特征值、特征向量计算
定义法 A α = λ α A\alpha=\lambda\alphaAα=λα ∣ λ E − A ∣ = 0 |\lambda E - A|=0∣λE−A∣=0P−1A P = B P^{-1}AP=BP−1AP=BAα=λα⇒B(P− 1 α)=λ(P− 1 α) A\alpha=\lambda\alpha\Rightarrow B(P^{-1}\alpha)=\lambda(P^{-1}\alpha) Aα=λα⇒B(P−1α)=λ(P−1α)Bα=λα⇒A(Pα)=λ(Pα) B\alpha=\lambda\alpha\Rightarrow A(P\alpha)=\lambda(P\alpha) Bα=λα⇒A(Pα)=λ(Pα)相似P−1A P = B P^{-1}AP=BP−1AP=B
性质∣A∣=∣B∣ |A|=|B| ∣A∣=∣B∣r(A)=r(B) r(A)=r(B) r(A)=r(B)r ( B ) = r (P−1A P ) = r ( A P ) = r ( A ) r(B)=r(P^{-1}AP)=r(AP)=r(A)r(B)=r(P−1AP)=r(AP)=r(A)∣λE−A∣=∣λE−B∣⇒λA=λB |\lambda E-A|=|\lambda E - B|\Rightarrow\lambda_A=\lambda_B ∣λE−A∣=∣λE−B∣⇒λA=λB∑ai i =∑bi i\sum a_{ii}=\sum b_{ii} ∑aii=∑bii∑λ i= ∑aii\sum\lambda_i=\sum a_{ii}∑λi=∑aii 求A nA^nAnA ∼ B A\sim BA∼B
⇒ A + k E ∼ B + k E \Rightarrow A+kE\sim B+kE⇒A+kE∼B+kE ⇒ ∣ A + k E ∣ = ∣ B + k E ∣ \Rightarrow |A+kE|=|B+kE|⇒∣A+kE∣=∣B+kE∣ ⇒ r ( A + k E ) = r ( B + k E ) \Rightarrow r(A+kE)=r(B+kE)⇒r(A+kE)=r(B+kE) ⇒λA+kE=λB+kE\Rightarrow \lambda_{A+kE}=\lambda_{B+kE}⇒λA+kE=λB+kEA ∼ B , B ∼ C ⇒ A ∼ C A\sim B,B\sim C\Rightarrow A\sim CA∼B,B∼C⇒A∼C
证明题（证明 A AA和 C CC相似，找一个中介 B BB，通常为对角矩阵）A AA可相似对角化
A ∼ Λ ⇔ ∃ A\sim \Lambda\Leftrightarrow \existA∼Λ⇔∃可逆矩阵 P PP使得P−1A P = Λ P^{-1}AP=\LambdaP−1AP=Λ A P = P Λ AP=P\LambdaAP=PΛ A(γ 1 γ 2 γ 3 ) =(γ 1 γ 2 γ 3 )[a1000a2000a3]A\left(\begin{array}{r} \gamma_1\ \gamma_2\ \gamma_3 \end{array}\right)=\left(\begin{array}{r} \gamma_1\ \gamma_2\ \gamma_3 \end{array}\right)\left[\begin{array}{cccc} a_{1} & 0 & 0 \\ 0 & a_{2} & 0\\ 0 & 0 & a_{3} \end{array}\right] A(γ1 γ2 γ3)=(γ1 γ2 γ3)⎣⎡a1000a2000a3⎦⎤ A(γ 1 γ 2 γ 3 ) =(a 1 γ 1 a 2 γ 2 a 3 γ 3 )A\left(\begin{array}{r} \gamma_1\ \gamma_2\ \gamma_3 \end{array}\right)=\left(\begin{array}{r} a_1\gamma_1\ a_2\gamma_2\ a_3\gamma_3 \end{array}\right) A(γ1 γ2 γ3)=(a1γ1 a2γ2 a3γ3) Aγ 1=a 1 γ 1, Aγ 2=a 2 γ 2, Aγ 3=a 3 γ 3A\gamma_1=a_1\gamma_1,A\gamma_2=a_2\gamma_2,A\gamma_3=a_3\gamma_3Aγ1=a1γ1,Aγ2=a2γ2,Aγ3=a3γ3对角矩阵是 A AA的三个特征值 P PP矩阵的列向量为特征向量（只有和对角矩阵相似的时候）A AA和对角矩阵相似
A ∼ Λ ⇔ A A\sim \Lambda\Leftrightarrow AA∼Λ⇔A有 n nn个无关的特征向量充分条件A A A有n n n个不同的特征值AT=A A^T=A AT=A若λi \lambda_i λi是k k k重特征值，那λi \lambda_i λi必有k k k个无关的特征向量r (λ iE − A ) = n − k r(\lambda_i E-A)=n-kr(λiE−A)=n−k (λ iE − A ) x = 0 (\lambda_i E-A)x=0(λiE−A)x=0特征值性质
不同特征值的特征向量线性无关 k kk重特征值最多有 k kk个线性无关的特征向量 ∣ A ∣ = ∏Λ i|A|=\prod\Lambda_i∣A∣=∏Λi ∑λ i= ∑aii\sum\lambda_i=\sum a_{ii}∑λi=∑aii实对称矩阵
A AA必与对角矩阵相似不同特征值的特征向量相互正交内积为0⇒ \Rightarrow ⇒齐次方程⇒ \Rightarrow ⇒求α \alpha α 用正交矩阵相似对角化Q−1A Q = Λ Q^{-1}AQ=\LambdaQ−1AQ=Λ特征值都是实数 A =[a11…11a1…1⋮⋮⋮ ⋮111…a]A=\left[\begin{array}{cccc} a & 1 & 1 & \dots & 1 \\ 1 & a & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & \ & \vdots \\ 1 & 1 & 1 & \dots & a \\ \end{array}\right] A=⎣⎢⎢⎢⎡a1⋮11a⋮111⋮1…… …11⋮a⎦⎥⎥⎥⎤A=[a − 10 0 … 0 0a − 10 … 0 ⋮⋮⋮ ⋮0 0 0 …a − 1] +[ 1 1 1 … 1 1 1 1 … 1 ⋮⋮⋮ ⋮1 1 1 … 1 ] A=\left[\begin{array}{cccc} a - 1 & 0 & 0 & \dots & 0 \\ 0 & a - 1 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ & \vdots \\ 0 & 0 & 0 & \dots & a-1 \\ \end{array}\right]+\left[\begin{array}{cccc} 1 & 1 & 1 & \dots & 1 \\ 1 & 1 & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & \ & \vdots \\ 1 & 1 & 1 & \dots & 1 \\ \end{array}\right]A=⎣⎢⎢⎢⎡a−10⋮00a−1⋮000⋮0…… …00⋮a−1⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡11⋮111⋮111⋮1…… …11⋮1⎦⎥⎥⎥⎤A A A是对角矩阵，A A A一定与对角矩阵相似B:n,0,0,…,0 B:n,0,0,\dots,0 B:n,0,0,…,0A:n+a−1,a−1,a−1,…,a−1 A:n+a-1,a-1,a-1,\dots,a-1 A:n+a−1,a−1,a−1,…,a−1求正交矩阵 Q QQ使Q−1A Q = Λ Q^{-1}AQ=\LambdaQ−1AQ=Λ
求特征值求特征向量改造特征向量λi≠λj \lambda_i \neq \lambda_j λi=λj 只需单位化λi=λj \lambda_i = \lambda_j λi=λj 若已正交，只单位化若不正交， S c h m i d t SchmidtSchmidt正交化 构造正交矩阵 Q QQQ− 1 AQ=QTAQ=Λ Q^{-1}AQ=Q^{T}AQ=\Lambda Q−1AQ=QTAQ=ΛS c h m i d t SchmidtSchmidt正交化
α 1,α 2,α 3\alpha_1,\alpha_2,\alpha_3α1,α2,α3无关β 1=α 1\beta_1=\alpha_1β1=α1β 2=α 2− (α2,β1)(β1,β1) β 1\beta_2=\alpha_2-\frac{(\alpha_2,\beta_1)}{(\beta_1,\beta_1)}\beta_1β2=α2−(β1,β1)(α2,β1)β1β 3=α 3− (α3,β1)(β1,β1) β 1− (α3,β2)(β2,β2) β 2\beta_3=\alpha_3-\frac{(\alpha_3,\beta_1)}{(\beta_1,\beta_1)}\beta_1-\frac{(\alpha_3,\beta_2)}{(\beta_2,\beta_2)}\beta_2β3=α3−(β1,β1)(α3,β1)β1−(β2,β2)(α3,β2)β2γ i= βi∣βi∣\gamma_i = \frac{\beta_i}{|\beta_i|}γi=∣βi∣βi特征值、特征向量定义
A − nA-n A−n阶， α\alpha α是 nn n维非 00 0列向量，若 A α = λ αA\alpha=\lambda\alpha Aα=λα，称 λ\lambda λ是 矩阵 AA A的特征值， α\alpha α是矩阵 AA A属于特征值 λ\lambda λ的特征向量。
若Aα=λα A\alpha=\lambda\alpha Aα=λα，α≠0 \alpha\neq 0 α=0，对∀k≠0 \forall k\neq 0 ∀k=0，有A(kα)=λ(kα) A(k\alpha)=\lambda(k\alpha) A(kα)=λ(kα)若α1,α2 \alpha_1,\alpha_2 α1,α2是A A A关于特征值λ \lambda λ的特征向量，k1α1+k2α2≠0 k_1\alpha_1+k_2\alpha_2\neq 0 k1α1+k2α2=0仍是A A A关于特征值λ \lambda λ的特征向量A α = λ α , α ≠ 0A\alpha=\lambda\alpha,\alpha\neq 0 Aα=λα,α=0
(λE−A)α=0,α≠0 (\lambda E-A)\alpha=0,\alpha\neq 0 (λE−A)α=0,α=0α \alpha α是(λE−A)x=0 (\lambda E-A)x=0 (λE−A)x=0的非零解∣λE−A∣=0 |\lambda E - A|=0 ∣λE−A∣=0求特征值λi \lambda_i λi（共n n n个）由(λiE−A)x=0 (\lambda_i E-A)x=0 (λiE−A)x=0求基础解系矩阵A A A属于特征值λi \lambda_i λi线性无关的特征向量∣ λ E − A ∣ =λ3 − (a11 +a22 +a33 )λ2 + S λ − ∣ A ∣|\lambda E - A|=\lambda^3-(a_{11}+a_{22}+a_{33})\lambda^2+S\lambda-|A| ∣λE−A∣=λ3−(a11+a22+a33)λ2+Sλ−∣A∣
S=∣a 11 a 12 a 21 a 22 ∣+∣a 22 a 23 a 32 a 33 ∣+∣a 11 a 13 a 31 a 33 ∣ S=\left|\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right|+\left|\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|+\left|\begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array}\right| S=∣∣∣∣a11a21a12a22∣∣∣∣+∣∣∣∣a22a32a23a33∣∣∣∣+∣∣∣∣a11a31a13a33∣∣∣∣若r(A)=1 r(A)=1 r(A)=1∣ λ E − A ∣ =λ 3− (a 11+a 22+a 33) |\lambda E - A|=\lambda^3-(a_{11}+a_{22}+a_{33})∣λE−A∣=λ3−(a11+a22+a33)λ 2=λ 2[ λ − (a 11+a 22+a 33) ] \lambda^2=\lambda^2[\lambda-(a_{11}+a_{22}+a_{33})]λ2=λ2[λ−(a11+a22+a33)]λ 1=a 11+a 22+a 33\lambda_1=a_{11}+a_{22}+a_{33}λ1=a11+a22+a33,λ 2=λ 3= 0 \lambda_2=\lambda_3=0λ2=λ3=0 上述结论可以推广到n n n阶 λ 1=a 11+a 22+ ⋯ +ann,λ 2= ⋯ =λ n= 0 \lambda_1=a_{11}+a_{22}+\dots+a_{nn},\lambda_2=\dots=\lambda_n=0λ1=a11+a22+⋯+ann,λ2=⋯=λn=0A A AA+kE A+kE A+kEA− 1A^{-1} A−1A∗ A^* A∗An A^n AnP− 1 AP P^{-1}AP P−1APλ \lambda λλ+k \lambda+k λ+k1λ \frac{1}{\lambda} λ11λ/A/ \frac{1}{\lambda}{/A/} λ1/A/λn \lambda^n λnλ \lambda λα \alpha αα \alpha αα \alpha αα \alpha αα \alpha αP− 1 α P^{-1}\alpha P−1αA ∗ A=∣A∣EA^*A=|A|EA∗A=∣A∣EA ∗ Aα=∣A∣αA^*A\alpha=|A|\alphaA∗Aα=∣A∣αA ∗ λα=∣A∣αA^*\lambda\alpha=|A|\alphaA∗λα=∣A∣αA ∗ α= ∣ A ∣λ αA^*\alpha=\frac{|A|}{\lambda}\alphaA∗α=λ∣A∣α
例题
α 1,α 2\alpha_1,\alpha_2α1,α2是 A AA不同特征值的特征向量，证明α 1+α 2\alpha_1+\alpha_2α1+α2不是 A AA的特征向量. A − 2 A-2A−2阶，α 1,α 2\alpha_1,\alpha_2α1,α2二维线性无关， Aα 1= 0 , Aα 2= 2α 1+α 2A\alpha_1=0,A\alpha_2=2\alpha_1+\alpha_2Aα1=0,Aα2=2α1+α2，求 A AA的特征值、特征向量.定义法相似