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?
 A: 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 
$$
\begin{bmatrix}
0\\
0 \\
1 \\
0\\
\end{bmatrix}.
$$
A: The geometric multiplicity (dimension of the eigenspace) of each of the eigenvalues of $A$ equals its algebraic multiplicity (root order of eigenvalue) if and only if the matrix $A$ is diagonalizable (i.e. for $A\in\Bbb K^{n\times n}$ there exists $P,D\in\Bbb K^{n\times n}$, where $P$ is invertible and $D$ is diagonal, such that $P^{-1}AP=D$). Just because the algebraic multiplicity of an eigenvalue is, in this case, equal to $2$, that does not mean that the corresponding geometric multiplicity is also $2$. The general relationship between the two multiplicities is that the algebraic multiplicity is greater than or equal to the geometric multiplicity. Furthermore, equality holds for all eigenvalues if and only if the matrix is diagonalizable. 
