I am having trouble with explaining my answer and I was wondering if I could receive some guidance on how to better phrase it.
Question: Consider a $2$x$2$ real symmetric matrix $M$. Suppose that is eigenvalues are $\lambda_1$ and $\lambda_2$ such that they are not equal with corresponding eigenvectors $w_1$ and $w_2$. (I apologize for not using proper vector notation with the arrows on top.
Explain why an arbitrary vector $a$ can be written as a linear combination of the eigenvectors, aka, $a=c_1w_1+c_2w_2$ where $c_1,c_2 \in R$.
b) what is the result of $Ma$?
c) result of $M^2a$? $M^3a$?
d) From this idea, explain how you can find one of the eigenvectors of $M$ by finding $M^nw$ as $n$ approaches infinity. Why does this work regardless of vector you pick? Which eigenvalue will the eigenvector correspond to?
I can show explicitly the linar combination and do the same thing for part b and c, but that is not what I want to do. For part d I know that for iterative matrix multiplication, the end vector will have the same angle from the origin as the eigenvector with the largest eigenvalue. However, how does one find one of the eigenvecors of $M$ by calculating $M^nw$? It works because of part a. Essentially the vector can be formed by the linear combination, hence...
the reason why $a=c_1w_1+c_2w_2$, a linear combination of eigenvectors , is because for distinct eigenvalues we got two linearly independent eigenvectors, which in turns spans R^2.
in general,$ M^na$ is still a vector in $R^2$ so the eigenvectors can still span $M^na$.
now for part (d), take an arbitary vector $a$ in R^2. write $a$ as $a=c_1w_1+c_2w_2$. Assume $c_1$ and $c_2$ are both nonzero, (if $c_1$ or $c_2$ is zero, $a$ is already an eigenvector, which we are done)
$ A^na= A^n(c_1w_1+c_2w_2) = c_1A^nw_1+c_2A^nw_2 = c_1\lambda_1^nw_1 +c_2\lambda_2^nw_2 $
now divide by the bigger eigenvalue n times, say $\lambda_1$
$\frac{A^na}{\lambda_1^n}= c_1w_1 +c_2(\frac{\lambda_2}{\lambda_1})^nw_2 \rightarrow c_1w_1$
so this way you found an eigenvector $w_1$, corresponding to the bigger eigenvalue