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Let T be a nilpotent linear operator on the vector space $\mathbb{R}^5$. Let $d_i$ denote the dimension of the kernel $T^i$.Which of the following can possibly occur as a value of $(d_1,d_2,d_3)$?

(a)$(1,2,3)$

(b)$(2,3,5)$

(c)$(2,2,4)$

(d)$(2,4,5)$

I have tried this problem to solve using the inequality $$rank(A)+rank(B)-n\le r(AB) \leq \min\{rank(A),rank(B)\},$$where A,B are matrices of order n. I found all options are correct. It does not match the answer. Please help. I studied a similar question Nilpotent matrix and relation between its powers and dimension of kernels. When I applied the inequality here it matches answer.

Tulip
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  • It is enough to consider block diagonal matrices in which each block is an upper shift matrix. These are the Jordan forms and the dimensions of the kernels of the powers don't change after passing to the Jordan normal form. In the cases that the first dimension is $2$, you must have $2$ blocks. So, consider all cases: One of size $1$ and another of size $4$, one of size $2$ and another of size $3$. Next note what happens when you take powers of an upper shift matrix. The diagonal of ones shift one position up. So, the nullity increases by one. – plop May 28 '21 at 16:10
  • Well, in a $n\times n$ upper shift block the nullity stops increasing after the $n$ power, of course. – plop May 28 '21 at 16:12
  • Yes,I have done and understood after taking power of upper shift matrix. But I would like to ask "can I solve this problem using RANK-NULLITY theorem and the inequality mentioned in the question?". – Tulip May 29 '21 at 08:38

1 Answers1

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Edit: folks were thinking about this in terms of Jordan Normal Form, so I'll try and edit that in too.

In the answer of the question you cited, the lemma tells you that (b) and (c) cannot be correct since the sequences $(d_{i+1}-d_i)$ are not decreasing.

The first sequence $(1,2,3)$ can occur. For instance take $T$ to be represented by a matrix with zeros everyone except one diagonal above its main diagonal, where it is ones. Each time you take a power of $T$, the diagonal "moves up" and you gain one dimension in your kernel. The matrix that I'm describing here is the single Jordan block matrix $J_5(0)$.

The other sequence $(2,4,5)$ can also occur. I'll write out this example fully because it's harder to describe:

$$ \begin{bmatrix}0 & 1 & 0 & 0 & 0\\ 0 &0& 1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}^2 = \begin{bmatrix}0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix} $$

In the language of JNF, this matrix is $J_3(0)\oplus J_2(0)$.

quickquestion4
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  • Understood. But can I solve the problem using Rank-Nullity Theorem and above inequality mentioned in the question ? – Tulip May 29 '21 at 08:41