TITLE: $T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ has 2 distinct eigenvalues. Showing that $v$ or $T(v)− \lambda_1v$ are eigenvectors of $T$
QUESTION [5 upvotes]: $T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ which is diagonalizable with 2 distinct eigenvalues. Showing that either $v$ is an eigenvector for $\lambda_1$ or else $T(v)− \lambda_1v$ is an eigenvector for $\lambda_2$.
I tried to use the matrix representation of $T$ and the Cayley-Hamilton theorem, but I can't reach the conclusion. Is there a way to prove it without the matrix representation?
I would prefer to get guidance rather than a full solution.

REPLY [6 votes]: I will assume that $v\ne0$; otherwise, the statement is false.
Let $v_1$ be an eigenvector corresponding to the eigenvalue $\lambda_1$ and let $v_2$ be an eigenvector corresponding to the eigenvalue $\lambda_2$. If $v\in\Bbb R^2$, then $v$ can be written as $\alpha_1v_1+\alpha_2v_2$. There are two possibilites now:

$\alpha_2=0$: then $v=\alpha_1v_1$ and therefoe it is an eigenvector corresponding to the eigenvalue $\lambda_1$.
$\alpha_2\ne0$: then $T(v)-\lambda_1v=\alpha_2(\lambda_2-\lambda_1)v_2$ and therefore $T(v)-\lambda_1v$ is an eigenvector corresponding to the eigenvalue $\lambda_2$.

REPLY [5 votes]: In the second case, consider
$$
T \bigl( T(v)-\lambda_1 v \bigr) = T^2(v) - \lambda_1 T(v)
.
$$
By Cayley–Hamilton, $T^2(v)$ can be replaced with $(\lambda_1+\lambda_2)T(v) - \lambda_1 \lambda_2 v$.
Can you continue from there?