When do the eigenvectors of a $n\times n$ matrix span $\mathbb{R}^n$? (aka when is a matrix diagonalizable?) When learning Linear Algebra, I learned that rotation matrices don't have real eigenvalues or eigenvectors, because their characteristic polynomial does not factor over $\mathbb{R}$. Over the complex numbers it does, so an $n$-dimensional rotation matrix has $n$ linearly independent eigenvectors.
Somehow I concluded that over the complex numbers, all $n\times n$-matrices have $n$ linearly independent eigenvectors. But the following shear matrix does not:
$$
S = \left(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}\right)
$$
This matrix has two eigenvalues $\lambda_{1,2} = 1$, and the eigenvector $(1,0)$. So the question arises, what is the other eigenvector? Mathematica says $(0,0)$ is the other eigenvector, but I find that a cheap answer, since $(0,0)$ is trivially an eigenvector of every matrix.
So, my first question is, does one really consider $(0,0)$ an eigenvector of $S$?
Second question: Is there a condition for an $n\times n$-matrix to have $n$ linearly independent eigenvectors? Maybe the symmetry of the matrix? Maybe that the eigenvalues are nondegenerate?
 A: No, $(0,0)$ is not an eigenvector. The matrix $S$ you wrote down has only one eigenvector. It has one eigenvalue, $1$, which has a algebraic multiplicity (the number of times it appears as a root of the characteristic polynomial) of $2$ and a geometric multiplicity (the dimension of its eigenspace) of $1$.
In general, a $n\times n$ matrix does not have $n$ linearly independent eigenvectors. For a detailed explanation, I suggest you read about the Jordan canonical form of a matrix, and connected to that, algebraic and geometric multiplicity of eigenvalues, invariant subspaces and characteristic/minimal polynomials. The entire theory behind this is not very complicated (it's taught usually at the end of an introductory linear algebra class), but it does take a couple of lessons to fully cover.
However, there are cases when we can be sure that the matrix has $n$ eigenvectors (in other words, we say that the matrix is diagonalizable). One simple condition that assures this is if the matrix is symmetric.
A: The general criterion is this:

An $n\times n$ matrix with coefficients in a field $K$ is diagonalisable over $K$ if and only if its minimal polynomial splits as a product of distinct linear factors.

By the Hamilton-Cayley theorem, a matrix $A$ is a root of its characteristic polynomial, hence $\{p(x)\in K[x]\mid p(A)=0\}$ is a non-zero ideal in $K[x]$. A monic generator of this ideal is the minimal polynomial of $A$. It is a divisor of the  characteristic polynomial.
One  proves the minimal polynomial  and the characteristic polynomial have the same roots.
Example:
If $A=\begin{bmatrix}0& 1&0\\0&0&0\\0&0&0\end{bmatrix}$, the characteristic polynomial of $A$ is $-x^3$, but its minimal polynomial is $x^2$.
