Proof of the formula for the number of Jordan blocks

Question

here's what I need help with: Let A$\in M_{n\times n}(F)$ be a Jordanizable matrix. Let n$_k(\lambda)$ be the number of J$_k$($\lambda$) blocks. I'm trying to prove this formula: n$_k$($\lambda$)=rank($(A-\lambda)^{k+1}$)+rank($(A-\lambda)^{k-1}$)-2 $\cdot$ rank($(A-\lambda)^k$) for every 1$\leq$k

Answer 1

What you have stated is not true in general, but it is true for $\lambda=0$.

To prove this we may assume that $A$ is already in Jordan form, since similar matrices have equal rank. Write $\rho(B)$ for the rank of $B$. Suppose that $\chi_A(X)=X^m g(X)$ where $g(0)\not=0$ and the degree of $g$ is $(n-m)$.

Then $$ \rho(A^0)= (n-m)+ n_1(0)\cdot 1+n_2(0)\cdot 2 + n_3(0)\cdot 3+\dots $$ and $$ \rho(A^1)= (n-m) +n_2(0)\cdot 1 + n_3(0)\cdot 2+\dots $$ and $$ \rho(A^2)= (n-m) + n_3(0)\cdot 1+\dots. $$

In general we have that

$$ \rho(A^s)=(n-m)+ \sum_{r=s}^{n} n_r(0)\cdot (r-s) $$

From this last we can deduce the required $$ n_k(0)=\rho(A^{k+1}) - 2\rho(A^k) +\rho(A^{k-1}) $$ at once.

The true result for general $\lambda$ is that

$$ n_k(\lambda)=\rho((A-\lambda)^{k+1}) - 2\rho((A-\lambda)^k) +\rho((A-\lambda)^{k-1}). $$