Jordan-Chevalley 分解

$n\times n$ 階矩陣 $A$ 的 Jordan 典型形式為 $A=MJM^{-1}$，其中 Jordan 矩陣 $J$$n_i\times n_i$ 階 Jordan 分塊 $J_i$ 構成的分塊對角矩陣，或用直和 (direct sum) 表示為

$\displaystyle J=\begin{bmatrix} J_1&&\\ &\ddots&\\ &&J_k \end{bmatrix}=\bigoplus_{i=1}^k J_i$

$J_i=\begin{bmatrix} \lambda_i&1& &\\ &\ddots&\ddots&\\ &&\ddots&1\\ &&&\lambda_i \end{bmatrix}=\begin{bmatrix} \lambda_i&& &\\ &\ddots&&\\ &&\ddots&\\ &&&\lambda_i \end{bmatrix}+\begin{bmatrix} 0&1& &\\ &\ddots&\ddots&\\ &&\ddots&1\\ &&&0 \end{bmatrix}=D_i+T_i$

$\displaystyle J=\bigoplus_{i=1}^kJ_i=\bigoplus_{i=1}^k(D_i+T_i)=\bigoplus_{i=1}^kD_i+\bigoplus_{i=1}^kT_i=D+T$

$A=M(D+T)M^{-1}=MDM^{-1}+MTM^{-1}$

\begin{aligned} DT&=\bigoplus_{i=1}^k D_i\bigoplus_{i=1}^k T_i\\ &=\bigoplus_{i=1}^k D_iT_i\\ &=\bigoplus_{i=1}^k T_iD_i\\ &=\bigoplus_{i=1}^k T_i\bigoplus_{i=1}^k D_i\\ &=TD, \end{aligned}

\begin{aligned} SN&=(MDM^{-1})(MTM^{-1})\\ &=MDTM^{-1}\\ &=MTDM^{-1}\\ &=(MTM^{-1})(MDM^{-1})\\ &=NS.\end{aligned}

$A\begin{bmatrix} M_1&\cdots&M_k \end{bmatrix}=\begin{bmatrix} AM_1&\cdots&AM_k \end{bmatrix}$

$MJ=\begin{bmatrix} M_1&\cdots&M_k \end{bmatrix}\begin{bmatrix} J_1&&\\ &\ddots&\\ &&J_k \end{bmatrix}=\begin{bmatrix} M_1J_1&\cdots&M_kJ_k\end{bmatrix}$

$A\begin{bmatrix} \mathbf{x}_{i1}&\cdots&\mathbf{x}_{in_i} \end{bmatrix}=\begin{bmatrix} \mathbf{x}_{i1}&\cdots&\mathbf{x}_{in_i} \end{bmatrix}\begin{bmatrix}\lambda_i&1&&\\ &\ddots&\ddots&\\ &&\ddots&1\\ &&&\lambda_i \end{bmatrix},~~1\le i\le k$

\begin{aligned} A\mathbf{x}_{i1}=\lambda\mathbf{x}_{i1}~~&\Rightarrow~~(A-\lambda_i I)\mathbf{x}_{i1}=\mathbf{0}\\ A\mathbf{x}_{i2}=\mathbf{x}_{i1}+\lambda_i\mathbf{x}_{i2}~~&\Rightarrow~~(A-\lambda_i I)\mathbf{x}_{i2}=\mathbf{x}_{i1}\\ &\vdots\\ A\mathbf{x}_{in_i}=\mathbf{x}_{i,n_i-1}+\lambda_i\mathbf{x}_{in_i}~~&\Rightarrow~~(A-\lambda_i I)\mathbf{x}_{in_i}=\mathbf{x}_{i,n_i-1}\end{aligned}

\begin{aligned} AS'&=(S'+N')S'=S'S'+N'S'\\ &=S'S'+S'N'=S'(S'+N')=S'A.\end{aligned}

$(A-\lambda_iI)^{n_i}(S'\mathbf{x}_{ij})=S'(A-\lambda_iI)^{n_i}\mathbf{x}_{ij}=\mathbf{0},~~1\le j\le n_i$

$S-S'=MDM^{-1}-MD'M^{-1}=M(D-D')M^{-1}$

$\displaystyle (N'-N)^{2n}=\sum_{j=0}^{2n}\binom{2n}{j}(N')^jN^{2n-j}=0$

[1] 維基百科：Jordan–Chevalley decomposition