Flag of subspaces

Let $T$ be a $4\times 4$ real matrix such that $T^4=0$. Let $k_i =\dim \ker T^i$ for $i=1,2,3,4$.

Then, which of the following is not a possibility for the sequence $k_1\leq\ k_2\leq\ k_3\leq\ k_4$ ?

1. $3\leq4\leq4\leq4$

2. $1\leq3\leq4\leq4$

3. $2\leq4\leq4\leq4$

4. $2\leq3\leq4\leq4$

well, it was asked previously but didn't receive an elaborate answer.

• Can you link to where it was asked previously? May 1, 2017 at 13:08
• math.stackexchange.com/q/1578368/370778 May 1, 2017 at 13:11
• What did you try? Where did you get stuck? Did you understand the answer's point about Jordan canonical form? May 1, 2017 at 13:16
• well, the answer is there.. Answer 1 corresponds to the Jordan matrix with only one block of size 4, Answer 3 comes from one with two Jordan blocks with size 2 each, and Answer 4 from one with size 1 and one with size 3. However, you can't get Answer 2 that way. May 1, 2017 at 13:17
• Nullity of $T$ will imply the number of jordan blocks related to $0$. That's all I could know. I din't quite get that answer. The question was actually asked in a national level test for entrance to Phd in India in $2015$ and the answer was given to be the second option May 1, 2017 at 13:29

Hint: Suppose that $T$ is in Jordan canonical form (without loss of generality). Note that for all $i$, $$\dim \ker T^i - \dim \ker T^{i-1}$$ is the number of Jordan blocks with size at least $i$ (here, $T^0$ is the identity matrix).