Algebraic and geometric multiplicities of eigenvalues of a $3 \times 3$ matrix "Let A = \begin{pmatrix}-1&-1&-2\\ \:2&2&1\\ \:6&2&6\end{pmatrix}
Find the characteristic polynomial of A. 
Find the distinct eigenvalues of A and their respective algebraic and geometric multiplicities."
So I calculated the value of the characteristic polynomial and got x^3 -7x^2 + 16 x -12 or (x - 3)(x - 2)(x - 2) and got the eigenvalues of 2 and 3. I believe these are the correct answers. However, I am not sure how to give the geometric and algebraic multiplicities for the eigenvalues. I thought that for the eigenvalue 3 the multiplicity would have been 1 and for the eigenvalue 2 it would have been 2. But it seems like I need two multiplicity values for each.
Any help?

The algebraic multiplicity of an eigenvalue $\lambda$ of an endomorphism $A$ is its multiplicity as a root of the characteristic polynomial $\chi_A$ of the endomorphism, in other words it is the greatest exponent $m$ such that $(x-\lambda)^m$ divides $\chi_A$.
Its geometric multiplicity is the dimension of the eigenspace $E_\lambda$. 
One has the inequalities:
$$1\le \text{geometric multiplicity}(\lambda)\le\text{algebraic multiplicity}(\lambda). $$
A necessary and sufficient condition for an endomorphism to be diagonalisable is that all of its eigenvalues have geometric multiplicity equal to its algebraic multiplicity.
This condition is automatically satisfied for eigenvalues which are  simple roots of $\chi_A$. There results that an endomorphism with only simple eigenvalues is diagonalisable.
