ページ1:

(93章彰師資訊10%)
0
2-17
A= +3
2
date
No.
orthogonally diagonalizes A
Ax=λx
(AI)X=O
入 2 +71x17
232 2 =0
IA)II=-(+1)+(x-5)=0
1=12=-1,12=5
1+1=
2422=0
H₂
1
fita fel a fit
利用Gram-schmidt 正交化法,將xx化為正交向量中央
-K
<dd>
1/11-12=+, eigenvector
x-afo] 16 fi adot
当入3=5
取
1-5 2
H=
635
F
-5
-65
17412) -21-33=0
T 0
Do
0
+
。
。
A-SPs & sest
0.
5.
GEE-JUIMP
395

ページ2:

date
No.
(93k資訊10%)
A-
入2
]
21] verify that A is normal and find a
unitary
处
2
P such that p* AP=D is diagonal
* out (3) 1 ] + A
A不是Hamitian
2
AA
12 1
2 à
5
。
==
=
2 入2
0
5
5
。
AA*
入
AA-AA
2
2.
A is normal
AX-AX. CA-AI)X=0
=0
|A-11/= (2-1)==0
(2入-六)(2一入+六)=0.
-
5
A=SDS+ St=5* 12271
5N=1 SAS=12
入1=2-12=2+9
½\1=2-^ (^ à][4]
EU
= a[ 1 ] az
取一
GEE-JUMP
=0
29
07
D=
0
2+1
PAP =P A p* = pt
396-

ページ3:

date
F:50 SX44: Show that tollowing conditions are equivalent for
p² po The tour of pare athogenimal
PERMAN The columns of pare athonormal
Pap=
U
ER
①⇒日
hnn
当
or'
No
an in matrik P
193高应电子)
17ppt=1
[41]
Va
Un
U
10
[ 47.
U
Un
|=
• 4th 137
is that
=I
·chcht
λ=
ale orthonormal Lov
397-
GEE JUMP

ページ4:

I=4x4 T
→③
|{V, V... V₁₁] = I
| Vtv utv₂
74
LU&V V&V
*****
4th 1
I = 4+.
Vitun
fax
fax of /
[VVVn] are orthonormal columns
例:find T, for TH=F_
H=[! | | 1
TH
FFT. I
TELI
(of):H由正交行向量構成
1/2 1/2 1/2 1/2
½ -1/2 - 1/2 1/2
則A=A=AT,TH=F,TOA=F
TAA==FATT=SFAT
T= =F(H)==FHT
GEE-JUMPT
-1
J
1-1-
T
1-5-1
-398
開為解8
T
I
-1
J
date
No.
(94台大电机(0%)

ページ5:

: ) A * = A + <> NAX=1x1
PA* = A + <AX, Ay > = <x117
即公正転換,保長又保留
說明:
A為鏡射矩陣
經內鏡射
变笑
錢射転換保長,A为公正矩陣AYA
pf:
PACE
DATE
(93中山电机)
-M
⇒
X=AAA=I
A* = A. Anoth
MAXI-CA. AX = (AX) * (AX) A=SD51, 5t=5* Start mottix
= X**AX = X*1X = X*X=11x12²
⇐
XEC^ AXI-IXIP-
LAX, AX7=24x7
(A)*(X)=XXX
AX-XIX-0
X*(TA-I)X=0 for any vector X
AA-I=0
A*=AT
AA-I
:鏡射分旋轉都是保長变换,鏡射矩陣與轉矩陣都是公正矩陣
399-
IMAGE

ページ6:

PAGE
DATE
「鼔明:設A為鏡射矩陣
公正転換保角
d=e
AX
=>
AAAA=I
<AX, Ay> = (AY)* (AX) = **A* AX = x*1X = 4*X = 2x,47
* MAXIMA coc = Ill
be
IMAGE
<AX, Ay>=<X17
(A)*(AX)=X
**AX=*X
*CAA-1)X=0
A-I
A*-AT

ページ7:

date
No.
34-9 plate portrait (3)
烤1
迥
[X010)]
將A对角化 A=SPS+
[^107/X17
[ 1/1 = ^/, X = 4ext
$-1/2
定理-説明:
若A為6階方陳何对角化,IA-AII=
12-1717-2)-(2-3720
7=1.2.2.3.3.3
M(X)=(14-1719-21(2-3)為最小多項式
(^1 01 131: AE khen, if A-3 A³+ 2A=0 prove or
=
1 Gent]
X=Get thebt
代入大=0
160) =
0 λ) disprove that A is diagonalizable over real
fieldo
194中央权学10%)
pf flop=x³-3x²+2x
(A)=AS-2A+2A=D
X為A之零化多項式.
fix=x144-3x²+2)
=X1X-11(x2)
令m)為A之最小多項式
mix必為f(x)因式
∵∵x沒有重根,m(x)沒有重根,A必可
对角化。
A之所有可能相異特徵值入=0,1,圾
Ax=2x (A-1)x=6 AER is
(10)] - Gp,+SpGs real; 'A-71 Rep
(X210)
定義:冪零矩陣(nilpotent) /
可确保特徵向量都是実权向量.
A diagonalizable over real field
An階方陣,若存在正整數r,使得AD,稱A為冪零矩陣.
說明:取f(x)=xr:f(A)=A=0 f(x)為A之零化多項式,令MD)為A
之最小多項式max=x²,8033;入口為A之所有特徵值,最小多項式有重根
A.不可对角化。
為四階標準零
例如:
To o b
Alp
000
100
1000
A=3780
10000
1000
P375

ページ8:

date
No.
①当入20,如:-5,2=-3
x=49 +63
be* {GPHP] = lim Se² R=00
lim X = fom ciest p₁ = 0
P1=0
B
(0.0) stable attractor poat
如x=2=-3
②当=b<0
X=GRE+ + 2B*
Jin Xin Ho
"X-10)=4PH+632
op
(0.0) / stable attractor point
③当,220,如入1=5,12=3
X=45 +4
x=
Anx Clip Be = ±0
2=
lon X-lim Ge² = 100
GEE-JUMP
开
10,0) a unstable repelling point
- 374 -

ページ9:

①=270311=1=3
X=4pe² +63
373 date
No.
B3'1 Does A²=0 imply A=0,
A is non? why ? 15%7
lim X = lim (CP+GB) est = lim e* (10) = (2) find the most genera ax2.
matrix
A such that A0 (7
194或大微机电)
sol: (1) A=0 7-A=D
例如:A=o
A
(0.0) / unstable repelling paint 12A+0
⑤当<od,如入1=-3,2=5
X=44e2+ + Best
from X = fim spett + Best = lim apes=12
若62=0
tmx-pet-0
+79
(2)令A=550
6751
T's =0
o b
most general A
S為任意可逆矩陣
· ES =α b₁
P
B
1310, A-fab70 17/a
[cdoc
D
= 5697
-49-M-1
To c] ad-bc -c a
+
=
[-ac a²].
ad-bc -² ac
11: A is hen real matrix if A²^=0
D
(0.0) Saddle point () then A"=D prove it. (944/12/2310%)
Sal: A² = 0, IA") =0
正確為:令fo=x²n
A^=0== 0証)
A=0
舉反例:A=[0].
A=0==0 销
:(A)=0,f(x)為A之零化多項求,
令的內為最小多項式M120=10,8=2.3.4.
A.f=0,MA之根為特徵值,A之所有特徵值為零
11-211= (-1) (2-0)=0
依據fgyley - Hamilton、定理
A=0証單
GEE-JUMP

ページ10:

⑥入户,冲
外
°
date
No
AX 1/1
[]
est.
I Gent + Gent
ligent tibet
(4(cottisht) + (lost-int)
Tia (Wittigint)-ics (cost-ment)
[+] (kaust-kisint
(4.) (kcost +lent 1
147
Sing
么
lord +
Sint
K
Just + 1 Sint
妆
uscott).
= Sincott)
=Jk+ cos(+1)
-=-=-9+=+====x+x
--
GEE-JUMP
+
(0.0) // marginally stable point
376 -
k
k

ページ11:

2
32
1=2+3%
12:2-31
X=4pet Belt
得了
los+3),
x=ike*sin(+3)
X=√1776)
11²+15= (√k²+15 est
X1= √AR+ SINCO+31)
X2=UK*331)
(UK)
->
377
+6
dote
we
No.
(010) unstable spiral point
2

ページ12:

date
No.
d5特殊矩陣
§5-1厄米特矩陣与实对箱矩陣
定义:1.当A,稱A為厄米特矩陣
2.当AGRAM,但T=A,稱A為澳对稱矩陣
对锰矩車必為厄米特矩車(印厄米特定理,实对验必足)
= A*= A
or实特矩車為实对转矩車.
3.当AGA為公正(vity)矩車,实物论正矩陣物為交(athogonal)矩陣
即正交矩陣滿足AM=A+
4. 設 x7
X= tot complex + + te
GEE-JUMP
<為
$47
<ry>=*=
=+x+
13.
47177=27187
147
(X3J
<x₁/7=1+x=<y.x7 = x++
ER
- 378

ページ13:

date
No
定理5:厄米特矩陣之特徵值必為果板(a)中兴电%)(00瀆科电子)(文大电)
A=A, AX-XXto
pt
欲証入二人
(AX)* = (xx)*
X *A* = XX*
*A=x*,又不→线过不用乾置
(入-x)x*x=0
得入-x=0
H
枚入 is read
定理7:厄米特矩陣中,相異特徵值对应之特徵向量必正交(93电机)
AAA=XAX; = X; X; @#);
1 xxx; > = X; *- Xx = 0
hote: A= f 2
+ AA
(2-1. 3...)
***
A=1
12 3 5
x*A=gx*
30-6
A=A
*=*
15-671
Ax=72, 14-2117=0
(x-1)*x=0;ioreal,实对稳矩陣之特徵向量必為
<x,xx>=0.
实友向量
又实对稱矩陣必有厄米特矩陣,相與特徵值对应之特徵向量正交
- 379 -
GEE-JUMP

ページ14:

2
2 出门
A2
54| find athenormal
exporectory
Sel
-24 5
Ax=x (A-I)X=0.
LHILE
-254 20
1
24
< 13.1=Y
2 -4
EXY-S
1A-1=-(+1)=(x-10)=0
取入1=10.2=3=1
当入1=10
8
2
2
华
(if)
京
'YIL
HI+3=1A)=12
+91-93-0
-44
51120
245x3
[X]
L
11 2
2-4
-2
4
134
4
11
2
LIXI
· XTX BY EXT
XTY
5115-7
DEX=1)
利用Gram-christ 正交化法
GEE-JUMP
121
0
380-
date_
No.

ページ15:

<3>
Les, 937
当2=3=1
2
0
45
12
2
egenvedors tok || H5≤ 4
x3
M=10equector
0
5
dote
No.
1---2
{中..}為正交特徵向量
=
{v1.02.137為歸一正交特徵向量
11145
- 381-
GEE-JUMP

ページ16:

曰:
date
No.
1-33
A353
(9)米科自动化10%)
634
07
=
A=b=-2, egenvectors 12 =4|1|+51 =41+4.
[+]
N=4, eigenvector to = 43 1 = 63X3 +3+3
At+A
2
→EA=4)
EA=-2)
Gram-Schnitten..
<x,d>
do to
& 0x3
<dd>
Dear$7
<>
当2
当=4
1=14, +15 Tre
1-1sd Take 432 b
to
GEE-JUMP
382-

ページ17:

date
(台科化工20%)
Sal
2入
2
2
2
252
17 tind a set of orthonormal
egnvators
1
2 2
(AAI)X=0
25 X 2 * =0
T
1A-AII=-(2-1)(入-7)=0
取入=7,12=3=1
1-5 21714
222 12 =0
12-563
当0=3=1
12
2-4 2x=0
1 2 1
(17
。
eingervatory
<2>≠0
F
+4|1=611651
利用Gran-Shmidt正交化法,將xx化为正向量為3
<d, dz>
383-
No.
GEE-JUMP

ページ18:

1/2=13=1 eigenvectors
01631-605+603
t
L
{中子为正交特微回量
th=
d
F
11 JG
=
GEE-JUMP
-384-
date
No.

ページ19:

PAGE
......
DATE
11] Find arthonormal eigenvectory
(93輔仁电子18%)
(1
AIII
Sol
I I I
AX=入X(A-I)X=0
11-λ
1
+A
l L
IA-XII=-²(入-3)=0
入=32=3=0
2
1
XFO
D
X=4x+6x
^3=3
(D-12)-2X1+22=0
0
+
寺0=3=0
G
H
-2
]]
+6
-D
<xx>=0
2837=0
E(X=3)=E(入=0)
385-
IMAGE

ページ20:

PACE
DATE
利用Gran-shmidt 正交化法,將x,xx化為交量好,好
买好
0
<td>
取
办
23.627
2
43-11931-56
a set of athonormal egenuatory is {ui.ub.us}
最多項式 md)=(x-3)
MAGE
- 386-

ページ21:

有問題!
(93电机10%)
1-2
A=2
|-
date
No.
2 -37 find exeruaties, eigenetan
-6
-2 0
入==-3
equator ([2]-off + [+]=4x+6x=44+14
13=5
egative [2] 6 [6]=>
F
<> to <> to <\2>=0
7
Ecl=-3)
如不是特徵向量
並不是所有矩陣都有一组正交特徵向量
-387-

ページ22:

BA [cose Sp
L-ST40
6050
(1) show that A is a unitary matrix
131 find the eigendure and eigenvector
(3) shod that eigenvections
(4) find for A=11010*
are or
orthogonal
(94中央电机20%)
pf A*A=
Trase Sub]
STIO COSO
date
No
·· A*A = I, PA² = A1, Ais a unitary matry
(2) 44-7X7 (4-71)
5009560-7
5749 75X17=0
1A-21/= (-2)+ sin 19
=(cos(9-7-isin 19)
₁ = costisin1 = 10
5220
12 = Coso-istro = è
= cos + sin
(3) <XXXX
[A
=0,XLX
(cost)-2+isin 0/=0
XXXXX.112
= [1
=2
XXXX[42
(4) BRU = [Y/JE 1/5]
[-ising sino ] [×17=0
sine istine]
11/JE NJE
D=
AUDU
月ㄩˇ= U+為A之公正对角化,
= Cos-isin
Fising sin x]=0
|-sino sinx!
GEE-JUMP
[x]=a[]=xx
388-

ページ23:

date
No.
定理:厄米特矩陣中,特徵值重根板=特徵向量攻
AT algebinic multiplicity = geometric multiplicity
特徵向量不缺少,必可对角化
定理:厄米特矩陣中,必存在一组歸一正交特徵向量,將A对角化,A=SDS+,且=*
good
0
=
110....0
169
57=57
45*131*
=I
+
整理:当A*A,以第一正文持徵向量將A对角化,A=SD5
即5=5*,為公正矩陣,稱A可公正对角化
2.当A為奧對稱矩陣,以歸一正交向量,將A对角化,A=SDS+
則5=(4)為正交矩陣,稱A可正交对向化,正交矩陣,即如实长的公正
(orthogonal)
矩陣
121-22
3.当A=AY,稱A為正規(roital) matrix,必可公正对角化.
即相異特做值对应之特徵向量必正文
正規矩陣包含下列矩陣:
会証
a. A=A原米特矩陣,实厄米特矩陣,特徵值必為突技
b. A--A.反厄米特矩陣or反实米特矩陣,特徵值為純虛枚震
C.AA. 公正矩陣☆实公正矩陣(即正交矩陣),特徵值大小必為1
389-
`GEE-JUMP

ページ24:

姚
' date
A [2, 17] and a corbay matix's that degli A
Hλ
co
A-A
2-1.
HX 1-XJ2.
=0
/A-11/=(2-1) (1-1)-2=0
=3入=0
入=0,70=3
2
1
|-
Hλ
161
1417/6/177-64
</1X17= |**= {111}|
Hi..
=0
1-2
0
0
D=
T
0
3.
S=
A-5051 51-5*
GEE-JUMP
390 -
No.

ページ25:

date
No.
O 2-1
A= 23-2 find a matrix Ll such that Libet -A, DIs a diagonal mottik
120
A=UDLUT = UD(
乙為正交矩陣
∴T=A ACR,A可正交对角化
K-
(A-AI)X=0
入 2 +7147
23-1-2
-
[x] [x- =
(A-A1|=-(λ+1)-(1-5)=0
11=2=-1,73=5
当入1=2=1
112 +71417
2.4 2
x=0
=0
A=-1
2
ㄒㄧ
(9)北斜电机25%)
1-1+=+)=2, 13=5
127
+27= 0 5+ 1-74
Liff
利用Cron-schoist尘化法
0
0
127
== expectos 2 = -1 +1-2
[4.7
0
5
391.
152
<5>
GEEN JIVE

ページ26:

当13=5
15
date
No.
2
22
-|
U=
12:0
2.5x3
1417
12-13 2 = 1303 1773-641-613-0
十
0=4
47
。
|-
0
0
0
•A=UDUT But=ut
GEE-JUMP
392 -
0
0
5

ページ27:

PAGE
......
DATE
(93中央电机15%)
M-15 / 141 1
1-7] unitary diagonalizes M
Sol
M*=M, MX-XX (M-AI)X = 0
J
1147
|=0
Ità
討好
当入=1
(√) (+1)
2-1=0,入=±1
113-1147
+
H
=G
HA
2
A
AT
盒
以
14-651-53
Ha
(HJ)
今
145
=0
<> (115 th])
Ità
Itλ
士
-393-
İMAGE

ページ28:

PAGE
DATE
取
=
HJ
JEFU
tà
001697
1
564057
H+4
IMAGE
394 -