特征根求解

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

(1)有分解：

[{ m{A}} = alpha eta = left[ {egin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{...}&{{a_n}}
end{array}} ight]left[ {egin{array}{*{20}{c}}
{{b_1}}\
{{b_2}}\
{...}\
{{b_n}}
end{array}} ight]]

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

证明：

[Aalpha  = alpha eta alpha  = alpha (eta alpha ) = alpha tr(A) = tr(A)alpha  Rightarrow lambda  = tr(A),X = alpha ]

(3)$eta X=0$有n-1个无关解

证明：任取$eta X=0$的一个解，有$eta Y=0$：

[AY = (alpha eta Y) = alpha (eta Y) = 0Y]

所以$Y$为0根的特征向量，所以$eta X=0$恰有n-1个解

平移法则：

(1)$A pm cI$与A有相同的特征向量

[{ m{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：$A=A_{n imes p}$，$B=B_{p imes n}$，$AB in {C^{n imes n}},BA in {C^{p imes p}}$，有

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

(2)AB与BA的特征值只差n-p个0

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

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