Number of eigenvalues and their eigenspaces So Let matrix $A$ have eigenvalues as follows :
$$
e_1=0\\
e_2=0\\
e_3=2\\
e_4=2\\
$$
From here can we deduce that dimension of the eigenspace when the eigenvalues is $2$ is 2? can we deduce this? 
If we could deduce that we could also deduce that dimension of the nullspace is $2$ since $e_1=e_2=0$ two eigenvalues pointing at $0$
To clear the question  a bit :
Can we conclude that rank of the eigenspace of a specific eigenvalue is equal the number of repetition of the eigenvalue?

No, think of
$$
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & 2 \\
\end{bmatrix}.
$$
The eigenspace relative to the eigenvalue $0$ has dimension $1$, generated by 
$$
\begin{bmatrix}
1\\
0 \\
0 \\
0\\
\end{bmatrix}.
$$
The eigenspace relative to the eigenvalue $2$ has dimension $1$, generated by 
$$
\begin{bmatrix}
0\\
0 \\
1 \\
0\\
\end{bmatrix}.
$$
