I'm currently working on the question:
Suppose that $T:V\rightarrow V $ is a linear map on a 4 dimensional vector space $V$, and that $T$ has 3 different eigenvalues, one of which has a 2-dimensional eigenspace.
i) Must $T$ be diagonalizable?
ii) Can $T$ be diagonalizable?
My thoughts are yes, to both i) and ii). My justification is that the sum of the dimensions of the eigenspaces ($1+1+2$) is equal to the dimension of $V$ which is 4. Here I am assuming that the other two eigenspaces are of dimension 1.
Is my reasoning correct?
(It's not hard to prove the other eigenspaces have dimension 1, if you want)
The main thing is whether or not you've considered whether the different eigenspaces are linearly dependent or not.
For example, if you take the three one-dimensional subspaces of $\mathbb{R}^3$, the first having basis vector $[1,0,0]$, the second $[1,1,0]$, and the third $[0,1,0]$, then the sum of their dimensions is 1+1+1, but collectively they don't add up to a three-dimensional vector space!
If you've thought about this issue, and think in your problem that it's sufficiently obvious that it doesn't need explanation, then you're probably fine. But it's possible your professor may be skeptical and you should write on it.
If you haven't thought about this issue, then your proof isn't fine and you need to do so! Although you have a chance to get lucky and have your professor think you've thought of it.
Also, I assume you're invoking a theorem that having a complete set of eigenspaces means the matrix is diagonalizable? I'll leave it to your judgement whether the fact you are doing that needs to be said explicitly.