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(完整版)矩阵的特征值与特征向量-英文文献

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2021-02-17 01:51
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2021年2月17日发(作者:avoided)


6.2 Definitions and examples


DEF INITION6.1.1


(Eigenvalue,eigenvector) Let


A



be a complex square matrix. Then if


?


is a complex number and


X



a non



zero complex column vector satisfying


AX


?


?


X


,


we call


X



an eigenvector of


A


, while


?



is called an eigenvalue of


A


. We also say that


.


X



is an eigenvector corresponding to the eigenvalue ?


So in the above example


P


1



and


P


2



are eigenvectors corresponding to


?


1



and


?


2


,


respectively. We shall give an algorithm which starts from the



eigenvalues of


?


a


h


?


and constructs a rotation matrix


A



such that


P


t


AP



is diagonal.



A


?


?


?


?

< p>
h


b


?


As noted above, if


?


is an eigenvalue of an


n


?


n



matrix


A


, with corresponding


eigenvector


X


, then


(

< br>A


?


?


I


n


)


X


?


0


, with


X


?


0


, so


det(


A


?

< br>?


I


n


)


?


0



and there are at


most n distinct eigenvalues of


A


.


Conversely if


det(


A


?


?


I


n


)


?< /p>


0


, then


(

A


?


?


I


n


)


X


?


0< /p>



has a non



trivial solution


X



and


so




is an eigenvalue of


A



with


X



a corresponding eigenvector.


DEFINITION6.1.2


(Characteristicpolynomi al,equation)


The polynomial


det(


A


?


?


I


n


)



is called the characteristic polynomial of


A



and is often


denoted by


ch


A


(


?


)


. The equation


det(


A


?


?


I


n


)


?


0



is called the characteristic equation of


A. Hence the eigenvalues of


A



are the roots of the characteristic polynomial of


A


.


For a


2


?


2



matrix


A


?

< br>?


?


a


b


?


, it is easily verified that the characteristic polynomial is


?

?


c


d


?


?


2


?


(


tr aceA


)


?


?


det


A


, where


trac eA


?


a


?


d



is the sum of the diagonal elements of


A


.


?< /p>


2


1


?


EXAM PLE6.2.1



Find the eigenvalues of


A


?


?


?


and find all eigen- vectors.


1


2


?

< p>
?


Solution.



The characteristic equation of


A



is


?< /p>


2


?


4


?


?


3


?


0

< p>
, or


(


?


?


1)(


?


?


3)


?


0


.


Hence


?


?


1



or 3. The eigenvector equation

< br>(


A


?


?


I


n


)


X


?


0



reduces to


1


?


?


x


?


?


0


?

< br>?


2


?


?


?


?


?




?


1


?


?


?


2


?


?


?


?


y


?


?


0


?


?

< br>or


(


2


?

< br>?


)


x


?


y


?


0


.


x


?


(


2


?< /p>


?


)


y


?


0


Taking


?


?


1



gives


x


?

y


?


0


x


?


y


?


0


,


which has solution


x

< br>?


?


y


,


y



arbitrary. Consequently the eigenvectors corresponding to


?


?


1



are the vectors


?


?



with


y


?


0


. < /p>


?


y


?


Taki ng


?


?


3



gives


?


?

y


?


?


x


?


y


?


0


,


x


?


y


?


0


which has solution


x


?


?


y


,


y



arbitrary. Consequently the eigenvectors corre-


sponding to


?


?


3



are the vectors


?


?


, with


y


?


0


.


Our next result has wide applicability:


THEOREM6.2.1



Let


A



be a


2


?


2



matrix having distinct eigenvalues


?


1



and


?


2



and


corresponding eigenvectors


X


1



and


X


2


. Let


P



be the matrix whose columns are


X


1



and


?


y


?


?


y


?


?


?


1


0


?


X


2


, respectively. Then


P



is non



singular and



P


AP


?


?< /p>


?



0


?


?


2


?


?

< p>
1


Proof. Suppose


AX


1


?


?


1


X


1



and

< p>
AX


2


?


?


2


X


2


. We show that the system of homogeneous


equations


xX


1


?


yX


2


?

< p>
0



has only the trivial solution. Then by theorem 2.5.10 the matrix


P


?


?


x

< br>1


x


2


?



is non



singular.


So assume


























xX


1


?< /p>


yX


2


?


0


.





















(6.3)


Then


A


(


xX


1


?


yX


2


)


?

< br>A


0


?


0


, so


x


(


AX


1


)


?


y

(


AX


2


)


?


0


. Hence

























x


?


1


X


1


?


y


?


2


X


2


?


0


.

















(6.4)


Multiplying equation 6.3 by


?


1



and subtracting from equation 6.4 gives


(


?


2


?


?


1


)


yX


2


?


0


.


Hence


y


?


0


, as


(


?


2


?


?


1


)


?< /p>


0



and


X


2


?


0


. Then from equation 6.3,



xX


1


?


0



and hence


x


?


0


.


Then the equations


AX


1


?


?


1

X


1



and

< br>AX


2


?


?

2


X


2



give


AP


?

A


?


x


1


x


2


?


?


?< /p>


Ax


1


Ax


2< /p>


?


?


?


?


x


1


?


x

< p>
2


?



?


?


0


?


?

?


1


0


?


?


?


x


1


x< /p>


2


?


?


1


?


P


?


?

< p>
?


0


?


?


?


0


?


2

?


?


2


?


EXAMPLE6.2.2 Let


A


?

?


?


2


1


?


be the matrix of example 6.2.1. Then < /p>


?


?


1


2


?


?


?


1

< p>
?


?


1


?


X


1


?


?

?



and


X

< br>2


?


?


?



are eigenvectors corresponding to eigenvalues


?


1


?


?


1


?


1 and 3, respectively. Hence if


P


?< /p>


?


?


1


0


?


?


1


?

< p>
1


?


?


1


, we



have


There are two


P


AP


?


?


?


?


?


0


3


?


?


1


1


?


immediate applications of



theorem first is to the calculation of


A


n


: If


P


?


1


AP


?


diag


(


?


1


,


?


2


)


, then


A


?

< br>Pdiag


(


?


1


,


?


2


)

< br>P



?


?


1


n


?


?


1


0


?


?


1


n


?


?


1


0


?


?


1


n


and


A


?


(


P


?


?

< br>P


)


?


P


?


0


?


?


P


?


P


?


0


?


?


2


?


?


2


?


?


0


n


0


?

< br>?


1


?


P


.


?


1


n


?


The second application is to solving a system of linear differential equations


dx


dy


?


ax

?


by







?


xc


?


dy



dt


dt


where


A


?


?


?


a


b


?


is a matrix of real or complex numbers and


x



and


y



?


?


c


d


?


?


are functions of t. The system can be written in matrix form as


X


?


AX


,


?


x


?


whe re


X


?


?


?


and


?


y


?


X


?


?


?< /p>


?


?


dx


?


?


x


?


?


dt


?


?


?

< p>
?


?


?


?


?



dy


?

< br>y


?


?


?


?


?


?


?


d t


?


?


?


x< /p>


1


?


We make the substitution


X


?


PY


, where


Y


?


?

< br>?


. Then


x


1



and


y


1



are also functions


?


x


2


?


of


t


and


X


?


PY


?


AX


?


A


(


PY


)


, so

Y


?


(


P


?


1


AP


)


Y


?


?


?


?


?


?


?


1


0


?


Y



.



?


?


0


?


2


?

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