Clarification regarding the number of Jordan blocks of some specific order

Question

Good day! Last week at my algebra class I was asked to prove (and find) the following:

Given some nilpotent operator $T$ prove that for some number $k$, the number of Jordan normal blocks of order at least $k$ in the Jordan Form of $T$ is given by: $\dim \ker(T^k) - \dim \ker(T^{k-1})$.
In addition, find a formula for the exact number of Jordan blocks of order $k$.

The first part of the question is what got me a little bit confused, I believe of some wrong idea regarding the Jordan normal form. I shall try to describe my thought process at approaching this problem:
Suppose $T$ is a nilpotent operator and $k$ some number as described in the question.
As $T$ is nilpotent we know that $\ker T \subsetneq \ker T^2 \subsetneq \dots \subsetneq \ker T^{k-1} \subsetneq \ker T^k$.
Then, $\dim \ker(T^k) - \dim \ker(T^{k-1})$ simply equates to the amount of linearly independent vectors that zero $T^k$ and do not zero $T^{k-1}$ (and thus do not zero none of the other smaller powers of T).
Let $\{v_1,\dots,v_r\}$ be a set of such vectors. Let us also denote: \begin{align*} W_1 = \{T^{k-1}(v_1), T^{k-2}(&v_1), \dots, T(v_1), v_1\}\\ \vdots \\ W_r = \{T^{k-1}(v_r), T^{k-2}(&v_r), \dots, T(v_r), v_r\} \end{align*} Wouldn't each $W_i$ map to a Jordan block of order $k$ that must be present in the Jordan normal form of $T$? That is to say, for example take $W_1$, than we have the following coordinate vectors: \begin{gather*} [T(T^{k-1}(v_1))]_{W_1} = \begin{pmatrix} 0\\ \vdots\\ 0 \end{pmatrix}, [T(T^{k-2}(v_1))]_{W_1} = \begin{pmatrix} 1\\ \vdots\\ 0 \end{pmatrix}, \dots, [T(v_1)]_{W_1} = \begin{pmatrix} 0\\ \vdots\\ 1\\ 0 \end{pmatrix} \end{gather*} If that is the case, (which it is not and I am not sure why) how does $\dim \ker(T^k) - \dim \ker(T^{k-1})$ ($r$ in the above example) represent all the blocks of order $k$ and all those blocks with order greater than $k$?

I was not able to be as clear and concise as I wished to have been, so in summary, how come $\dim \ker(T^k) - \dim \ker(T^{k-1})$ doesn't simply equate to the number of Jordan blocks of order $k$ exactly in the Jordan form of $T$? Where do the other, bigger blocks, come from?
I am aware one can also look at this problem in terms of ranks instead of kernels but I would like to stick to this way of looking at things.

Sincerest thanks to any readers, may your day turn out marvellous.

Answer 1

Your approaching this question from the wrong end. A decomposition into Jordan blocks is given by a possible decomposition of the space that is not uniquely determined by the operator $T$ (i.e., there are some choices involved in the decomposition). But once one such decomposition is fixed you cannot hope to get a natural map from $\ker(T^k)\setminus\ker(T^{k-1})$ to the set of Jordan blocks (as you seem to claim), simply because $\ker(T^k)\setminus\ker(T^{k-1})$ is connected to there is no way to separate out the distinct blocks.

Instead you should, given a decomposition into Jordan blocks and $k$, ask how much each block contributes to $\def\dk{\operatorname{dim}\ker}$ $\dk(T^k)-\dk(T^{k-1})$. This question is meaningful, since the Jordan blocks define a direct sum decomposition into $T$-stable subspaces, so $T$ is determined by its restrictions to each of those subspaces (as operators on those subspaces), while $\dk(T^k)-\dk(T^{k-1})$ is simply the sum of the corresponding numbers for $T$ replaced by one of the restrictions. Now it is easy to see that for a Jordan block of size $s<k$ the contribution is $s-s=0$, while for a Jordan block of size $s\geq k$ the contribution is $k-(k-1)=1$.

Answer 2

@angryavian has given a good answer - not least because it indicates that the whole point of finding the Jordan form this way is that we don't have to chase around with vectors, we just compute some ranks/nullities in some standard way, and that gives the answer.

But I'll try to answer some of the detailed questions.

You ask "how come $\dim\ker(T^k)−\dim\ker(T^{k−1})$ doesn't simply equate to the number of Jordan blocks of order $k$ exactly in the Jordan form of $T$?"

Work out an example to see what is going on.

The easiest one to look at first is $$ T=\begin{pmatrix} 0&1&0\\0&0&0\\0&0&0\end{pmatrix} $$ consisting of one Jordan block of size $2$ and one of size $1$.

Then with $k=1$ we have that the nullity of $T^1$ is $2$; the nullity of $T^0$ is $0$, so that $\dim\ker(T^1)−\dim\ker(T^{1−1})=2$. That is indeed the number of blocks of size $\geqslant 1$ and not the number of blocks of size exactly $1$.

I believe that part of your confusion is when you write things like "equates to the amount of linearly independent vectors that zero $T^k$ and do not zero $T^{k−1}$". I don't really know precisely what you mean by this, and maybe if you clarified this your problem would go away. I suspect that your argument is going wrong because you seem to be constructing bases for subspaces $X\geqslant Y$ by starting at $X$ and working downwards, which is isn't a good idea: you need to start with $Y$ and work upwards.