Finding dim(Ker(A)) according to a given characteristic polynomial I was given this question:

1) a) is simple, the sum of the algebraic multiplicities is 6, so the matrix 6X6.
2) My problem is with b), how am i supposed to find the dimKerA?
What is the general idea? really stuck..
Thanks is advance
 A: Hint: $\lambda=0$ is an eigen value of the matrix with multiplicity $2$. $\dim \ker(A)$ is same as nullity of $A$. What can you say about the nullity of $A$?
A: That $A$ is diagonalizable is key here. For instance, the characteristic polynomial of the matrix $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is $\lambda^2$, but the dimension of its nullspace is 1, not 2. 
Depending on how much theory you have covered, you can use the fact that a matrix is diagonalizable if and only if the dimension of the eigenspace is equal to the algebraic multiplicity of the corresponding eigenvalue in the characteristic polynomial. For $\lambda = 0$, the eigenspace is the same as the nullspace. 
Or here is a more detailed explanation if you have not covered that theorem. Since $A$ is diagonalizable, there is a diagonal matrix $D$ such that $D = P^{-1}AP$. You probably know that the diagonal entries of $D$ are exactly the eigenvalues of $A$. It is clear, then, that $dim(Ker(D)) = 2$. So we just want to prove that $dim(Ker(D)) = dim(Ker(A))$.
Well, suppose the column vectors $\textbf{v},\textbf{w} \in Ker(D)$, are two basis vectors of $ker(D)$. So they are nonzero and linearly independent. Also
$$
P^{-1}AP\textbf{v} = \textbf{0} \Longleftrightarrow AP\textbf{v} = P\textbf{0} = \textbf{0},
$$
and similarly for $\textbf{w}$. Thus $P\textbf{v},P\textbf{w}  \in Ker(A),$ and since $P$ is invertible $P\textbf{v}$ and $P\textbf{w}$ are nonzero and linearly independent. So we know that $dim(Ker(A)) \geq 2$. Suppose $dim(Ker(A)) > 2$. Then there exists $\textbf{b} \in Ker(A)$, nonzero and linearly independent from $P\textbf{v}$, and $P\textbf{w}$. Since $P$ is invertible, $P^{-1}\textbf{b}, \textbf{v}, \textbf{w}$ are also linearly independent, and you can check by a similar calculation as above that $P^{-1}\textbf{b} \in Ker(D)$, which is a contradiction, since $dim(Ker(D)) = 2$. 
