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)?

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
$$\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}
   \begin{pmatrix} x \\ y \end{pmatrix}=  \begin{pmatrix} 3x \\ 3y \end{pmatrix}$$
$$\mathbf{x}_2 =  \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
