6.2 Definitions and examples
DEFINITION6.1.1(Eigenvalue,eigenvector) Let A be a complex square matrix. Then if λis a complex number and X a non –zero complex column vector satisfying AX X λ=, we call X an eigenvector of A , while λ is called an eigenvalue of A . We also say that
X is an eigenvector corresponding to the eigenvalue ¸
. So in the above example 1P and 2P are eigenvectors corresponding to 1λ and 2λ, respectively. We shall give an algorithm which starts from the eigenvalues of
a h A h
b ⎡⎤
=⎢⎥⎣⎦
and constructs a rotation matrix A such that t P AP is diagonal. As noted above, if λis an eigenvalue of an n n ⨯ matrix A , with corresponding
eigenvector X , then ()0n A I X λ-=, with 0X ≠, so det()0n A I λ-= and there are at most n distinct eigenvalues of A .
Converly if det()0n A I λ-=, then ()0n A I X λ-= has a non –trivial solution X and so ， is an eigenvalue of A with X a corresponding eigenvector. DEFINITION6.1.2(Characteristicpolynomial,equation)
The polynomial det()n A I λ- is called the characteristic polynomial of A and is often denoted by ()A ch λ. The equation det()0n A I λ-= is called the characteristic equation of A. Hence the eigenvalues of A are the roots of the characteristic polynomial of A . For a 22⨯ matrix a b A c d ⎡⎤
=⎢
⎥
⎣⎦
, it is easily verified that the characteristic polynomial is ()det traceA A λλ-+2, where traceA a d =+ is the sum of the diagonal elements of A . EXAMPLE6.2.1 Find the eigenvalues of 2112A ⎡⎤
=⎢⎥⎣⎦and find all eigen-vectors.
Solution. The characteristic equation of A is 2430λλ-+=, or
(1)(3)0λλ--=.
Hence 1λ= or 3. The eigenvector equation ()0n A I X λ-= reduces to
x y λ
λ-⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥-⎣
⎦⎣⎦⎣⎦210120， or
()()x y x y λλ-+=+-=20
20
.
乙脑疫苗不良反应Taking λ=1 gives
x y x y +=+=00
,
which has solution x y =-, y arbitrary. Conquently the eigenvectors corresponding to
1λ= are the vectors y y -⎡⎤
⎢⎥⎣⎦
，with 0y ≠.
Taking 3λ= gives
x y x y -+=-=,
which has solution x y =-, y arbitrary. Conquently the eigenvectors corre- sponding to 3λ= are the vectors y y ⎡⎤⎢⎥⎣⎦
, with 0y ≠. Our next result has wide applicability:
THEOREM6.2.1 Let A be a 22⨯ matrix having distinct eigenvalues 1λ and 2λ and corresponding eigenvectors 1X and 2X . Let P be the matrix who columns are 1X and
2X , respectively. Then P is non –singular and 11
200P AP λλ-⎡⎤
=⎢⎥⎣⎦
Proof. Suppo 111AX X λ= and 222AX X λ=. We show that the system of homogeneous equations
120xX yX +=许拉斯
has only the trivial solution. Then by theorem 2.5.10 the matrix []12P x x = is non –singular. So assume
120xX yX +=. (6.3)
Then 12()00A xX yX A +==, so 12()()0x AX y AX +=. Hence
11220x X y X λλ+=. (6.4)
Multiplying equation 6.3 by 1λ and subtracting from equation 6.4 gives
下一页的我
212()0yX λλ-=.
Hence 0y =, as 21()0λλ-≠ and 20X ≠. Then from equation 6.3, 10xX = and hence 0x =.
Then the equations 111AX X λ= and 222AX X λ= give
[][][][]1212121112220000AP A x x Ax Ax x x x x P λλλλλλ===⎡⎤⎡⎤
===⎢⎥⎢⎥
⎣⎦⎣⎦
EXAMPLE6.2.2 Let 2112A ⎡⎤
=⎢
⎥⎣⎦
be the matrix of example 6.2.1. Then 111X -⎡⎤=⎢⎥⎣⎦ and 211X ⎡⎤
=⎢⎥⎣⎦
are eigenvectors corresponding to eigenvalues
1 and 3, respectively. Hence if 1111P -⎡⎤=⎢
⎥
⎣⎦
, we have 1
1003P AP -⎡⎤=⎢⎥⎣⎦There are two immediate applications of theorem 6.1.1.The first is to the calculation of n A : If
112(,)P AP diag λλ-=, then 21(,)A Pdiag P λλ=
and 11111
1221000()000
n
n n
n n A P P P P P P λλλλλλ---⎡⎤⎡⎤⎡⎤===⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦
. The cond application is to solving a system of linear differential equations
dx ax by dt =+ dy xc dy dt
=+ where a b A c d ⎡⎤
=⎢
⎥
⎣⎦
is a matrix of real or complex numbers and x and y are functions of t. The system can be written in matrix form as X AX •
=,
where x X y ⎡⎤
=⎢⎥⎣⎦
and
dx x dt dy
y dt X
••
•⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦
We make the substitution X PY =, where 12x Y x ⎡⎤
=⎢⎥⎣⎦
独身者. Then 1x and 1y are also functions
of t and ()X PY AX A PY •
•
===, so 1120()0Y P AP Y Y λλ•
-⎡⎤
==⎢
⎥
⎣⎦
.
Hence 111x x λ•= and 111y y λ•
=.
The differential equations are well –known to have the solutions 111(0)t
x x e λ=and
111(0)t y y e λ=, where 1(0)x is the value of 1x when 0t =.
[If
dx
kx dt
=, where k is a constant, then ()0kt kt kt kt kt d dy e x ke x e ke x e kx dt dt
-----=-+=-+=. Hence kt e x - is constant, so 0(0)(0)kt k e x e x x --==. Hence (0)kt x x e -=.]
However 111(0)(0)(0)(0)x x P y y -⎡⎤⎡⎤=⎢⎥⎢⎥⎣⎦⎣⎦
, so this determines 1(0)x and 1(0)y in
terms of (0)x and (0)y . Hence ultimately x and y are determined as explicit functions of t , using the equation X PY = .
EXAMPLE6.1.3 Let 2345A -⎡⎤
=⎢⎥-⎣⎦
. U the eigenvalue method to
derive an explicit formula for n A and also solve the system of differential
equations
23dx x y dt =- 45dy x y dt
=-, given 7x = and 13y = when 0t =.
Solution. The characteristic polynomial of A is 332λλ++ which has distinct roots 11λ=- and 22λ=-. We find corresponding eigenvectors 111X -⎡⎤
=⎢
⎥⎣⎦
¸ and 234X ⎡⎤=⎢⎥⎣⎦. Hence if 1314P ⎡⎤
=⎢⎥
⎣⎦
, we have 1(1,2)P AP diag -=--.
Hence 11
((1,2))((1),(2))1343(1)014110
(2)131043(1)14021343132(1)11142432332(1)442
342n n n n n
n n n n
n n n
n n
n
n A Pdiag P Pdiag P --=--=---⎡⎤-⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥
--⎣⎦⎣⎦⎣⎦-⎡⎤⎡⎤⎡⎤
=-⎢⎥⎢⎥⎢
⎥-⎣⎦⎣⎦⎣⎦
-⎡⎤⨯⎡⎤=-⎢⎥⎢⎥-⨯⎣⎦⎣⎦⎡⎤
吉祥话成语
-⨯-+⨯=-⎢⎥
-
⨯-+⨯⎣⎦
To solve the differential equation system, make the substitution X PY = . Then 11113,4x x y y x y =+=+. The system then becomes
1111
2x x y y •
•
=-=-so 11(0)t x x e -=and 211(0)t y y e -=. Now .
111(0)(0)43711(0)(0)11136x x P y y ---⎡⎤⎡⎤
⎡⎤⎡⎤⎡⎤===⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦
.so 111t x e -=-and 212t y e -=, Hence
222211113(6)1118,114(6)1124t t t t t t t t x e e e e y e e e e --------=-+=-+=-+=-+For a
more complicated example we solve a system of inhomogeneous recurrence relations.
EXAMPLE6.2.4 Solve the system of recurrence relations
12122
n n n n n n x x y y x y +=--=-++,
为什么软件打不开given that 00x = and 00y =.
Solution. The system can be written in matrix form as
1n n X AX B +=+,
Where 2112A -⎡⎤=⎢
⎥-⎣⎦ and 12B -⎡⎤
=⎢⎥
⎣⎦
. It is then an easy induction to prove that
1
02()n n n X A X A A I B -=++++L . (6.5)
Also it is easy to verify by the eigenvalue method that车厘
1313113222
1313n n n
n
端午节手抄报
n n A U V ⎡⎤+-==+⎢⎥-+⎣⎦
