Hi I was wondering whether I could get some help on eigenvectors using maths models:
So I have this problem to solve:
Two Italian restaurant chains Pizza Express and Prezzo are in competition for the pizza market in Portsmouth. Assume that people going to Pizza Express for a pizza have 70 % chance of going back and 30 % chance of switching to Prezzo. Similarly people who go to Prezzo have a 80 % chance of returning and 20 % chance of switching to Pizza Express for their next pizza. Suppose that x represents the number of Pizza Express customers, y represents the number of Prezzo customers and n represents the number of pizza visits to Pizza Express or Prezzo.
I have 2 equations that are modelled using the information above:
$$x_{n+1}= 0.7x_n+0.2y_n$$ $$y_{n+1}= 0.3x_n+0.8y_n$$
The smaller eigenvector of the coefficient matrix $=a$, which has a corresponding eigenvector $=(b, 0.71)^T$.
The larger eigenvector of the coefficient matrix $=c$, which has a corresponding eigenvector $=(d, -0.83)^T.$
I've worked out using the 2 equations above, that $a=0.5$ and $c=1$. From my University lecture notes it says use the 2 equations in the form:
$$(0.7-λ_{1,2})A+(0.2)B=0$$ $$(0.3)A+(0.8-λ_{1,2})B=0$$
In using these equations, i have found my $b=0.2$ and $d=0.3$, but that's wrong.
Can anyone help me with understanding how I'm meant to get the eigenvectors.
Let us name the matrix $$E=\begin{bmatrix} 0.7&0.2\\ 0.3&0.8\end{bmatrix}.$$ For the eigenvalue $0.5$, an eigenvector $v$ will satisfy $$\tag1 Ev=0.5v.$$ So if they tell you that $v=[b\ \ \ 0.71]^T$ is an eigenvector for $0.5$ you have, from the first row of $(1)$, that $$ 0.7b+0.2\times0.71=0.5b. $$ Solving, $b=-0.71$ (you would have obtained the same value for $b$ had you used the second row).
I'll let you find $d$ on your own.