特征根求解

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秩1方阵公式：若方阵$A=A_{n \times n}, rank(A)=1$，则有如下性质

(1)有分解：

\[{\rm{A}} = \alpha \beta = \left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{...}&{{a_n}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{b_1}}\\
{{b_2}}\\
{...}\\
{{b_n}}
\end{array}} \right]\]

(2)$\lambda (A) = \{ tr(A),0,...,0\} $（n-1个0），$\lambda_{1}=tr(A)$ 且$A\alpha = \lambda_{1} \alpha$

证明：

\[A\alpha  = \alpha \beta \alpha  = \alpha (\beta \alpha ) = \alpha tr(A) = tr(A)\alpha  \Rightarrow \lambda  = tr(A),X = \alpha \]

(3)$\beta X=0$有n-1个无关解

证明：任取$\beta X=0$的一个解，有$\beta Y=0$：

\[AY = (\alpha \beta Y) = \alpha (\beta Y) = 0Y\]

所以$Y$为0根的特征向量，所以$\beta X=0$恰有n-1个解

平移法则：

(1)$A \pm cI$与A有相同的特征向量

\[{\rm{AX}} = \lambda X \Rightarrow AX \pm cX = \lambda X \pm cX \Rightarrow (A \pm cI)X = (\lambda  \pm c)x\]

(2)$\lambda (A \pm cI) = \{ {\lambda _1} \pm c,{\lambda _2} \pm c,...,{\lambda _n} \pm c\} $与$\lambda (A) = \{ {\lambda _1},{\lambda _2},...,{\lambda _n}\} $

(3)$\lambda (kA) = \{ k{\lambda _1},k{\lambda _2},...,k{\lambda _n}\} $与$\lambda (A) = \{ {\lambda _1},{\lambda _2},...,{\lambda _n}\}$

换位公式：$A=A_{n \times p}$，$B=B_{p \times n}$，$AB \in {C^{n \times n}},BA \in {C^{p \times p}}$，有

(1)$\left| {\lambda I - AB} \right| = {\lambda ^{n - p}}\left| {\lambda I - BA} \right|$

(2)AB与BA的特征值只差n-p个0

\[\begin{array}{l}
\lambda (BA) = \{ {\lambda _1},{\lambda _2},...,{\lambda _n}\} \\
\lambda (AB) = \{ {\lambda _1},{\lambda _2},...,{\lambda _n},0,...,0\}
\end{array}\]

(3)$tr(AB) = tr(BA) = {\lambda _1} + {\lambda _2} + ... + {\lambda _n}$

特征根求解

标签：begin   分解   bsp   alpha   mes   amp   rank   公式   time   

原文地址：https://www.cnblogs.com/codeDog123/p/10206782.html