Why does factoring out eigenvector and result in the identity scaled by the eigenvalue? In deriving a method to find the eigenvalues (and corresponding eigenvectors of a linear mapping), we start with the definition:
$$Av = \lambda v$$
$$Av - \lambda v = 0$$
$$(A - \lambda I)v = 0$$.
I am confused about how to arrive at the third step. Whenever I've seen this, the author seems to take this fact for granted and not explain its significance.
My question is $\textbf{twofold}$ and perhaps the second will follow easily from an answer to the first. 
1) What definitions leads to the the result on the third line?
2) What is the geometric intuition for the action of what is happening in to the linear mapping by subtracting the eigenvalue the $i$th element of each $i$th column vector?
Thanks in advance.

Say $V$ is a vector space over a field $\mathbb F$ and $A, B : V \to V$ are linear maps. Then, for any $\lambda,\mu \in \mathbb F$, the linear map $\lambda A + \mu B : V \to V$ is defined by
$$(\lambda A + \mu B) (v) =  \lambda A(v) + \mu B(v)$$
for all $v \in V$.
