Let $N\in \text{Mat}(10 \times 10,\mathbb{C})$ be nilpotent. Furthermore let $\text{dim} \ker N =3 $, $\text{dim} \ker N^2=6$ and $\text{dim} \ker N^3=7$. What is the Jordan Normal Form?

The only thing I know is that there have to be three blocks, since $\text{dim} \ker N = 3$.

Thank you very much in advance for your help.

Travis Willse
  • 83,154
  • 8
  • 100
  • 207
  • 754
  • 4
  • 12

3 Answers3


Since $N$ is nilpotent, it has $0$ as unique eigenvalue. Since $\dim\ker N=3$, by the rank nullity theorem, the rank of $N$ is $7$. Hence the JNF of $N$ has seven 1's. Assume that there are $k$ blocks, and denote by $n_i$ the size of the $i$-th block. Of course, $n_1+\cdots+n_k=10$. Now, the rank of $N$ is $(n_1-1)+(n_2-1)+\cdots+(n_k-1)=10-k$, and we know that this rank is 7 (by rank nullity theorem). Hence k=3. You already got to that point.

Now, let $p$ be the number of blocks of size $1$. Then, the rank of $N^2$ is $(n_1-2)+\cdots+(n_k-2)+p=10-2k+p$. Since $\dim\ker N^2=6$, we can conclude (by the rank–nullity theorem again) that the rank of $N^2$ is $4$. Hence (since $k=3$) we conclude that $p=0$. So all blocks have a size at least 2.

Finally, let $q$ be the number of blocks of size $2$. Then, the rank of $N^3$ is $(n_1-3)+\cdots+(n_k-3)+q=10-3k+q$. Since $\dim\ker N^3=7$, we conclude that the rank of $N^3$ is $3$. Since $k=3$, we conclude that $q=2$. Hence $N$ has 2 blocks of size $2$ and one block of size $6$: we hence conclude that the JNF is (up to reordering of the blocks): $$\left(\begin{array}{cc|cc|cccccc}0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0\\\hline0&0&0&0&0&1&0&0&0&0\\0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&0&0&0\end{array}\right).$$

  • 3,960
  • 14
  • 19
  • Where does the equation for the rank of N^2( or N^3) come from? – Hans Feb 08 '16 at 12:40
  • 1
    @Jan: Take a block $J$ of size $n$ (with eigenvalue 0). Clearly, the rank of $J$ is $n-1$. Now, $J^2$ will have one less $1$'s than $J$ has (work out the multiplication by hand at least once; it's very enlightening). Hence the rank of $J^2$ is $n-2$. But wait, this can't be true if $n=1$, so the rank of $J^2$ is $n-2$ only if the block is of size $\geq2$. Otherwise, if $J$ is of size $1$, the rank of $J^2$ is $0$. In the answer, for the rank of $N^2$, the sizes of the blocks is taken care of by adding to $(n_1-2)+\cdots+(n_k-2)$ the number $p$ of blocks of size $1$. – gniourf_gniourf Feb 08 '16 at 12:46
  • I think I get it! Very enlightening indeed! Thanks!! – Hans Feb 08 '16 at 13:41

This can be seen in term of partitions.

An $n \times n$ nilpotent matrix $N$ can be described via a partition $$ p = (n_{1}, n_{2}, \dots, n_{k}) $$ of $n$, with $n_{1} \ge n_{2} \ge \dots \ge n_{k} > 0$, which records the size of the nilpotent Jordan block in a Jordan normal form.

Now one can show (it is really straightforward) that for the dual partition $q$ of $p$ one has $$ q = (\dim(\ker(N)), \dim(\ker(N^{2})) - \dim(\ker(N)), \dim(\ker(N^{3})) - \dim(\ker(N^{2})), \dots). $$

In your case $$ q = (3, 6-3, 7-6, \dots) = (3, 3, 1, \dots), $$ and thus $$ q = (3, 3, 1, 1, 1, 1). $$ The dual partition is $$ p = (6, 2, 2). $$

Andreas Caranti
  • 65,777
  • 4
  • 59
  • 123

Hint Since the matrix $N$ is nilpotent, the only eigenvalue is zero, and as the question points out, the number of blocks is equal to the geometric multiplicity of the eigenvalue zero, which by definition is the dimension of the kernel (in this case, $3$). Denote the sizes of the three Jordan blocks by $k_1, k_2, k_3$.

Computing gives that the Jordan block $J_k$ of eigenvalue $0$ and size $k$ satisfies $\dim \ker (J_1^2) = 1$ and $\dim \ker (J_k^2) = 2$ for $k > 1$. So, the Jordan normal form $J = J_1 \oplus J_2 \oplus J_3$ of $N$ satisfies $$6 = \dim \ker (N^2) = \dim \ker (J^2) = \dim \ker (J_{k_1}^2) + \dim \ker (J_{k_2}^2) + \dim \ker (J_{k_3}^2),$$ and hence we must have $\dim \ker (J_{k_a}^2) = 2$, and therefore $k_a \geq 2$, for all $a$.

We can similarly analyze the kernels of $J_k^3$, which in this case turns out to be enough to determine the Jordan normal form $J$.

Travis Willse
  • 83,154
  • 8
  • 100
  • 207