Are all $v \in N(A - \lambda I)$ eigenvectors or are just some $v \in N(A - \lambda I)$ eigenvectors Let's say I have an arbitrary matrix $A$. 
An eigenvector of $A$ would be a vector $\vec{x}$, such that 


*

*$A\vec{x} = \lambda \vec{x}$

*$\vec{x} \in N(A-\lambda I)$ i.e. $(A - \lambda I)\vec{x} = \vec{0}$


where $\lambda$ is an eigenvalue of $\vec{x}$

Now are all $\vec{x} \in N(A-\lambda I)$ eigenvectors of $A$, i.e. do all $\vec{x}$ in the nullspace of $A - \lambda I$ also automatically satisfy condition $(1)$, that $A\vec{x} = \lambda \vec{x}$?
Written more concretely is the following statement true:
$$\vec{x} \in N(A-\lambda I) \implies A\vec{x} = \lambda \vec{x} \ \ \ \ \ \forall   \vec{x} \in N(A-\lambda I)$$
If that was the case then the eigenvectors, would $span$ the entire nullspace. 
Furthemore what relation does $N(A-\lambda I)$ have to $N(A)$ and the eigenvectors? If $\vec{x} \in N(A-\lambda I)$, does that imply that $\vec{x} \in N(A)$? Can we deduce anything about the eigenvectors of $A$ from $N(A)$ or can we only deduce them from $N(A - \lambda I)$?
 A: The answer to your title question is "just some", because there is a single exception: the null vector.
By definition, an eigenvalue of $A$ relative to $\lambda$ is a nonzero vector $x$ such that $Ax=\lambda x$ (forgive me if I don't use arrows for vectors). Now
$$
Ax=\lambda x
\iff
Ax-(\lambda I)x=0
\iff
(A-\lambda I)x=0
\iff
x\in N(A-\lambda I)
$$
so the only exception is the vector that is not an eigenvector by definition, that is the zero vector.
Of course, if your definition allows $0$ to be an eigenvector, then the answer to the title question is "all". However, most books on the subjects don't consider $0$ an eigenvector, so check carefully yours.

About relationships between the eigenvalues/eigenvectors and $N(A)$, I'm afraid there's none: consider simply the case of an invertible matrix, which can have any set of nonzero eigenvalues compatible with its order, but has $N(A)=0$.
A: I'll to answer one question at a time.
First of all, your claims 1 and 2 are the same.
if $Ax =\lambda x$, then note that $Ax = \lambda I x$ and thus we can factor out the $x$ on both sides and then subtract the RHS from the LHS, getting $(A - \lambda I)x =0.$ This means that $x \in N(A - \lambda I).$
In other words, the claims $x \in N(A - \lambda I)$ and $Ax =\lambda x$ are equivalent.
By inspecting the equation $(A - \lambda I)x =0,$ evidently $\alpha x$ also works for any $\alpha \in \mathbb{R}$. Thus you are correct in saying that any vector in $N(A - \lambda I)$ is also an eigenvector corresponding to the eigenvalue $\lambda.$
In other words, any vector in the subspace $N(A - \lambda I)$ is an eigenvector of $A$ corresponding to $\lambda.$
So you could say that $(\lambda, N(A - \lambda I))$ is an eigenvalue-eigensubspace pair for $A$, but nobody does.
Typically we use normalized eigenvectors, i.e., we use the eigenvector such that $||x||=1.$
This narrows it down to $\alpha = \pm 1,$
so there's still a bit of abiguity there.
When you say that the eigenvectors span the nullspace, you are sort of correct. Be more precise: "any eigenvector of $A$ corresponding to $\lambda$ spans the nullspace of $A - \lambda I$"
Last paragraph of the question: any eigenvector corresponding to an eigenvalue of 0 does lie in the null space of $A$ (why?).
You might remember reading somewhere that the rank of a matrix is related to the dimension of its null space and the number of eigenvalues equal to zero.  I'll let you draw the rest of the connection here.
A: The first statement is correct by the definition of nullspace, as $x\in N(A-\lambda I)$, then all those vectors should fulfill the relationship $(A-\lambda I)x=0$, from where we can derive easily the relation $Ax=\lambda x$.
For the statement you make that if $x\in N(a-\lambda I)$ then $x\in N(A)$, it is false as it can be seen in the next counterexample. Consider matrix A:
$\left( \begin{array}{ccc}
2 & 0 \\
0 & 1 \\
 \end{array} \right)$, it can be clearly seen that the nullspace of this matrix is zero dimensional. However, it has eigenvectors $a=(1,0),b=(0,1)$, which form the nullspace of the matrix $A-\lambda I$, that is the zero matrix, so being in that nullspace does not imply to be in the nullspace of $A$
