I am studying for a final in linear algebra and a practice problem that was posed is the following:
Let $a_1,a_2\in R$ (R is the set of reals). For $n\geq 3$, set $a_n = \frac{5}{2}a_{n-1}-a_{n-2}$. Show that the sequence $(a_n)$ converges if and only if $(a_1,a_2)\in span\{(2,1)\}$.
I just want help for the 'only if' part, so trying to show it converges then $(a_1,a_2)\in span\{(2,1)\}$ portion. I have solved these kind of problems but they usually give a base case like saying what the first term or two is and then set up a series of linear equations to solve for the closed form, but we aren't given what $a_1$ is so I am not really sure what to do. I can set $n=3$ and $n=4$ and get the following:
$a_3 = \frac{5}{2}a_2-a_1$ and $a_4=\frac{11}{4}a_2-\frac{3}{2}a_1$
but I am not sure what to do from here. Do I need to actually figure out the closed form cause how do I do that without a base case. Anyways, any help would be greatly appreciated!
You can formulate this question in the framework of linear algebra by writing
$$\left(\begin{matrix}a_{n}\\a_{n-1}\end{matrix}\right)= \left(\begin{matrix}\frac{5}{2}&-1\\1&0\end{matrix}\right) \left(\begin{matrix}a_{n-1}\\a_{n-2}\end{matrix}\right)= \left(\begin{matrix}\frac{5}{2}&-1\\1&0\end{matrix}\right)^{n-2} \left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)$$
By studying the matrix
$$A=\left(\begin{matrix}\frac{5}{2}&-1\\1&0\end{matrix}\right)$$
and looking at its eigenvectors and eigenvalues
$$\lambda_{1}=2\Longrightarrow v_{1}=\left(\begin{matrix}2\\1\end{matrix}\right)$$
$$\lambda_{2}=\frac{1}{2}\Longrightarrow v_{2}=\left(\begin{matrix}1\\2\end{matrix}\right)$$
you can see that for this multiplication to remain finite, it must contain only $v_{2}$.