# Linear Algebra: Finding a steady state matrix

Here is the problem:

And here is what I tried to do:

I tried doing it in my calculator and got that the answer is 40 and 160 as you keep multiplying PX. Can anyone point out what I'm doing wrong or help me?

Thanks.

Problem Statement:

A college dormitory houses 200 students. Those who watch an hour or more of TV on any day always watch for less than an hour the next day. One-fourth of those who watch TV for less than an hour one day will watch an hour or more the next day. Half of the students watch TV for an hour or more today.

a) How many will watch TV for an hour or more tomorrow?
b) In 2 days?
c) Find the steady state matrix for populations described in this problem.

Work:

$P = \stackrel{>1 \hphantom{XX} <1}{\begin{bmatrix} 0 & 0.25\\ 1 & 0.75 \end{bmatrix}} {>1 \atop <1} \quad x = \begin{bmatrix} 100\\ 100 \end{bmatrix} {>1 \atop <1}$

$Px = \begin{bmatrix} 0 & 0.25\\ 1 & 0.75 \end{bmatrix} \begin{bmatrix} 100\\ 100 \end{bmatrix} = \begin{bmatrix} 25\\ 175 \end{bmatrix} {>1 \atop }$

a) 25.

$P^{\,2}x = \begin{bmatrix} 0 & 0.25\\ 1 & 0.75 \end{bmatrix} \begin{bmatrix} 25\\ 175 \end{bmatrix} = \begin{bmatrix} 43.75\\ 156.25 \end{bmatrix} {>1 \atop <1}$

b) $43.75 \sim 44$.

Steady state $P\,\overline{x} = \overline{x}$

$\begin{bmatrix} 0 & 0.25\\ 1 & 0.75 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} \Rightarrow \begin{bmatrix} 0.25 x_{2}\\ x_{1}+0.75 x_{2} \end{bmatrix} \Rightarrow \begin{bmatrix} 25 \\ 175 \end{bmatrix}$

\boxed{\begin{align*} 0.25 x_{2} &= x_{1}\\ x_{1} + 0.75 x_{2} &= x_{2} \end{align*}}

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The showing of your work was very helpful, it made it possible to know the exact issue. I wish everybody did this, it would make it much easier for responses to be relevant. Identifying the exact type of course is very good. That way, responses can be more or less at the right level. –  André Nicolas Jun 15 '11 at 16:01
Yes, +1 for a good question :) –  Chris Taylor Jun 15 '11 at 16:43

The steady state is the population split that is an eigenvector of the transition matrix with eigenvalue 1, that is:

$$\left(\begin{matrix} 0 & 1/4 \\ 1 & 3/4 \end{matrix}\right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} x \\ y \end{matrix} \right)$$

This leads you to the equation

$$y = 4x$$

and you also know that

$$x + y = 200$$

and solving these jointly gives $x=40$, $y=160$.

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great. I didn't realize I could use x+y=200 to solve for x1 and x2. Thank you. –  Virtuoso Jun 15 '11 at 17:55

I assume that what you did with the calculator is to keep multiplying by the matrix until things appeared to stabilize.

Very streamlined versions of what you did are often used in practice. So the idea was a good one, and shows that you understand the meaning of long-term distribution. The idea is not wrong, but it is almost certainly not what this part of the problem expects you to do.

In a first Linear Algebra course, with a small ($2 \times 2$) matrix, you would be expected to calculate an eigenvector for eigenvalue $1$, as described by Chris Taylor.

Possibly your approach would get some part marks. But you have been taught about eigenvalues and eigenvectors. This part of the question is probably designed to see whether you are aware of the connection between eigenvectors for eigenvalue $1$ and the long-term distribution, and can do the requisite eigenvector calculation.

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Just doing the matrix multiplies is the appropriate way to get the first two parts, and the answers are correct. –  Ross Millikan Jun 15 '11 at 16:39