# 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.

- The matrix $A$ is symmetric, hence diagonalizable.
- Diagonalize $A$.
- 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

- the sum of row entries of $A$ is $1$
- 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.

Add to solve later

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