Find the Limit of a Matrix

Nagoya University Linear Algebra Exam Problems and Solutions

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)

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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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