\(
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\require{physics}
\require{ams}
\require{mathtools}
\)
What is the population of Australia and New Zealand after 100 years of
the following migration pattern? Starting with 1 million people in Australia
and no one in New Zealand.
Continue on the reverse. How does \( XVX^{-1} \) come up?
The answer is the result of the following expression:
\[
\begin{bmatrix}
0.8 & 0.3 \\
0.2 & 0.7
\end{bmatrix}^{100}
\begin{bmatrix}
1,000,000 \\
0
\end{bmatrix}
\]
To calculate the result, use eigenvalues and eigenvectors. The transition
matrix has eigenvalues 1.0 and 0.5 corresponding to eigenvectors:
\[
\begin{bmatrix}
0.6 \\
0.4 \\
\end{bmatrix}
\text{ and }
\begin{bmatrix}
-1 \\
1 \\
\end{bmatrix}
\]
We can express the starting population as a linear combination of the eigenvectors:
\[
\begin{bmatrix}
1,000,000 \\
0
\end{bmatrix}
=
\begin{bmatrix}
600,000 \\
400,000
\end{bmatrix}
+
\begin{bmatrix}
400,000 \\
-400,000
\end{bmatrix}
\]
Then,
\[
\begin{bmatrix}
0.8 & 0.3 \\
0.2 & 0.7
\end{bmatrix}^{100}
\begin{bmatrix}
1,000,000 \\
0
\end{bmatrix}
\]
becomes:
\[
1.0^{100}
\begin{bmatrix}
600,000 \\
400,000
\end{bmatrix} +
0.5^{100}
\begin{bmatrix}
400,000 \\
-400,000
\end{bmatrix}
\]
Which is
\[
\begin{bmatrix}
600,000 \\
400,000
\end{bmatrix}
\]
General approach
The above intuitive approach is an application of the somewhat opaque
symbolic expression \( XVX^{-1} \). The steps are:
- Express the starting population as a linear combination of eigenvectors.
- Apply the transition matrix to the eigenvectors.
- Express the result as a linear combination of eigenvectors.
Repeat the solution from this perspective, letting \( A \) be the transition matrix,
and \( X \) be the matrix of eigenvectors and \( V \) be the diagonal matrix of eigenvalues.
We will set \( X \) to be:
\[
\begin{bmatrix}
600,000 & 400,000 \\
400,000 & -400,000
\end{bmatrix}
\]
We are trying to calculate:
\[
\text{result} = A^{100} \begin{bmatrix} 1,000,000 \\ 0 \end{bmatrix}
\]
By eigenvector decomposition, we have:
\[
\text{result} = X V^{100} X^{-1} \begin{bmatrix} 1,000,000 \\ 0 \end{bmatrix}
\]
First, calculate \( X^{-1} \begin{bmatrix} 1,000,000 \\ 0 \end{bmatrix} \):
\[
X^{-1} \begin{bmatrix} 1,000,000 \\ 0 \end{bmatrix} =
\begin{bmatrix}
1 \\
1
\end{bmatrix}
\]
Then multiply by \( V^{100} \):
\[
V^{100} \begin{bmatrix} 1 \\ 1 \end{bmatrix} =
\begin{bmatrix}
1.0^{100} \\
0.5^{100}
\end{bmatrix}
=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
\]
Finally, get back to the original basis:
\[
X \begin{bmatrix} 1 \\ 0 \end{bmatrix} =
\begin{bmatrix}
600,000 & 400,000 \\
400,000 & -400,000
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\begin{bmatrix}
600,000 \\
400,000
\end{bmatrix}
\]
