Showing that either $v$ is an eigenvector for $2\times 2$ matrix $A$ or else $(A − \lambda \operatorname{Id})v$ is an eigenvector for $A$. Suppose the $2 \times 2$ matrix $A$ has repeated eigenvalues $\lambda$. Let $v \in \mathbb{R}^{2}$. Using the fact that a $2\times 2$ matrix $A$ always satisfies its own characteristic equation. (That is, if $\lambda^2 + \alpha\lambda + \beta = 0$ is the characteristic equation associated to $A$, then the matrix $A^2 + \alpha A + \beta I$ is the $0$ matrix), I must show that either $v$ is an eigenvector for $A$ or else $(A − \lambda \operatorname{Id})v$ is an eigenvector for $A$.
I know that if the matrix has repeated eigenvalues the characteristic polynomial should look like $(a-\lambda)^2 - bc = 0$ for $A = \begin{bmatrix}a & b \\ c & a \end{bmatrix}$. I am stuck on using the fact that $A$ satisfies its own polynomial to show the final result. Also once I have shown $(A-\lambda \operatorname{Id})v$ is an eigenvector if $v$ is not do I need to show the other direction, i.e. that $v$ is an eigenvector if $(A-\lambda \operatorname{Id})v$ is not?
 A: Are you given that $a=d$? Because that neither implies, nor is implied by, the fact that $A$ has a repeated eigenvalue. Rather, your characteristic polynomial will be $(x-\lambda)^2$, so you know $(A-\lambda I)^2=O$.
So, for any $\mathbf{v}$ you have $(A-\lambda I)(A-\lambda I)\mathbf{v}=(A-\lambda I)^2 \mathbf{v}=O\mathbf{v}=\mathbf{0}$.
Therefore, either $(A-\lambda I)\mathbf{v}$ is already zero, or it's an eigenvector. 
A: Let $\lambda$ be the single repeated eigenvalue of $A$; then the characteristic polynomial $\chi_A(x)$ is well-known to be
$\chi_A(x) = x^2 - 2\lambda x + \lambda^2, \tag 1$
which may be written as
$\chi_A(x) = x^2 - 2 \lambda x + \lambda^2 = (x - \lambda)^2, \tag 2$
since
$\text{trace}(A) = \lambda + \lambda = 2\lambda, \; \det(A) = \lambda \lambda = \lambda^2; \tag 3$
by Cayley-Hamilton, $A$ satisfies $\chi_A(x)$, hence
$(A - \lambda)^2 = 0; \tag 4$
this may be written in the form
$A(A - \lambda) - \lambda(A - \lambda) = (A - \lambda)^2 = 0, \tag 5$
or
$A(A - \lambda) = \lambda(A - \lambda); \tag 6$
suppose $V$ is not an eigenvector of $A$; then we have
$AV \ne \lambda V, \tag 7$
or
$W = (A - \lambda)V \ne 0; \tag 8$
then via (6),
$AW = A(A - \lambda)V = \lambda (A - \lambda)V = \lambda W, \tag 9$
and since as has been mentioned in (8)
$W \ne 0, \tag{10}$
we see that
$(A - \lambda)V = W \tag{11}$
is an eigenvector of $A$.
