# Find the Limit of a Matrix

## Problem 50

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)

Contents

## 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.

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