Preliminary

Jordan Matrix

Jordan Normal Form Theorem

\(A\in \mathbb{C}^{n\times n}\), \(P_A(\lambda)=\prod_{i=1}^s(\lambda-\lambda_i)^{n_i}\), then it is similar to Jordan normal form \(diag(J_1,J_2,\cdots,J_s)\), \(J_i\in \mathbb{n_i\times n_i}\), where \(n_i\) is algebraic multiplicity of \(A\),

\[ J_i=diag(J_{i_1}, J_{i_2},\cdots, J_{i_{m_i}}), \quad m_i \text{ is the geometric multiplicity of } A \]

\[ J_{ij}=J_{l_{ij}}(\lambda_i)=\left[\begin{array}{cccc} \lambda_i & 1\\ &\lambda_i &\ddots\\ &&\ddots&1\\ &&&\lambda_i \end{array} \right]\in \mathbb{C}^{l_{ij}\times l_{ij}} \]

And the Jordan normal form is unique if we do not consider the sequence of small blocks.

Minimal polynomial

Matrix \(A\) has a minimal polynomial

\[ d_{A}(\lambda)=d_{J}(\lambda)=\prod_{i=1}^s(\lambda-\lambda_i)^{k_i} \]

where

\[ k_i=\max_{1\leq j\leq m_i}\{l_{ij}\} \]

\(m_i\) determines how much blocks compose the whole Jordan normal form. For a Jordan normal form \(J_i\), with \(n_i\) algebraic multiplicity of \(\lambda_i\), has \(m_i\) number of Jordan small blocks. We can easily see that the prime diagonal has \(n_i-m_i\) number of \(1\), which means rank\((J-\lambda_i I)=n_i-m_i\). Then to confirm the composition of these small blocks, we have to write \(n_i-m_i\) as a summation of \(m_i\) number non-negative integers, each of which corresponds to a Jordan matrix.

Not complicatedly speaking, we have to calculate the power of \(J_i-\lambda_i\) to get the result.

Denote \(\delta_p^i\) as the number of \(p\)th Jordan small blocks corresponding to eigenvalue \(\lambda_i\), where \(p=1,2\cdots, n_i\), so

\[ \begin{align*} \sum_{p=1}^{n_i}\delta_p^i=m_i, \quad \text{total number of blocks}\\ \sum_{p=1}^{n_i}p\delta_p^i=n_i\quad \text{total rank of Jordan} \end{align*} \]

Introduce

\[ r_k^i=r(A-\lambda_i I)^k \]

so

\[ \begin{align*} r_1^i&=r(A-\lambda_i I)\\ &=r(J-\lambda_i I)\\ &=n-n_i+r(J_i-\lambda_i I)\\ &=n-n_i + \sum_{p=1}^{n_i}(p-1)\delta_p^i \quad \text{1 order block becomes 0 when substraction} \end{align*} \]

So after \(k\) times power

\[ r_k^i=n-n_i+\sum_{p=k}^{n_i}(p-k)\delta_p^i \quad \text{Why?} \]

and

\[ \begin{align*} r_{k-1}^i&=n-n_i+\sum_{p=k-1}^{n_i}(p-k+1)\delta_p^i\\ &=n-n_i+\sum_{p=k}^{n_i}(p-k+1)\delta_p^i \end{align*} \]

If we define

\[ d_k^i:=r_{k-1}^i-r^i_k=\sum_{p=k}^{n_i}\delta_p^i \]

and

\[ d_{k+1}^i=\sum_{p=k+1}^{n_i}\delta_p^i \]

Substract the aboce two equaiton, we get

\[ \delta_p^i=d_k^i-d_{k+1}^i \]

So if a Jordan small block has \(\delta_p^i\) number, then the rank decline from power \(p-1\) to power \(p\) of \((A-\lambda_i I)\) would display.