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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.

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  • $\begingroup$ Can you link to where it was asked previously? $\endgroup$
    – Joppy
    May 1, 2017 at 13:08
  • $\begingroup$ math.stackexchange.com/q/1578368/370778 $\endgroup$
    – Hirakjyoti Das
    May 1, 2017 at 13:11
  • $\begingroup$ What did you try? Where did you get stuck? Did you understand the answer's point about Jordan canonical form? $\endgroup$
    – Ben Grossmann
    May 1, 2017 at 13:16
  • $\begingroup$ 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. $\endgroup$
    – Sebastian Schulz
    May 1, 2017 at 13:17
  • $\begingroup$ 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 $\endgroup$
    – Hirakjyoti Das
    May 1, 2017 at 13:29

1 Answer 1

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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).

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