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.
 A: The third line comes from applying the distributive property of multiplication over addition. If $A$ and $B$ are matrices and $v$ is a vector then $Av-Bv=(A-B)v$ always holds. The case for vectors follows from the case where $v$ is a matrix by thinking of it as a $n\times 1$ matrix.
Geometrically, this means that an eigenvector lies in the kernel of $A-\lambda I$ where $\lambda$ is the corresponding eigenvalue.
A: The transition from the second to the third line depends on the observation that the linear transformations themselves form a vector space - you can add them and multiply them by scalars by using those operations on the vectors in the range. (That's essentially what @Stefano 's answer says.)
The third line tells you that the eigenvector $v$ is in the kernel of the linear transformation $A - \lambda I$. I don't know whether that qualifies as geometric information for you.
A: 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 $v \in V$.
A: This is not (quite) an answer, but what I hope is a helpful notational hint.  
I think the OP's confusion may stem from the fact that the left-hand side is a vector and the right-hand side seems like a scalar.  To avoid such ambiguity, in my publications I always use bold face for vectors and matrixes and simple italics for scalars.  
As such, I would have written the second equation as:
${\bf A}{\bf \nu} - \lambda {\bf \nu} = {\bf 0}$,
clarifying that ${\bf \nu}$ is a vector, that $\lambda$ is a scalar, and that ${\bf 0}$ is a vector of $0$s.
