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 method is correct.

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 $\lambda_1I_2-A$.

The signs were also wrong for the other eigenvector. You must have made the same mistake.

  • $\begingroup$ 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. $\endgroup$ – Kamil Jul 18 '15 at 19:19
  • $\begingroup$ @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. $\endgroup$ – KittyL Jul 18 '15 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.