# 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

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• 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.