Dimension of set of commutable matrices

Let $A$ be an $n \times n$ complex matrix and $V = \{B\mid AB=BA\}$.

I've proved that $V$ is a vector space. How can I prove that $\dim V \ge n$ for any $A$?

Some hints:

1. Start by considering the case that $A$ is a Jordan block and work out the result for this case (this can be done by an explicit calculation).

2. Generalize this to a matrix in Jordan normal form (find "basis matrices" for each Jordan block, note they are linear independent and show that you get at least $n$ of them).

3. Generalize again to all complex matrices, by making use of the fact that if $J$ is the Jordan normal form of $A$, $J = M^{-1} A M$ for some invertible matrix $M$, and conjugating the basis matrices from step 2 with $M$.