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Let $T$ be a $4\times 4$ real matrix such that $T^4=0$ .Let $k_i=\dim \ker T^i;1\le i\le 4$.

Is $1\le 3\le 4\le 4$

a possibility for the sequence $k_1\le k_2\le k_3\le k_4$?

My try:

Consider the map $T$ given by $T(e_1)=0;T(e_2)=T(e_3)=e_1;T(e_4)=e_2$ where $\{e_i:1\le i\le 4\}$ is the standard basis of $\Bbb R^4$.Here $k_1=1$

Then $T^2(e_2)=T^2(e_3)=0;T^2(e_4)=e_1$.Here $k_2=3$

Also $T^3(e_i)=T^4(e_i)=0\forall i$ .Here $k_3=k_4=4$

Hence such a possibility is there.

But unfortunately the answer is given as NO.

Where am I going wrong in the given one?Please help.

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  • $\begingroup$ dim ker T=2 not 1. As well as $T(e_1)=0$ you have $T(e_2-e_3)=0$. $\endgroup$
    – almagest
    Jun 16, 2016 at 6:49
  • $\begingroup$ This will help math.stackexchange.com/questions/1709475/… $\endgroup$
    – Kushal Bhuyan
    Jun 16, 2016 at 7:13
  • $\begingroup$ I got your point ;But can it be proven that such a possibility never exists @almagest $\endgroup$
    – Learnmore
    Jun 16, 2016 at 7:23
  • $\begingroup$ @learnmore Have you looked at the question KushaiBhuyan suggested? Are you familiar with the Jordan form? $\endgroup$
    – almagest
    Jun 16, 2016 at 7:26
  • $\begingroup$ Another one math.stackexchange.com/questions/1711949/…. Jordan forms is the easiest way to understand this problem. Are u familiar with this form? $\endgroup$
    – Kushal Bhuyan
    Jun 16, 2016 at 7:28

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