & 0 \\
0 &-2/7 &0 \\
0 & 0 & -2/7
\end{bmatrix}.\] Hence
\[A=P\begin{bmatrix}
1 & 0 & 0 \\
0 &-2/7 &0 \\
0 & 0 & -2/7
\end{bmatrix}P^{-1}.\] Therefore we have
\[ A^n=P\begin{bmatrix}
1^n & 0 & 0 \\
0 & (-2/7)^n &0 \\
0 & 0 & (-2/7)^n
\end{bmatrix}P^{-1}
\text{ and } \lim_{n \to \infty} A^n=P\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 &0 \\
0 & 0 & 0
\end{bmatrix}P^{-1}. \] Find the inverse $P^{-1}$ by your favorite method
\[P^{-1}=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt] \frac{2}{3} &\frac{-1}{3} &\frac{-1}{3} \\[6pt] \frac{1}{3} & \frac{1}{3} & \frac{-2}{3}
\end{bmatrix}.\] Then the answer is
\[\lim_{n \to \infty} A^n=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt] \frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\[6pt] \frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{bmatrix}.\]

Comment.

You might come up with the idea to use the diagonalization as it is an exam problem in linear algebra.
But you soon realize that computing the characteristic polynomial for $A$ is a bit messy.
So sometimes it is important to find eigenvalues without finding the roots of the characteristic polynomial.

We observed that

1. the sum of row entries of $A$ is $1$
2. if we add $(2/7)I$ to $A$ then we obtain a matrix whose entries are all identical.

These observation yields the eigenvalues of $A$ without using the characteristic polynomial.

If the sum of entries of each row of a matrix is $1$, then the matrix is called called a stochastic matrix (or Markov matrix, probability matrix).
The matrix $A$ in the problem is an example of a stochastic matrix.

Stochastic matrices have always $1$ as an eigenvalue.
see the post
 Stochastic matrix (Markov matrix) and its eigenvalues and eigenvectors
for an example of a $2\times 2$ stochastic matrix.

The post Find the Limit of a Matrix 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.