TITLE: Diagonalizable vs full rank vs nonsingular (square matrix)
QUESTION [14 upvotes]: There are many discussions of such type problems (comparison), for example:
Diagonalizable vs Normal 
Today, I want to clearly understand the  topic.
Suppose the matrix $A\in \mathbb{R}^n$.  

Since the multiplication of all eigenvalues is equal to the determinant of the matrix, $A$ full rank is equivalent to $A$ nonsingular. 
The above also implies $A$ has linearly independent rows and columns.  So $A$ is invertible.         
$A$ is diagonalizable iff $A$ has $n$ linearly independent eigenvectors. ($A$ is nondefective).
Note: $A$ is defective if geo. multiplicity $<$ alge. multiplicity.    
A diagonalizable matrix does not imply full rank (or nonsingular).

My problem is 

Does full rank matrix (nonsingular) imply it is diagonalizable?   

Equivalently:  

Does a matrix with all its columns or rows linear independently imply all its eigenvectors linear independently?

REPLY [20 votes]: Full rank and diagonalizability are independent from one another.

A diagonalizable matrix with full rank:
$$
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
$$
A non diagonalizable matrix with full rank:
$$
\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}
$$
A diagonalizable matrix without full rank:
$$
\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}
$$
A non diagonalizable matrix without full rank:
$$
\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}
$$