A simple linear algebra question. Seems that I forgot quite many of eigenvalues, vectors, and spaces stuff. 
The question is, assuming we have $3 \times 3$ matrix, namely $A$. If $A$ has a single eigenvalue, what is the dimension of corresponding eigenspace?
The answer turned out to be 1, 2, or 3. Could someone please explain me why?
 A: It seems like it should be $1$, but instead it's $1, 2$ or $3$ because one eigenvalue may have multiple eigenvectors.
For a a minimal example, let $A = I$, the identity matrix. $1$ is an eigenvalue of $I$ because $I\mathbf{x} = \mathbf{x} = 1\mathbf{x}$ for all vectors $\mathbf{x}$. It's the only eigenvalue of $I$.
All vectors are eigenvectors of $I$, and they all have eigenvalue $1$. So the eigenspace of $1$ (set of all eigenvectors having that eigenvalue) is the whole space.
If only one eigenvector was associated with that eigenvalue, the dimension would be $1$. If two eigenvectors, the dimension would be $2$.
A: This really depends on the matrix. The following 3 matrices, which have eigenvalue $\lambda$, have an eigenspace of dimensions 1,2, and 3 respectively:
$A=\begin{bmatrix}
\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda 
\end{bmatrix}, \qquad \begin{bmatrix}
\lambda & 1 & 0\\
0 & \lambda & 0\\
0 & 0 & \lambda 
\end{bmatrix}, \qquad \begin{bmatrix}
\lambda & 0 & 0\\
0 & \lambda & 0\\
0 & 0 & \lambda 
\end{bmatrix}$
Note that without loss of generality and for simplicity, we can use Jordan canonical form representation. 
For any given matrix, either you transform to Jordan form, or if dimension permits, compute the eigenvectors.
A: It solely depends on the given matrix A. As for example, if A is a null matrix of order n then zero is the only eigenvalue of A. However, corresponding to it, the homogeneous system
| A-0I | x=0 where x is a vector
gives exactly n linearly independent eigenvectors.
