Eigenvalues with no eigenvectors Can a matrix (example 2x2) with 2 distinct eigenvalues have no eigenvetors?
(since, the A-3I as well as A-1I both are coming out as invertible with -3 and -3 with only 0 vector in the nullspace)?
 A: If $\lambda$ is an eigenvalue of $A$, then that means (by definition) $A\vec{v} = \lambda\vec{v}$ for some $\vec{v}$.
If such a $\vec{v}$ exists, then it's an eigenvector. If such a $\vec{v}$ doesn't exist, then $\lambda$ isn't an eigenvalue (again by definition).
A: As Draconis wrote, the usual definition of an eigenvalue already states that there's an eigenvector. But let's assume you do an unusual definition that says “an eigenvalue is a root of the characteristic polynomial” (normally that's a theorem, but let's assume that this is your definition).
The characteristic polynomial of $A$ is $p(x) = \det(xI - A)$ (sometimes it is defined with the opposite sign, but that's irrelevant for the roots), where $I$ is the identity matrix.
Now if $\lambda$ is a root of the polynomial, that means $p(\lambda) = \det(\lambda I-A) = 0$. But that implies that $\lambda I - A$ is not invertible, that is, its null space is of positive dimension.
Be $v$ a non-zero element of that null space. Then we have, by definition, $(\lambda I - A)v = 0$. Basic algebra then shows that this is equivalent to $Av = \lambda v$. In other words, $v$ is an eigenvector.
A: If a matrix has an eigenvalue, then pretty much by definition (or an easy theorem), it has an eigenvector. 
A: By definition, a matrix has eigenvalue $\lambda$ and eigenvector $\mathbf{x}$ when $A \mathbf{x}=\lambda \mathbf{x}$ holds for some $\lambda$ and a non-zero vector $\mathbf{x}$. 
To address the matrix example in the comments ...
$$A=\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}$$
The eigenvalues are $\lambda_1 =1$ and $\lambda_2=3$.
The eigenvectors are given by 
$$\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}
   \begin{pmatrix} x \\ y \end{pmatrix}=  \begin{pmatrix} x \\ y \end{pmatrix}$$
$$\mathbf{x}_1 =  \begin{pmatrix} \phantom{-}1 \\ -1 \end{pmatrix}$$
and
$$\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}
   \begin{pmatrix} x \\ y \end{pmatrix}=  \begin{pmatrix} 3x \\ 3y \end{pmatrix}$$
$$\mathbf{x}_2 =  \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
