Eigenspace and Eigenvalues (Comprehension Question) 
I'm a bit confused on a couple linear algebra topics..
Say I have a $5 \times  5$ matrix, with 3 eigenvalues     With One of
  the corresponding eigen-spaces being 1 dimensional Another
  corresponding eigen-spaces dimension being 2 dimensional 
Would it be possible for that matrix to NOT be diagonalizeable?

I would say YES because if the $3rd$ eigenvalue
and it's associated eigen-space dimension is not $2$
then it can't possibly be diagonalizeable.
Is this logical? Also would it be possible to have an eigen-space
with dimension $0$?
Also if anyone could help me understand how this relates to the geometric multiplicities and algebraic multiplicities it would be greatly appreciated!
 A: Yes you are right. To answer your other question, yes the third eigenvalue must have eigenspace with dimension $1$, since it does not make sense to have an eigenspace of dimension $0$.
More concretely, you are given that the geometric multiplicities (dimension of eigenspace) of the first two eigenvalues are $1$ and $2$ respectively. 
Since you are in a $5$-dimensional space, the geometric multiplicity of the third eigenvalue must be $1$ or $2$ (since the sum of the geometric multiplicities is $\le 5$).


*

*If it is $2$, then the sum of the geometric multiplicities is $5$, so the matrix would be diagonalizable, a contradiction.

*Thus the geometric multiplicity of the third eigenvalue must be $1$, as you reasoned.


On the other hand, the algebraic multiplicities will always add up to $5$ and each must be at least their geometric multiplicity. So the three eigenvalues' algebraic multiplicities must be $(1,2,2)$ or $(2,2,1)$ or $(1,3,1)$.

You can explicitly construct examples using a Jordan matrix. For example, the following matrix has algebraic multiplicities $1,2,2$.
$$\begin{bmatrix}
\lambda_1 \\
& \lambda_2 \\
&& \lambda_2 \\
&&& \lambda_3 & 0\\
&&&1&\lambda_3
\end{bmatrix}.$$
