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