Suppose $T- \lambda I$ is a nilpotent linear transformation on vector space $V$ with index of nilpotency $m < \mathrm{dim}V$.
Assume we know $n_k = \mathrm{dim}(T – \lambda I)^k$ for $k = 1, 2, \cdots m$.
I know that the number of Jordan sub-blocks is $n_ 1 = \mathrm{dim}(T – \lambda I)$.
I also understand that the maximum length of a chain or cycle will be $ m $ , in other words, max size of the Jordan sub-blocks is $m$.
In many places, additional formulae are stated that give the distribution of Jordan blocks of various sizes if all the $n_k$ are known. What is the proof of such results ?
In other words, can someone explain with a proof what information about the Jordan block does $n_2 = \mathrm{dim}(T – \lambda I)^2$ carry?
To me, that only means that there are $n_2$ basis vectors of $\mathrm{dim}(T – \lambda I)^2$, of which $n_1$ are basis vectors of $\mathrm{dim}(T – \lambda I)$.
But how does all this tie up with the number of Jordan blocks of size $2 \times 2$ ?
Best Answer
If you have a single Jordan block $A$ with eigenvalue 0, namely the diagonal entries are zero, then $dim A^k = n-1-k$ for $k=0,...,n-1$ and $A^k=0$ for $k>n-1$. In other words, every time you increase the power by one the dimension goes down by one, unless you already got to the minimal dimension (which is zero) and then it doesn't change.
If you have several block in an $n\times n$ matrix, then $n-\dim A$ is equal the number of blocks (including $1\times 1$ blocks which are just zero. From $\dim A$ to $\dim A^2$ you decrease by 1 for every block which is at least of size 2. It follows that $dim A - dim A^2$ is the number of block of dimension at least 2. You also get that $(n- \dim(A))-(\dim(A)-\dim(A))=n-2\dim(A)+\dim(A^2)$ is the number of block of size 1. A similar argument will show that $\dim(A^{k-1})-\dim(A^k)$ is the number of block of at least size k.
EDIT:
Lets consider an example. Suppose we have a matrix of the form $$A=\left(\begin{array}{ccccc} J_{2} & 0 & 0 & 0 & 0\\ 0 & J_{2} & 0 & 0 & 0\\ 0 & 0 & J_{3} & 0 & 0\\ 0 & 0 & 0 & J_{4} & 0\\ 0 & 0 & 0 & 0 & J_{4} \end{array}\right)$$ where $J_i$ is a Jordan block of size $i\times i$ with zeros on the diagonal. In particular $rank(J_i)=\dim(J_i)=i-1$. We therefore have $\dim(A)=(2-1)+(2-1)+(3-1)+(4-1)+(4-1)$ is the sum of dimensions of the Jordan blocks, while the size of the matrix $A$ is $n=2+2+3+4+4$ so we conclude that $n-\dim(A)=5$ is the number of Jordan blocks.
Taking the square of $A$ means taking the square of each Jordan block. For each block in our matrix we now have $\dim(J_i^2)=\dim(J_i)-1$ so that $\dim(A)-\dim(A^2)=5$ because each block contribute 1 (this is because $\dim(A^2)$ is just the sum of $\dim(J_i^2)$). Consider now $A^3$ - the blocks of the form $J_2^2$ are already 0 so that $\dim(J_2^2)=\dim(J_2^3)$. For $i>2$ we again have $\dim(J_i^3)=\dim(J_i^2)-1$ so that overall we get that $\dim(A^2)-\dim(A^3)=3$ because the blocks $J_2$ contribute nothing, while each of $J_3,J_4,J_4$ contribute one. We get that $\dim(A^2)-\dim(A^3)=3$ counts the number of blocks of size at least 3. Similarly, for the fourth power, the block $J_3$ stop to contribute so that $\dim(A^3)-\dim(A^4)=2$ counts the number of blocks of size at least 4. Finally $A^n=0$ for $4\geq 0$ so that $\dim(A^n)-\dim(A^{n+1})=0$ for all such $n$.