【线性代数基础进阶】特征值和特征向量

特征值、特征向量

定义：设$A$是$n$阶矩阵，$\alpha$是$n$维非$0$列向量，且
$$A\alpha =\lambda \alpha$$
则称$\lambda$是矩阵$A$的特征值，$\alpha$是矩阵$A$对应于特征值$\lambda$的特征向量

由$A\alpha =\lambda \alpha ,\alpha \ne 0\Rightarrow (\lambda E-A)\alpha =0,(\lambda E-A)x=0\Rightarrow \alpha$是齐次方程组$(\lambda E-A)x=0$的非$0$解

1. 由$|\lambda E-A|=0$求特征值${\lambda }_i$，共$n$个（含重根）
2. 对$({\lambda }_{i}E-A)x=0$，求基础解系，即特征值${\lambda }_i$的线性无关的特征向量，写通解得${\lambda }_i$所有的特征向量

定理：如果${\alpha }_1,{\alpha }_2$都是矩阵$A$对应于特征值$\lambda$的特征向量，则当$k_1{\alpha }_1+k_2{\alpha }_2\ne 0$时，$k_1{\alpha }_1+k_2{\alpha }_2$仍是矩阵$A$关于特征值$\lambda$的特征向量

定理：如果${\lambda }_1$与${\lambda }_2$是$A$不同的特征值，对应的特征向量分别是${\alpha }_1$和${\alpha }_2$，则${\alpha }_1,{\alpha }_2$必定线性无关

定理：设$A$是$n$阶矩阵，特征值是${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$，则有

• $\sum {\lambda }_i=\sum a_{ii}$
• $|A|=\prod {\lambda }_i$

例：求 A = ( 17 − 2 − 2 − 2 14 − 4 − 2 − 4 14 ) A=

A=⎝ ⎛​17−2−2​−214−4​−2−414​⎠ ⎞​特征值，特征向量

三阶行列式主对角线元素都含有未知数，直接展开要解三次方程，一般考虑通过行列加减，找出某一行或某一列所有的元素都含有含未知数的公因式或$0$

由$A$的特征多项式
∣ λ E − A ∣ = ∣ λ − 17 2 2 2 λ − 14 4 2 4 λ − 14 ∣ = ∣ λ − 17 2 2 2 λ − 14 4 0 18 − λ λ − 18 ∣ = ∣ λ − 17 4 2 2 λ − 10 4 0 0 λ − 18 ∣ = ( λ − 18 ) ∣ λ − 17 4 2 λ − 10 ∣ = ( λ − 18 ) ( λ 2 − 27 λ + 162 ) = ( λ − 18 ) 2 ( λ − 9 )

∣λE−A∣​=∣ ∣​λ−1722​2λ−144​24λ−14​∣ ∣​=∣ ∣​λ−1720​2λ−1418−λ​24λ−18​∣ ∣​=∣ ∣​λ−1720​4λ−100​24λ−18​∣ ∣​=(λ−18)∣ ∣​λ−172​4λ−10​∣ ∣​=(λ−18)(λ2−27λ+162)=(λ−18)2(λ−9)​
因此${\lambda }_1={\lambda }_2=18,{\lambda }_3=9$
当$\lambda =18$时，$(18E-A)x=0$
( 1 2 2 2 4 4 2 4 4 ) → ( 1 2 2 0 0 0 0 0 0 )

(122244244)→(122000000)" role="presentation" style="position: relative;">⎛⎝⎜122244244⎞⎠⎟→⎛⎝⎜100200200⎞⎠⎟$$\begin{array}{r}\left(\begin{array}{ccc}1& 2& 2\\ 2& 4& 4\\ 2& 4& 4\end{array}\right)\to \left(\begin{array}{ccc}1& 2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\end{array}$$

⎝ ⎛​122​244​244​⎠ ⎞​→⎝ ⎛​100​200​200​⎠ ⎞​​
得基础解系：${\alpha }_1=(-2,1,0{)}^T,{\alpha }_2=(-2,0,1{)}^T$，因此特征向量$k_1{\alpha }_1+k_2{\alpha }_2,k_1,k_2$不同时为$0$
当$\lambda =9$时，$(9E-A)x=0$
( − 8 2 2 2 − 5 4 2 4 − 5 ) → ( 2 0 − 1 0 1 − 1 0 0 0 )

(−8222−5424−5)" role="presentation" style="position: relative;">⎛⎝⎜−8222−5424−5⎞⎠⎟$$\left(\begin{array}{ccc}-8& 2& 2\\ 2& -5& 4\\ 2& 4& -5\end{array}\right)$$

\rightarrow

(20−101−1000)" role="presentation" style="position: relative;">⎛⎝⎜200010−1−10⎞⎠⎟$$\left(\begin{array}{ccc}2& 0& -1\\ 0& 1& -1\\ 0& 0& 0\end{array}\right)$$

⎝ ⎛​−822​2−54​24−5​⎠ ⎞​→⎝ ⎛​200​010​−1−10​⎠ ⎞​
得基础解系${\alpha }_3=(1,2,2{)}^T$，因此特征向量$k_3{\alpha }_3,k_3\ne 0$

对于一个矩阵若，秩等于$1$即存在一行能表示其他所有行，秩等于$2$即存在两行能表示其他所有行，之后同理，类似最大线性无关组

( − 8 2 2 2 − 5 4 2 4 − 5 )

(−8222−5424−5)" role="presentation" style="position: relative;">⎛⎝⎜−8222−5424−5⎞⎠⎟$$\left(\begin{array}{ccc}-8& 2& 2\\ 2& -5& 4\\ 2& 4& -5\end{array}\right)$$

⎝ ⎛​−822​2−54​24−5​⎠ ⎞​求基础解系，本题题解省略的化简步骤，其实化简步骤并不容易，但此处可以用其他思路。
由于已知$|\lambda E-A|=0$，即该矩阵的秩一定小于$3$；随便选两行，这里选二三行，发现二者线性无关（不成比例），因此该矩阵秩大于$1$，可得该矩阵秩为$2$。
由于这两行能线性表示另一行，因此可以构造新的矩阵
( 2 − 5 4 2 4 − 5 0 0 0 )

(2−5424−5000)" role="presentation" style="position: relative;">⎛⎝⎜220−5404−50⎞⎠⎟$$\left(\begin{array}{ccc}2& -5& 4\\ 2& 4& -5\\ 0& 0& 0\end{array}\right)$$

⎝ ⎛​220​−540​4−50​⎠ ⎞​
可以理解为由于这两行能线性表示另一行，因此另一行一定能被全消为$0$
此时只需对新的矩阵行化简即可

设$A\alpha =\lambda \alpha ,\alpha \ne 0$，则有

• $(A+kE)\alpha =(\lambda +k)\alpha$
• $A^n\alpha ={\lambda }^n\alpha$

例：$A$为$3$阶矩阵，特征值是$-1,0,4$，如果$A+B=2E$，则$B$的特征值为（）

设$A\alpha =\lambda \alpha ,\alpha \ne 0$
由$A+B=2E$，有$B=2E-A$，则
$$B\alpha =(2E-A)\alpha =2\alpha -A\alpha =(2-\lambda )\alpha$$
因此$B$的特征值为$3,2,-2$

例：$A$是$3$阶矩阵，$A^2+2A-3E=0$，证明矩阵$A$的特征值只能是$1$或$-3$

设$\lambda$是$A$的任一特征值，对应的特征向量是$\alpha$，即$A\alpha =\lambda \alpha ,\alpha \ne 0$，那么
$$A^2\alpha ={\lambda }^2\alpha$$
由$A^2+2A-3E=0$
有
A 2 α + 2 A α − 3 α = 0 ( λ 2 + 2 λ − 3 ) α = 0 , α ≠ 0 λ 2 + 2 λ − 3 = 0

A2α+2Aα−3α(λ2+2λ−3)αλ2+2λ−3​=0=0,α=0=0​
因此$\lambda =1$或$\lambda =-3$

从题目条件只能推出矩阵特征值只能是$1$或$-3$
例如： ( 1 1 1 ) , ( 3 3 3 ) , ( 1 0 0 0 − 1 − 2 0 − 2 − 1 )

(111)" role="presentation" style="position: relative;">⎛⎝⎜111⎞⎠⎟$$\left(\begin{array}{ccc}1& & \\ & 1& \\ & & 1\end{array}\right)$$

,

(333)" role="presentation" style="position: relative;">⎛⎝⎜333⎞⎠⎟$$\left(\begin{array}{ccc}3& & \\ & 3& \\ & & 3\end{array}\right)$$

,

(1000−1−20−2−1)" role="presentation" style="position: relative;">⎛⎝⎜1000−1−20−2−1⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& -2\\ 0& -2& -1\end{array}\right)$$

⎝ ⎛​1​1​1​⎠ ⎞​,⎝ ⎛​3​3​3​⎠ ⎞​,⎝ ⎛​100​0−1−2​0−2−1​⎠ ⎞​

例：已知$\alpha =(1,1,-1{)}^T$是 A = ( 2 − 1 2 5 a 3 − 1 b − 2 ) A=

A=⎝ ⎛​25−1​−1ab​23−2​⎠ ⎞​的一个特征向量，则$a=(),b=()$

设$A\alpha =\lambda \alpha ,\alpha \ne 0$
( 2 − 1 2 5 a 3 − 1 b − 2 ) ( 1 1 − 1 ) = λ ( 1 1 − 1 )

=\lambda ⎝ ⎛​25−1​−1ab​23−2​⎠ ⎞​⎝ ⎛​11−1​⎠ ⎞​=λ⎝ ⎛​11−1​⎠ ⎞​
有
{ 2 − 1 − 2 = λ 5 + a − 3 = λ − 1 + b + 2 = − λ

{2−1−2=λ5+a−3=λ−1+b+2=−λ" role="presentation" style="position: relative;">⎧⎩⎨2−1−2=λ5+a−3=λ−1+b+2=−λ$$\{\begin{array}{l}2-1-2=\lambda \\ 5+a-3=\lambda \\ -1+b+2=-\lambda \end{array}$$

⎩ ⎨ ⎧​2−1−2=λ5+a−3=λ−1+b+2=−λ​
可得$\lambda =-1,a=-3,b=0$

相似矩阵

设$A,B$都是$n$阶矩阵，如果存在可逆矩阵$P$使
$$P^{-1}AP=B$$
就称矩阵$A$相似与矩阵$B$，$B$是$A$的相似矩阵，记作$A\sim B$

相似的基本性质

• $A\sim A$
• 如果$A\sim B$，则$B\sim A$
• 如果$A\sim B,B\sim C$，则$A\sim C$

如果$A\sim B$

• $|\lambda E-A|=|\lambda E-B|\Rightarrow {\lambda }_A={\lambda }_B$
  证明：
  ∣ λ E − B ∣ = ∣ λ E − P − 1 A P ∣ = ∣ P − 1 ( λ E − A ) P ∣ = ∣ P − 1 ∣ ∣ λ E − A ∣ ∣ P ∣ = ∣ λ E − A ∣
  |λE−B|=|λE−P−1AP|=|P−1(λE−A)P|=|P−1||λE−A||P|=|λE−A|" role="presentation" style="position: relative;">|λE−B|=|λE−P−1AP|=|P−1(λE−A)P|=|P−1||λE−A||P|=|λE−A|$$\begin{array}{rl}|\lambda E-B|& =|\lambda E-P^{-1}AP|\\ & =|P^{-1}(\lambda E-A)P|\\ & =|P^{-1}||\lambda E-A||P|\\ & =|\lambda E-A|\end{array}$$
  ∣λE−B∣​=∣λE−P−1AP∣=∣P−1(λE−A)P∣=∣P−1∣∣λE−A∣∣P∣=∣λE−A∣​
• $r(A)=r(B)$
  证明：
  如果$A$可逆，有$r(AB)=r(B),r(BA)=r(B)$，则
  r ( B ) = r ( P − 1 A P ) = r ( A P ) = r ( A )
  r(B)=r(P−1AP)=r(AP)=r(A)" role="presentation" style="position: relative;">r(B)=r(P−1AP)=r(AP)=r(A)$$\begin{array}{r}r(B)=r(P^{-1}AP)=r(AP)=r(A)\end{array}$$
  r(B)=r(P−1AP)=r(AP)=r(A)​
• $|A|=|B|$
  $$|B|=|P^{-1}AP|=|P^{-1}||A||P|=|A|$$
• $\sum a_{ii}=\sum b_{ii}$
• $A^n\sim B^n$
  $A+kE\sim B+kE$
  $A^{-1}\sim B^{-1}$

相似对角化

如果$A\sim \Lambda$，则称矩阵$A$可相似对角化

定理：$A\sim \Lambda \iff A$有$n$个线性无关的特征向量
证明：
如果$A{\alpha }_1={\lambda }_1{\alpha }_1,A{\alpha }_2={\lambda }_2{\alpha }_2,A{\alpha }_3={\lambda }_3{\alpha }_3$，且${\alpha }_1,{\alpha }_2,{\alpha }_3$线性无关，则
A ( α 1 , α 2 , α 3 ) = ( A α 1 , A α 2 , A α 3 ) = ( λ 1 α 1 , λ 2 α 2 , λ 3 α 3 ) = ( α 1 , α 2 , α 3 ) ( λ 1 λ 2 λ 3 )

A(α1​,α2​,α3​)​=(Aα1​,Aα2​,Aα3​)=(λ1​α1​,λ2​α2​,λ3​α3​)=(α1​,α2​,α3​)⎝ ⎛​λ1​​λ2​​λ3​​⎠ ⎞​​
令$P=({\alpha }_1,{\alpha }_2,{\alpha }_3)$，由于${\alpha }_1,{\alpha }_2,{\alpha }_3$线性无关，则$P$可逆，有
A P = P Λ P − 1 A P = Λ = ( λ 1 λ 2 λ 3 )

AP=PΛP−1AP=Λ=(λ1λ2λ3)" role="presentation" style="position: relative;">APP−1AP=PΛ=Λ=⎛⎝⎜λ1λ2λ3⎞⎠⎟$$\begin{array}{rl}AP& =P\mathrm{\Lambda }\\ P^{-1}AP& =\mathrm{\Lambda }=\left(\begin{array}{ccc}{\lambda }_1& & \\ & {\lambda }_2& \\ & & {\lambda }_3\end{array}\right)\end{array}$$

APP−1AP​=PΛ=Λ=⎝ ⎛​λ1​​λ2​​λ3​​⎠ ⎞​​
必要性得证
如果$P^{-1}AP=\Lambda$，有
A P = P Λ A ( α 1 , α 2 , α 3 ) = ( α 1 , α 2 , α 3 ) ( a 1 a 2 a 3 ) = ( a 1 α 1 , a 2 α 2 , a 3 α 3 ) ( A α 1 , A α 2 , A α 3 ) = ( a 1 α 1 , a 2 α 2 , a 3 α 3 )

AP=PΛA(α1,α2,α3)=(α1,α2,α3)(a1a2a3)=(a1α1,a2α2,a3α3)(Aα1,Aα2,Aα3)=(a1α1,a2α2,a3α3)" role="presentation" style="position: relative;">APA(α1,α2,α3)(Aα1,Aα2,Aα3)=PΛ=(α1,α2,α3)⎛⎝⎜a1a2a3⎞⎠⎟=(a1α1,a2α2,a3α3)=(a1α1,a2α2,a3α3)$$\begin{array}{rl}AP& =P\mathrm{\Lambda }\\ A({\alpha }_1,{\alpha }_2,{\alpha }_3)& =({\alpha }_1,{\alpha }_2,{\alpha }_3)\left(\begin{array}{ccc}{a}_1& & \\ & a_2& \\ & & a_3\end{array}\right)\\ & =(a_1{\alpha }_1,a_2{\alpha }_2,a_3{\alpha }_3)\\ (A{\alpha }_1,A{\alpha }_2,A{\alpha }_3)& =(a_1{\alpha }_1,a_2{\alpha }_2,a_3{\alpha }_3)\end{array}$$

APA(α1​,α2​,α3​)(Aα1​,Aα2​,Aα3​)​=PΛ=(α1​,α2​,α3​)⎝ ⎛​a1​​a2​​a3​​⎠ ⎞​=(a1​α1​,a2​α2​,a3​α3​)=(a1​α1​,a2​α2​,a3​α3​)​
因此有
$$A{\alpha }_1=a_1{\alpha }_1,A{\alpha }_2=a_2{\alpha }_2,A{\alpha }_3=a_3{\alpha }_3$$
又因为$P=({\alpha }_1,{\alpha }_2,{\alpha }_3)$可逆，则${\alpha }_1,{\alpha }_2,{\alpha }_3$线性无关，充分性得证、

推论：如果$A$有$n$个不同的特征值，则$A\sim \Lambda$

定理：$A\sim \Lambda \iff \lambda$是$A$的$k$重特征值，则$\lambda$有$k$个线性无关的特征向量

相似对角化解题步骤：

1. 求特征值${\lambda }_1,{\lambda }_2,{\lambda }_3$
2. 求特征向量${\alpha }_1,{\alpha }_2,{\alpha }_3$
3. 构造可逆矩阵$P=({\alpha }_1,{\alpha }_2,{\alpha }_3)$，则 P − 1 A P = ( λ 1 λ 2 λ 3 ) P^{-1}AP=
   (λ1λ2λ3)" role="presentation" style="position: relative;">⎛⎝⎜λ1λ2λ3⎞⎠⎟$$\left(\begin{array}{ccc}{\lambda }_1& & \\ & {\lambda }_2& \\ & & {\lambda }_3\end{array}\right)$$
   P−1AP=⎝ ⎛​λ1​​λ2​​λ3​​⎠ ⎞​

例：已知 A = ( 0 0 1 0 1 0 1 0 0 ) A=

A=⎝ ⎛​001​010​100​⎠ ⎞​，求可逆矩阵$P$，使$P^{-1}AP=\Lambda$

由特征多项式
∣ λ E − A ∣ = ∣ λ 0 − 1 0 λ − 1 0 − 1 0 λ ∣ = ( λ − 1 ) 2 ( λ + 1 ) |\lambda E-A|=

=(\lambda-1)^{2}(\lambda+1) ∣λE−A∣=∣ ∣​λ0−1​0λ−10​−10λ​∣ ∣​=(λ−1)2(λ+1)
则$A$的特征值$1,1,-1$
当$\lambda =1$时，由$(E-A)x=0$
( 1 0 − 1 0 0 0 − 1 0 1 ) → ( 1 0 − 1 0 0 0 0 0 0 )

(10−1000−101)" role="presentation" style="position: relative;">⎛⎝⎜10−1000−101⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right)$$

\rightarrow

(10−1000000)" role="presentation" style="position: relative;">⎛⎝⎜100000−100⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$$

⎝ ⎛​10−1​000​−101​⎠ ⎞​→⎝ ⎛​100​000​−100​⎠ ⎞​
特征向量${\alpha }_1=(0,1,0{)}^T,{\alpha }_2=(1,0,1{)}^T$
当$\lambda =-1$时，由$(-E-A)x=0$
( − 1 0 − 1 0 − 2 0 − 1 0 − 1 ) → ( 1 0 1 0 1 0 0 0 0 )

(−10−10−20−10−1)" role="presentation" style="position: relative;">⎛⎝⎜−10−10−20−10−1⎞⎠⎟$$\left(\begin{array}{ccc}-1& 0& -1\\ 0& -2& 0\\ -1& 0& -1\end{array}\right)$$

\rightarrow

(101010000)" role="presentation" style="position: relative;">⎛⎝⎜100010100⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$$

⎝ ⎛​−10−1​0−20​−10−1​⎠ ⎞​→⎝ ⎛​100​010​100​⎠ ⎞​
特征向量${\alpha }_3=(-1,0,1{)}^T$
令
P = ( α 1 , α 2 , α 3 ) = ( 0 1 − 1 1 0 0 0 1 1 ) P=(\alpha_{1},\alpha_{2},\alpha_{3})=

(01−1100011)" role="presentation" style="position: relative;">⎛⎝⎜010101−101⎞⎠⎟$$\left(\begin{array}{ccc}0& 1& -1\\ 1& 0& 0\\ 0& 1& 1\end{array}\right)$$

P=(α1​,α2​,α3​)=⎝ ⎛​010​101​−101​⎠ ⎞​
有
P − 1 A P = ( 1 1 − 1 ) P^{-1}AP=

(11−1)" role="presentation" style="position: relative;">⎛⎝⎜11−1⎞⎠⎟$$\left(\begin{array}{ccc}1& & \\ & 1& \\ & & -1\end{array}\right)$$

P−1AP=⎝ ⎛​1​1​−1​⎠ ⎞​

例：已知 A = ( 2 0 0 0 0 1 0 1 x ) , B = ( 2 0 0 0 y 0 0 0 − 1 ) A=

,B= A=⎝ ⎛​200​001​01x​⎠ ⎞​,B=⎝ ⎛​200​0y0​00−1​⎠ ⎞​相似

• 求$x,y$
• 求可逆矩阵$P$使$P^{-1}AP=B$

由于$A\sim B$，有
∑ a i i = ∑ b i i ⇒ 2 + 0 + x = 2 + y + ( − 1 ) ∣ A ∣ = ∣ B ∣ ⇒ − 2 = − 2 y ∣ λ E − A ∣ = ∣ λ E − B ∣ ⇒ ∣ λ − 2 0 0 0 λ − 1 0 − 1 λ − x ∣ = ∣ λ − 2 0 0 0 λ − y 0 0 0 λ + 1 ∣ ⇒ ( λ − 2 ) ( λ 2 − x λ − 1 ) = ( λ − 2 ) [ λ 2 + ( 1 − y ) λ − y ] ⇒ { − x = 1 − y − 1 = − y

∑aii​=∑bii​∣A∣=∣B∣∣λE−A∣=∣λE−B∣​⇒2+0+x=2+y+(−1)⇒−2=−2y⇒∣ ∣​λ−200​0λ−1​0−1λ−x​∣ ∣​=∣ ∣​λ−200​0λ−y0​00λ+1​∣ ∣​⇒(λ−2)(λ2−xλ−1)=(λ−2)[λ2+(1−y)λ−y]⇒{−x=1−y−1=−y​​(1)(2)(3)​

可以选$(1),(2)$或单独选$(3)$，一般选$(1),(2)$，计算较简单

或
$B$的特征值$2,y,-1$，则$\lambda =-1$是$A$的特征值，有
∣ − E − A ∣ = ∣ − 3 0 0 0 − 1 − 1 0 − 1 − 1 − x ∣ = − 3 ∣ − 1 − 1 − 1 − 1 − x ∣ = − 3 x = 0 |-E-A|=

=-3=-3x=0 ∣−E−A∣=∣ ∣​−300​0−1−1​0−1−1−x​∣ ∣​=−3∣ ∣​−1−1​−1−1−x​∣ ∣​=−3x=0
若用$\lambda =2$，有
∣ 2 E − A ∣ = ∣ 0 0 0 0 2 − 1 0 − 1 2 − x ∣ = 0 |2E-A|=

|00002−10−12−x|" role="presentation" style="position: relative;">∣∣∣∣00002−10−12−x∣∣∣∣$$\left|\begin{array}{ccc}0& 0& 0\\ 0& 2& -1\\ 0& -1& 2-x\end{array}\right|$$

=0 ∣2E−A∣=∣ ∣​000​02−1​0−12−x​∣ ∣​=0
若用$\lambda =y$，有
∣ y E − A ∣ = ∣ y − 2 0 0 0 y − 1 0 − 1 y − x ∣ = ( y − 2 ) ( y 2 − x y − 1 ) = 0 |yE-A|=

|y−2000y−10−1y−x|" role="presentation" style="position: relative;">∣∣∣∣y−2000y−10−1y−x∣∣∣∣$$\left|\begin{array}{ccc}y-2& 0& 0\\ 0& y& -1\\ 0& -1& y-x\end{array}\right|$$

=(y-2)(y^{2}-xy-1)=0 ∣yE−A∣=∣ ∣​y−200​0y−1​0−1y−x​∣ ∣​=(y−2)(y2−xy−1)=0

该式需要结合上面的$(1),(2),(3)$使用，或者选择两个不相关的使用，例如$\lambda =0,\lambda =y$搭配
该方法当出现本题$\lambda =2$时$|2E-A|$本身有一行就为$0$或两行成比例，则需要换$\lambda$或用$(1),(2),(3)$方法

解得$x=0,y=1$，代入题设有
A = ( 2 0 0 0 0 1 0 1 0 ) ∼ ( 2 1 − 1 ) = B A=

\sim =B A=⎝ ⎛​200​001​010​⎠ ⎞​∼⎝ ⎛​2​1​−1​⎠ ⎞​=B
此处省略步骤，可得
P = ( 1 0 0 0 1 1 0 1 − 1 ) P=

(10001101−1)" role="presentation" style="position: relative;">⎛⎝⎜10001101−1⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& -1\end{array}\right)$$

P=⎝ ⎛​100​011​01−1​⎠ ⎞​

实对称矩阵

向量的内积

定义：设$\alpha =(a_1,a_2,\cdots ,a_n{)}^T,\beta =(b_1,b_2,\cdots ,b_n{)}^T$，有
$$(\alpha ,\beta )=a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n={\alpha }^T\beta ={\beta }^T\alpha$$

如果$(\alpha ,\beta )=0$，称$\alpha$与$\beta$正交

$(\alpha ,\alpha )=a_1^2+a_2^2+\cdots +a_n^2$，称$\sqrt{a_1^2+a_2^2+\cdots +a_n^2}$为向量$\alpha$的长度，记作$||\alpha ||$

内积的性质

• $(\alpha ,\beta )=(\beta ,\alpha )$
• $(k\alpha ,\beta )=(\alpha ,k\beta )=k(\lambda ,\beta )$
• $(\alpha +\beta ,\gamma )=(\alpha ,\gamma )+(\beta ,\gamma )$
• $(\alpha ,\alpha )\ge 0$，等号当且仅当$\alpha =0$时成立

例：求与${\alpha }_1=(-1,-1,1{)}^T,{\alpha }_2=(1,2,1{)}^T$都正交的向量

设$\alpha =(x_1,x_2,x_3{)}^T$与${\alpha }_1,{\alpha }_2$都正交，则${\alpha }^T{\alpha }_1=0,{\alpha }^T{\alpha }_2=0$
{ − x 1 − x 2 + x 3 = 0 x 1 − 2 x 2 − x 3 = 0

{−x1​−x2​+x3​=0x1​−2x2​−x3​=0​
有
( − 1 − 1 1 1 − 2 − 1 ) → ( 1 0 − 1 0 1 0 )

(−1−111−2−1)" role="presentation" style="position: relative;">(−11−1−21−1)$$\left(\begin{array}{ccc}-1& -1& 1\\ 1& -2& -1\end{array}\right)$$

\rightarrow

(10−1010)" role="presentation" style="position: relative;">(1001−10)$$\left(\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\end{array}\right)$$

(−11​−1−2​1−1​)→(10​01​−10​)
基础解系$(1,0,1{)}^T$，因此$\alpha =k(1,0,1{)}^T$

正交矩阵

定义：$A$是$n$阶矩阵，若$A{A}^T=A^{T}A=E$，则称$A$是正交矩阵

如果$A$是正交矩阵
$\iff A^T=A^{-1}$
$\iff A$的列向量都是单位向量且两两正交

如果$A$是正交矩阵$\Rightarrow |A|$为$1$或$-1$
证明：
A A T = E ⇒ ∣ A A T ∣ = ∣ E ∣ ∣ A ∣ ⋅ ∣ A T ∣ = 1 ∣ A ∣ 2 = 1

AAT=E⇒∣AAT∣∣A∣⋅∣AT∣∣A∣2​=∣E∣=1=1​
因此$|A|$为$1$或$-1$

实对称矩阵

定理：实对称矩阵必可相似对角化

定理：实对称矩阵不同特征值所对应的特征向量相互正交

定理：实对称矩阵必存在正交矩阵$Q$，使$Q^{-1}AQ=Q^{T}AQ=\Lambda$

例：$A$为$3$阶实对称矩阵，满足$A^2=A$，如果$r(A-E)=2$，则$A$的特征值（）

设$A\alpha =\lambda \alpha ,\alpha \ne 0$，则
$$A^2\alpha ={\lambda }^2\alpha$$
由$A^2=A$，有
λ 2 α = λ α ( λ 2 − λ ) α = 0 ⇒ λ 2 − λ = 0

λ2α(λ2−λ)α​=λα=0⇒λ2−λ=0​
因此$\lambda$是$1$或$0$，又由于$A$是实对称矩阵，$r(A-E)=2$，有
$$A\sim \Lambda \Rightarrow A-E\sim \Lambda -E\Rightarrow r(\Lambda -E)=2$$
$\Lambda$与$A$相似，即特征值相同，又有$r(\Lambda -E)=2$，则可构造
Λ = ( 1 0 0 ) \Lambda=

(100)" role="presentation" style="position: relative;">⎛⎝⎜100⎞⎠⎟$$\left(\begin{array}{ccc}1& & \\ & 0& \\ & & 0\end{array}\right)$$

Λ=⎝ ⎛​1​0​0​⎠ ⎞​
因此$A$的特征值为$1,0,0$

求$Q^{-1}AQ=Q^{T}AQ=\Lambda$

1. 求出$A$的特征值${\lambda }_1,{\lambda }_2,{\lambda }_3$
2. 求出对应的特征向量${\alpha }_1,{\alpha }_2,{\alpha }_3$
3. 该特征向量为${\gamma }_1,{\gamma }_2,{\gamma }_3$
   1. 如果特征值不同，只需单位化（实对称矩阵不同特征值所对应的特征向量相互正交）
   2. 若特征值有重根
      1. 如果特征向量已正交，只需单位化
      2. 如果特征向量不正交，需施密特（Schmidt）正交化
4. 构造正交矩阵 Q = ( γ 1 , γ 2 , γ 3 ) , Q − 1 A Q = Λ = ( λ 1 λ 2 λ 3 ) Q=(\gamma_{1},\gamma_{2},\gamma_{3}),Q^{-1}AQ=\Lambda=
   (λ1λ2λ3)" role="presentation" style="position: relative;">⎛⎝⎜λ1λ2λ3⎞⎠⎟$$\left(\begin{array}{ccc}{\lambda }_1& & \\ & {\lambda }_2& \\ & & {\lambda }_3\end{array}\right)$$
   Q=(γ1​,γ2​,γ3​),Q−1AQ=Λ=⎝ ⎛​λ1​​λ2​​λ3​​⎠ ⎞​

例：已知 A = ( 1 a − 1 a 3 1 − 1 1 1 ) A=

A=⎝ ⎛​1a−1​a31​−111​⎠ ⎞​，若$r(A)=2$，求$a$，求正交矩阵$Q$和对角矩阵$\Lambda$，使$Q^{-1}AQ=\Lambda$

求$a$可以用行变换，这里用秩的定义，即行列式不为$0$。由于$A$中有二阶子式
∣ 3 1 1 1 ∣ ≠ 0

\ne 0 ∣ ∣​31​11​∣ ∣​=0
于是$r(A)=0\iff |A|=0$
∣ 1 a − 1 a 3 1 − 1 1 1 ∣ = ∣ 0 a − 1 a + 1 3 1 0 1 1 ∣ = − ( a + 1 ) 2

|1a−1a31−111|" role="presentation" style="position: relative;">∣∣∣∣1a−1a31−111∣∣∣∣$$\left|\begin{array}{ccc}1& a& -1\\ a& 3& 1\\ -1& 1& 1\end{array}\right|$$

=

|0a−1a+131011|" role="presentation" style="position: relative;">∣∣∣∣0a+10a31−111∣∣∣∣$$\left|\begin{array}{ccc}0& a& -1\\ a+1& 3& 1\\ 0& 1& 1\end{array}\right|$$

=-(a+1)^{2} ∣ ∣​1a−1​a31​−111​∣ ∣​=∣ ∣​0a+10​a31​−111​∣ ∣​=−(a+1)2
因此$a=-1$，有
A = ( 1 − 1 − 1 − 1 3 1 − 1 1 1 ) A=

(1−1−1−131−111)" role="presentation" style="position: relative;">⎛⎝⎜1−1−1−131−111⎞⎠⎟$$\left(\begin{array}{ccc}1& -1& -1\\ -1& 3& 1\\ -1& 1& 1\end{array}\right)$$

A=⎝ ⎛​1−1−1​−131​−111​⎠ ⎞​
由$A$的特征多项式
∣ λ E − A ∣ = ∣ λ − 1 1 1 1 λ − 3 − 1 1 − 1 λ − 1 ∣ 找除主对角线外对应位置两数字和为 0 = ∣ λ 0 λ 1 λ − 3 − 1 1 − 1 λ − 1 ∣ = λ ( λ − 1 ) ( λ − 4 )

|λE−A|=|λ−1111λ−3−11−1λ−1|找除主对角线外对应位置两数字和为0=|λ0λ1λ−3−11−1λ−1|=λ(λ−1)(λ−4)" role="presentation">|λE−A|=∣∣∣∣λ−1111λ−3−11−1λ−1∣∣∣∣找除主对角线外对应位置两数字和为0=∣∣∣∣λ110λ−3−1λ−1λ−1∣∣∣∣=λ(λ−1)(λ−4)$$\begin{array}{rl}|\lambda E-A|& =\left|\begin{array}{ccc}\lambda -1& 1& 1\\ 1& \lambda -3& -1\\ 1& -1& \lambda -1\end{array}\right|\\ & 找除主对角线外对应位置两数字和为0\\ & =\left|\begin{array}{ccc}\lambda & 0& \lambda \\ 1& \lambda -3& -1\\ 1& -1& \lambda -1\end{array}\right|=\lambda (\lambda -1)(\lambda -4)\end{array}$$

∣λE−A∣​=∣ ∣​λ−111​1λ−3−1​1−1λ−1​∣ ∣​找除主对角线外对应位置两数字和为0=∣ ∣​λ11​0λ−3−1​λ−1λ−1​∣ ∣​=λ(λ−1)(λ−4)​
因此$A$的特征值为$0,1,4$
对$\lambda =1$
( E − A ) = ( 0 1 1 1 − 2 − 1 1 − 1 0 ) → ( 1 0 1 0 1 1 0 0 0 ) ⇒ α 1 = ( − 1 , − 1 , 1 ) T (E-A)=

(0111−2−11−10)" role="presentation" style="position: relative;">⎛⎝⎜0111−2−11−10⎞⎠⎟$$\left(\begin{array}{ccc}0& 1& 1\\ 1& -2& -1\\ 1& -1& 0\end{array}\right)$$

\rightarrow

(101011000)" role="presentation" style="position: relative;">⎛⎝⎜100010110⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)$$

\Rightarrow \alpha_{1}=(-1,-1,1)^{T} (E−A)=⎝ ⎛​011​1−2−1​1−10​⎠ ⎞​→⎝ ⎛​100​010​110​⎠ ⎞​⇒α1​=(−1,−1,1)T
对$\lambda =4$
( 4 E − λ ) = ( 3 1 1 1 1 − 1 1 − 1 3 ) → ( 1 0 1 0 1 − 2 0 0 0 ) ⇒ α 2 = ( − 1 , 2 , 1 ) T (4E-\lambda)=

(31111−11−13)" role="presentation" style="position: relative;">⎛⎝⎜31111−11−13⎞⎠⎟$$\left(\begin{array}{ccc}3& 1& 1\\ 1& 1& -1\\ 1& -1& 3\end{array}\right)$$

\rightarrow

(10101−2000)" role="presentation" style="position: relative;">⎛⎝⎜1000101−20⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& -2\\ 0& 0& 0\end{array}\right)$$

\Rightarrow \alpha_{2}=(-1,2,1)^{T} (4E−λ)=⎝ ⎛​311​11−1​1−13​⎠ ⎞​→⎝ ⎛​100​010​1−20​⎠ ⎞​⇒α2​=(−1,2,1)T
对$\lambda =0$
( 0 E − A ) = ( − 1 1 1 1 − 3 − 1 1 − 1 − 1 ) → ( 1 0 − 1 0 1 0 0 0 0 ) ⇒ α 3 = ( 1 , 0 , 1 ) T (0E-A)=

(−1111−3−11−1−1)" role="presentation" style="position: relative;">⎛⎝⎜−1111−3−11−1−1⎞⎠⎟$$\left(\begin{array}{ccc}-1& 1& 1\\ 1& -3& -1\\ 1& -1& -1\end{array}\right)$$

\rightarrow

(10−1010000)" role="presentation" style="position: relative;">⎛⎝⎜100010−100⎞⎠⎟$$\left(\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$$

\Rightarrow \alpha_{3}=(1,0,1)^{T} (0E−A)=⎝ ⎛​−111​1−3−1​1−1−1​⎠ ⎞​→⎝ ⎛​100​010​−100​⎠ ⎞​⇒α3​=(1,0,1)T
实对称矩阵，特征值不同，特征向量已经正交，只需单位化
γ 1 = 1 3 ( − 1 − 1 1 ) , γ 2 = 1 6 ( − 1 2 1 ) , γ 3 = 1 2 ( 1 0 1 ) \gamma _{1}=\frac{1}{\sqrt{3}}

(−1−11)" role="presentation" style="position: relative;">⎛⎝⎜−1−11⎞⎠⎟$$\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right)$$

,\gamma _{2}=\frac{1}{\sqrt{6}}

(−121)" role="presentation" style="position: relative;">⎛⎝⎜−121⎞⎠⎟$$\left(\begin{array}{c}-1\\ 2\\ 1\end{array}\right)$$

,\gamma _{3}=\frac{1}{\sqrt{2}}

(101)" role="presentation" style="position: relative;">⎛⎝⎜101⎞⎠⎟$$\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$$

γ1​=3 ​1​⎝ ⎛​−1−11​⎠ ⎞​,γ2​=6 ​1​⎝ ⎛​−121​⎠ ⎞​,γ3​=2 ​1​⎝ ⎛​101​⎠ ⎞​
令
Q = ( γ 1 , γ 2 , γ 3 ) = ( − 1 3 − 1 6 1 2 − 1 3 2 6 0 1 3 1 6 1 2 ) Q=(\gamma_{1},\gamma_{2},\gamma_{3})=

(−13−1612−13260131612)" role="presentation" style="position: relative;">⎛⎝⎜⎜⎜−13√−13√13√−16√26√16√12√012√⎞⎠⎟⎟⎟$$\left(\begin{array}{ccc}-\frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{3}}& \frac{2}{\sqrt{6}}& 0\\ \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}\end{array}\right)$$

Q=(γ1​,γ2​,γ3​)=⎝ ⎛​−3 ​1​−3 ​1​3 ​1​​−6 ​1​6 ​2​6 ​1​​2 ​1​02 ​1​​⎠ ⎞​
则
Q − 1 A Q = Λ = ( 1 4 0 ) Q^{-1}AQ=\Lambda=

(140)" role="presentation" style="position: relative;">⎛⎝⎜140⎞⎠⎟$$\left(\begin{array}{ccc}1& & \\ & 4& \\ & & 0\end{array}\right)$$

Q−1AQ=Λ=⎝ ⎛​1​4​0​⎠ ⎞​