# Eigenvalues and Eigenvectors in Models

I have a question on finding the eigenvalues and eigenvectors, and to then find the general formula of the 2 equations given.

I've found the eigenvalues of both equations, but i'm stuck on how to find the eigenvectors as I haven't been given any starting values. How would I go about solving the eigenvectors without the starting values?

Question is to solve the following:

$$a_n+_1=a_n+2b_n$$ $$b_n+_1=2a_n+b_n$$

I have found the eigenvalues to be -1 and 3. But the eigenvectors i can't find as I don't have any starting values and the eigenvectors say to write it in the form:

eigenvalue = -1, corresponding eigenvector = (number, $2^T$) eigenvalue = 3, corresponding eigenvector = (number, $2^T$)

Can anyone help or point me in the write direction, thanks.

$\begin{bmatrix} a_{n+1}\\b_{n+1}\end{bmatrix} = A\begin{bmatrix} a_{n}\\b_{n}\end{bmatrix}\\ \begin{bmatrix} a_{n+1}\\b_{n+1}\end{bmatrix} = A^n\begin{bmatrix} a_{1}\\b_{1}\end{bmatrix}\\ A = PDP^{-1}\\ A^n = PD^nP^{-1}$

Does not depend on the initial values: $a_1,b_1$

$A = \begin{bmatrix} 1&2\\2&1\end{bmatrix}\\ D = \begin{bmatrix} 3&\\&-1\end{bmatrix}\\ P = \begin{bmatrix} \frac {1}{\sqrt 2}&-\frac {1}{\sqrt 2}\\\frac {1}{\sqrt 2}&\frac {1}{\sqrt 2}\end{bmatrix}\\ A^n=\begin{bmatrix} \frac {1}{\sqrt 2}&-\frac {1}{\sqrt 2}\\\frac {1}{\sqrt 2}&\frac {1}{\sqrt 2}\end{bmatrix}\begin{bmatrix} 3^n&\\&(-1)^n\end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt 2}&\frac {1}{\sqrt 2}\\-\frac {1}{\sqrt 2}&\frac {1}{\sqrt 2}\end{bmatrix}\\ A^n=\begin{bmatrix} \frac {3^n + (-1)^n}{2}&\frac {3^n - (-1)^n}{2}\\\frac {3^n - (-1)^n}{2}&\frac {3^n + (-1)^n}{2}\end{bmatrix}\\ a_{n+1} = \frac {3^n + (-1)^n}{2} a_1 + \frac {3^n - (-1)^n}{2} b_1\\ b_{n+1} = \frac {3^n + (-1)^n}{2} b_1 + \frac {3^n - (-1)^n}{2} a_1\\$

Based on the comments you need something more basic...

Suppose there exits a $v$ such that

$Av = \lambda v\\ (A - \lambda I) v = 0\\$

$v$ is the kernel of $A - \lambda I$

It looks like you have found your eigenvalues.

$\lambda = 3\\ (A - 3 I) = \begin{bmatrix} -2 & 2\\2&-2\end{bmatrix}\\ \begin{bmatrix} -2 & 2\\2&-2\end{bmatrix}\begin{bmatrix} 1\\1 \end{bmatrix} = 0$

And that would be fine to use $\begin{bmatrix} 1\\1 \end{bmatrix}$ as your eigenvector. I chose to normalize my eigenvectors above.

When you have your eigenvalues then $P = [v_1,v_2]$

$AP = \lambda_1 v_1 + \lambda_2 v_2 = P \begin{bmatrix} \lambda_1\\&\lambda_2\end{bmatrix}$

• Thank you, i'm really stuck on what you've put there though...How would I use that information to be able to find the 2 eigenvectors? – The Statistician Feb 1 '18 at 23:38
• How do you find the eigenvectors of any matrix? $Av = \lambda v, \det (A-\lambda I) = 0$ find $\lambda_1, \lambda_2$ the eigenvector is the kernel of $A-\lambda I$ – Doug M Feb 1 '18 at 23:40
• So I just do $det(A-ƒI)$ where ƒ is lambda – The Statistician Feb 1 '18 at 23:42