Find the Jordan canonical form [closed]

$$N$$ 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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Calvin Khor, John B, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hi. Please show us what you've tried and any partial progress you've made. – Yuval Gat Nov 6 '18 at 10:53
• 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 – Raghav Maheshwari Nov 6 '18 at 10:56

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.