What is the largest possible for $\dim(V)$ Let $L:V \to V$ be a linear transformation and let $V$ be a subspace of $\mathbb{R}^n$ for some $n\in \mathbb{N}$. Assume that $L^5=0$ and let $\dim (\ker(L))=7$. What is the largest possible value of $\dim(V)$?
Since rank and nullity theorem
$$\dim(V)=\dim(\ker L)+\dim(\operatorname{Im}(L)) =7+4=13$$
because $\dim (\ker(L))=7$.
Am I right?
 A: Unless I'm misreading the question badly, I have no idea why we should assume that $L$ is rank 4. Here's an example to think about:
Let $I_7$ denote the $7 \times 7$ identity matrix, and consider the block matrix
$$L = \begin{pmatrix} 0 & I_7 & 0 & 0 & 0 \\ 0 & 0 & I_7 & 0 & 0 \\ 0 & 0 & 0 & I_7 & 0 \\ 0 & 0 & 0 & 0 & I_7 \\ 0 & 0 & 0 & 0 & 0\end{pmatrix}$$
(by $0$ I again mean a $7 \times 7$ block of zeroes, so this is a $35 \times 35$ matrix). It's easy to see that $L^5 = 0,$ and that the kernel of $L$ is exactly the first seven basis vectors.
Can you convince yourself that this is the largest example of this behavior? A Rank-Nullity argument may still be helpful.
Edit: To be quite clear, my $L$ acts on a 35-dimensional space, and I am claiming this is the largest dimension you can have. One approach would be to prove (it's not hard) that the nullity of $L^k$ is at most $k$ times the nullity of $L.$ Thus in your example, $\text{null}(L^5) = \dim V \leq 5\cdot7.$
A: $L^5 = 0$ implies that $L$ is nilpotent and the minimal polynomial of $L$ is $\mu_L(x) = x^k$ where $1 \le k \le 5$.
Recall that $k$ is the size of the largest Jordan block in the Jordan form for $L$.
On the other hand, $\dim\ker L = 7$ is precisely the number of Jordan blocks.
Since $\dim V$ is the sum of sizes of all Jordan blocks, we get:
$$\dim V = k \cdot \dim\ker L\le 7 \cdot 5 =35$$
To see that the maximum is indeed attained, consider the Jordan matrix with seven $5 \times 5$ blocks.
