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$N$ is a nilpotent $15\times15$ matrix over $\mathbb{R}$ such that $$\dim(\ker N) = 5, \quad \dim (\ker{N^2}) =8, \quad \dim(\ker{N^3})= 11,$$ $$\dim (\ker{N^4}) = 13, \quad \dim(\ker{N^5}) =15$$ Find the Jordan form.

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closed as off-topic by Namaste, Calvin Khor, John B, Leucippus, Cesareo Nov 7 '18 at 0:59

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  • $\begingroup$ Hi. Please show us what you've tried and any partial progress you've made. $\endgroup$ – Yuval Gat Nov 6 '18 at 10:53
  • $\begingroup$ So i calculated the characteristic polynomial that is x^15 , minimal polynomial that is x^5 , no of jordan blocks that is equal to 5( dim of the only eigen value) and the dimension of the first jordan block that will be 5x5. How should i proceed now $\endgroup$ – Raghav Maheshwari Nov 6 '18 at 10:56
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Since $\dim\text{Ker}(N^4)= 13$ and $\dim\text{Ker}(N^5)= 15$ the maximal Jordan block is of size 5, and you have exactly two Jordan blocks $J_1$ and $J_2$ of size 5, indeed, if there are no such blocks the $N^4=0$ otherwise if there is just one such block then $\dim\text{Ker}(N^4)= 14$. So it remains to calculate other blocks. Since $\dim\text{Ker}(N)= 5$ and the blocks $J_1$ and $J_2$ give you 2 vector of the kernel, then the other block must have in total 3 vector of the kernel, so the third jordan block cannot be of size 5 neither 4. Assume that $J_3$ has size 3 hence it gives you one vector of the kernel, and so the only other possibility are $J_4=J_5=0$ the $1\times 1$ $0-$matrix. Actually in this case you have $\dim\text{Ker}(N)= 5$ $\dim\text{Ker}(N^2)= 2+2+2+1+1=8$,$\dim\text{Ker}(N^3)= 3+3+3+1+1= 11$ and $\dim\text{Ker}(N^4)= 4+4+3+1+1=13$. Moreover it is easy to check that other combinations do not satisfie all condition.

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  • $\begingroup$ That was really helpful. Thanks $\endgroup$ – Raghav Maheshwari Nov 6 '18 at 11:21

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