# Application of diagonalization of matrix - Markov chains

Problem: Suppose the employment situation in a country evolves in the following manner: from all the people that are unemployed in some year, $1/16$ of them finds a job next year. Furthermore, from all the people that were employed in some year, $1/8$ of them loses their jobs the next year, while the rest keeps working. Suppose at this moment there are $4$ million people working and half a million people are unemployed. How will the employment situation look like over a year from now? In $2$ years? In $100$ years?

Attempt at solution: I know this is an application of diagonalization, where I have to use the fact that $A^k = P D^k P^{-1}$, with $D$ being some diagonal matrix.

I was thinking of setting up a difference equation. Let $x_t$ denote the number of working people in year $t$, and let $y_t$ denote the number of unemployed people in year $t$. Then should I set up a relation of the form \begin{align*} v_t = \begin{pmatrix} x_t \\ y_t \end{pmatrix} = A v_{t-1} ? \end{align*} This would relate the employment situation at year $t$ to some previous year. I found the matrix $A$ as $A = \begin{pmatrix} 7/8 & 1/16 \\ 1/8 & 15/16 \end{pmatrix}$ by writing above the column the two options 'from a job' and 'from no job' and left of the rows I wrote 'to a job' and 'to no job'. The characteristic polynomial of this matrix is \begin{align*} \det(xI_2 - A) = x^2 - \frac{29}{16} x + \frac{13}{16} = \frac{1}{16} (16x - 13)(x-1) \end{align*} So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{13}{16}$. An eigenvalue $v_1$ corresponding to $\lambda_1$ is a non-zero solution to \begin{align*} (\lambda_1 I_2 - A) v_1 = 0 \end{align*} or \begin{align*} \begin{pmatrix} 1/8 & 1/16 \\ 1/8 & 1/16 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{align*} This can be rowreduced. A possible eigenvector is $v_1 = \begin{pmatrix} -1/2 \\ 1 \end{pmatrix}$. Similarly I found an eigenvector $v_2$ corresponding to $\lambda_2$ as $v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. If I let $P = \begin{pmatrix} -1/2 & 1 \\ 1 & 1 \end{pmatrix}$, then $P^{-1} = \begin{pmatrix} -2/3 & 2/3 \\ 2/3 & 1/3 \end{pmatrix}$. But then if I want to write $D = P^{-1}A P$, it doesn't work ( I don't get a diagonal matrix).

Can someone point out where I went wrong please, and if this is the correct strategy to solve this problem?

The eigenvector corresponding to $\lambda_1=1$ should be
$$v_1 = \begin{pmatrix} 1/2 \\ 1 \end{pmatrix}$$
The signs were wrong in the matrix $\lambda_1I_2-A$.
• Ah right, stupid mistake. Should I continue in the manner I did? They asking how the employment situation looks like in a year from now. Does that correspond to $A^2$ ? If we let $A^1$ be the situation right now. – Kamil Jul 18 '15 at 19:19
• @Kamil: You should let your first vector, say, $(x_0,y_0)$ be $(4,1/2)$ (in millions). Applying $A$ on it gives you the data after 1 year, applying $A^n$ gives you the data after $n$ years. – KittyL Jul 18 '15 at 19:32