In $2\times 2$ matrices, why is eigenvector for $\lambda_1$ contained in column space of $A-\lambda_2 I$? Suppose we have the matrix
$$A=\begin{bmatrix}4 & -3 \\ 2 & -1 \end{bmatrix} $$
which has eigenvalues $\lambda_1=1$ and $\lambda_2 = 2$.
Then
$$A-\lambda_1 I=\begin{bmatrix}3 & -3 \\ 2 & -2 \end{bmatrix}\quad \text{and} \quad A-\lambda_2 I=\begin{bmatrix}2 & -3 \\ 2 & -3 \end{bmatrix}.$$
Why do vectors in the column space for $A-\lambda_2 I$ act as eigenvectors corresponding to $\lambda_1$, and vice versa?
(E.g. $(2,2)^T$ is an eigenvector for $\lambda_1$.)
 A: Note that by Cayley-Hamilton theorem
$$(A-\lambda_1 I)(A-\lambda_2 I)= A^2 -(\lambda_1 + \lambda_2) A+\lambda_1 \lambda_2 I=0$$
taking $\vec u_2, \vec v_2$ as the column vectors of $A-\lambda_2 I$
We must have $(A-\lambda_1 I)\vec u_2= \vec 0$
and $(A-\lambda_1 I) \vec v_2 = \vec 0$
So the column vectors of $A-\lambda_2$ are in the null space of $A-\lambda_1 I$, so they are eigenvectors of $A$ corresponding to $\lambda_1$
A: Assuming that $A\in\mathbb R^{2\times2}$ has distinct eigenvalues $\lambda_1$ and $\lambda_2$, we can write any vector as a linear combination $a\mathbf v_1+b\mathbf v_2$ of eigenvectors corresponding to these eigenvalues. Since $A\mathbf v_2=\lambda_2\mathbf v_2$, we then have $$(A-\lambda_2I)(a\mathbf v_1+b\mathbf v_2) = a(\lambda_1-\lambda_2)\mathbf v_1,$$ that is, the image (column space) of $A-\lambda_2I$ is spanned by the eigenvector $\mathbf v_1$. Similarly, $(A-\lambda_1I)(a\mathbf v_1+b\mathbf v_2)=b(\lambda_2-\lambda_1)\mathbf v_2$.  
Observe that this suggests a decomposition of $A$ into a sum of projections $P_1$ and $P_2$ onto its eigenspaces, with $$P_1={A-\lambda_2I\over\lambda_1-\lambda_2}\text{ and }P_2={A-\lambda_1I\over\lambda_2-\lambda_1}.$$  
N.B.: This is essentially the same as Nick Liu’s answer. 
A: View  $A$ as a linear operator from $\Bbb R^2$ to $\Bbb R^2$,  
Note  $\Bbb R^2=E_{\lambda_1}\oplus E_{\lambda_1}$, where $E_{\lambda_i}$ denotes the eigenspace of $\lambda_i$.
and $E_{\lambda_1}=N(A-\lambda_1)=A-\text{col}(A-\lambda_1I)=\text{col}(A-\lambda_2I$), and similar case for $\lambda_2$.
Therefore any $v\in\text{col}(A-\lambda_2I)$ is an eigenvector of $\lambda_1$, and similar case for $\lambda_2$.
A: I am not quite satisfied with this answer, so I hope others can give better ones.
$A$ has distinct eigenvalues so it is diagonalizable with an eigenbasis $\{v_1,v_2\}$ corresponding to $\lambda_1$ and $\lambda_2$.
If you apply $A-\lambda_2 I$ on some vector $v=c_1v_1+c_2v_2$, then
$$(A-\lambda_2 I)(c_1v_1+c_2 v_2) = (A-\lambda_2 I)(c_1 v_1) = c_1(\lambda_1-\lambda_2)v_1 \in \operatorname{span}\{v_1\}.$$
A: Not the best answer; correct me if I'm wrong about this.
All you need to show is that the column space A−λ1I is invariant under A. Because what that'd mean for a 2x2 matrix is that the columns of A−λ1I span a line (if there are a more than one eigenvalues) and the invariant property means that the vectors in its column space only get scaled after applying A. Now since A−λ1I maps the eigenvectors corresponding to eigenvalue λ1 to 0 then the columns of A−λ1I must be on the line spanned by the eigenvector corresponding to eigenvalue λ2, for that's the only other invariant subspace under A.
To show that the column space of A−λ1*I is invariant under A:
suppose u is in the column space of A−λ1*I
then,
Au = Au
=Au + (λ1)u - (λ1)u
= (A−λ1I)u + λ1u
both of which by definition are in the column space of A−λ1*I.
