Matrix

Matrix
紋的筆記-應用數學
⎛a a a ... a ⎞
11 12 13 1n
⎜ ⎟
⎜
a21a22 a23 ... a2n⎟
A= ( aij
) =
⎜ ⎟
⎜ ⎟
a a a ... a ⎟
( ij
⎝ m1 m2 m3 mn⎠
a :matrix element,i:row,j:column)
⎧2x+
3y= 1
ex: ⎨
⎩x−
y = 2
matrix addition
A B C
⇒
a + b = c
+ = ⇒ ij ij ij
Scalar multiplication
kA ⇒ ka ( ij )
zero matrix
(0) ,每一個元素都是0
row matrix
( ... ... ) , m = 1
column matrix
⎛⎞
⎜⎟,
n = 1
⎝⎠
⎛2 3 ⎞⎛x⎞ ⎛1⎞ ⎜ ⎟⎜ ⎟= ⎜ ⎟
⎝1 −1⎠⎝y⎠
⎝2⎠ square matrix
⎛... ... ⎞
⎜ ⎟,
m= n(只有方陣才有單位矩陣)
⎝... ... ⎠
unit matrix
⎛1 ⎜
⎜
⎜
⎜
⎝
1
⎞
⎟
⎟
⎟
⎟
1⎟
⎠
( ↔ :row列, :column行)
應用數學筆記

⎧ aij ≠ 0
diagonal matrix: ⎨
⎩aij
= 0
i = j
i ≠ j
⎛ ⎜
⎜
0
⎜
⎝ 0
0
0
0 ⎞
⎟
0
⎟
⎟
⎠
A 的相關矩陣:
Transpose(轉置)
Symmetric(對稱)
Antisymmtric
⎛1 2 3⎞
⎜ ⎟
ex:
⎜
2 1 4
⎟
⎜3 4 1⎟
⎝ ⎠
⎛0 −2 −3⎞
⎜ ⎟
ex:
⎜
2 0 4
⎟
⎜3 4 0 ⎟
⎝ ⎠
紋的筆記-應用數學
symmetric
a = a
T
ij ji
T
ij = ji = ij ( T
T
ij = ji =− ij ( T
a a a
a a a
antisymmetric
A = A)
Adjoint(伴隨) a
T
= ( a ) = ( a ) = a
+
Self-adjoint ; hermitian a
∗
= ( a ) = a
⎛ 4 2+
i ⎞
ex: A = ⎜ ⎟
⎝2−i−3 ⎠
A = − A)
+ ∗ ∗ ∗
ij ij ji ji
ij ji ij
+ ⎛ 4
⇒ A = ⎜
⎝2+ i
⇒ hermitian
2− i⎞ ⎟
−3 ⎠
⎛ 4
= ⎜
⎝2−i 2+
i⎞
⎟=
A
−3
⎠
Matrix multiplication
C AB
c = ∑ a b
= ⇒ ij ik kj
k
⎛ 2 3⎞
⎛ 1 −2⎞
ex: A = ⎜ ⎟,
B = ⎜ ⎟
⎝−1 1⎠
⎝−3 1 ⎠
∗
應用數學筆記

⇒
紋的筆記-應用數學
⎛ 2
AB = ⎜
⎝−1 3⎞⎛ 1
⎟⎜
1⎠⎝−3 −2⎞ ⎛−7 ⎟= ⎜
1 ⎠ ⎝−4 −1⎞
⎟=
C
3 ⎠
c = a b = a b + a b )
∑
( 11 1k k1
11 11 12 21
k
⎛ 2
ex: A = ⎜
⎝−3 1⎞
⎛−7 ⎟,
B = ⎜
4⎠
⎝ 5
3
0
2⎞
⎟
1⎠
⎛−9 ⇒ AB = ⎜
⎝41 6
−9 9 ⎞
⎟
−2⎠
⎛a1⎞ ⎛b1⎞
⎜ ⎟ ⎜ ⎟
AB ⋅ =∑ab,
A= ⎜
a2⎟
, B =
⎜
b2
⎟
⎜a⎟ ⎜
⎝ 3 ⎠ b ⎟
⎝ 3 ⎠
T
⇒ AB ⋅ = AB= ( a a
⎛b1⎞ ⎜ ⎟
a ) ⎜
b
⎟
= ab+ ab
⎜b⎟ ⎝ 3 ⎠
+ ab
ex: i i
i
A ⋅ B=∑a b
ex:
∗ ∗
i i
i
a a a
⎛b⎞ b a b
1
∗ ∗ ∗ ⎜ ⎟ ∗
1 2 3 ⎜ 2⎟
= ∑ i i
⎜ i
b ⎟
3
⇒ ( )
⎝ ⎠
1 2 3 2 1 1 2 2 3 3
∗ ∗
( A ⋅ A= a a = a
∑ ∑
2
i i i
i i
ex: ( AB) C= ABC ( )
( AB≠ BA)
左 ⇒ [( AB) C] = ∑ ( AB) C ( AB ) C
ij ik kj
k
右 ⇒ [ ABC ( )] = ∑ A( BC)
( )
ij ik kj
k
∴ Associativity(結合性)
∑ ∑
( AB) ij = Aik Bkj = AilBlj k l
ij:free index k:dummy index
T
ex: ( ABC) T T T
= C B A
T
左 ⇒ ( ABC) ( ABC)
)
應用數學筆記
= ∑ ∑ il lk kj ABC il lk kj
k l
k l
= ∑ Aik ∑ BklClj= ∑∑ ABC ik kl lj = ∑∑ ABC il lk kj
k l
k j
l k
= ∑∑ ABC
= ∑∑
ij = ji
jl lk ki
l k
T T T
= ∑∑
T
il
T
lk
T
kj = ∑∑ CB li kl Ajk
l k
l k
= ∑∑CkiBlk Ajl
= ∑∑
ABC jl lk ki
l k
l k
右 ⇒ C B A ( C ) ( B ) ( A )

ex:trace: trA = ∑ aii
i
ex: tr( ABC) = tr( BCA) = tr( CAB)
⇒ cyclic permutation(輪換)
pf: tr( ABC) ( abc)
ii = ∑∑∑ ABC
i
i l k
= ∑∑∑ B C A = tr( BCA)
ex: T
trA = trA
紋的筆記-應用數學
= ∑ il lk ki
i l k
i l k
lk ki il
= ∑∑∑ C A B = tr( CAB)
ki il lk
pf: trA = ∑ aii
i
T
∑
T
ii ∑ ii
i i
Pauli matrix
⇒ trA = ( A ) = A = trA
⎛0 1⎞
σ1
= ⎜ ⎟
⎝1 0⎠
, σ 2
⎛0−i⎞ = ⎜ ⎟
⎝i0⎠ , σ 3
⎛0 σ1σ2 = ⎜
⎝1 σ σ = iσ
1⎞⎛0 ⎟⎜
0⎠⎝i −i⎞
⎛i ⎟= ⎜
0 ⎠ ⎝0 0 ⎞
⎟=
iσ3
−i⎠
2 3 1
σ σ = iσ
3 1 2
⎛0 −i⎞⎛0 1⎞ ⎛−i 0⎞
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝i 0 ⎠⎝1 0⎠ ⎝ 0 i⎠
⎛1 0 ⎞
= ⎜ ⎟
⎝0 −1⎠
⇒ σ 2σ1 = = =− iσ3
=−σ1σ2
σ iσ j = δij + iεijkσk
⎧1
δij
= ⎨
⎩0
i = j
(kronecker delta)
i ≠ j
2 ⎛0 ( σ1)
= ⎜
⎝1 1⎞⎛0 ⎟⎜
0⎠⎝1 1⎞ ⎛1 ⎟= ⎜
0⎠ ⎝0 0⎞
⎟=
I
1⎠
2 2
( σ ) = ( σ ) = I
εijk
2 3
⎧ 1
⎪
= ⎨−1
⎪
⎩ 0
(ijk)是(123)的cyclic permutation
(ijk)是(123)的置換奇數次
其他
應用數學筆記

紋的筆記-應用數學
⇒
σ σ + σ σ = 2Iδ
i j j i ij
σ = ( σ1, σ2, σ3)
ex: i
e
σ ⋅ϑ , ϑ = ϑ1 ϑ2 ϑ3
∵
⇒
⎧ε112
= ε212
= ...... = 0
⎪
⎨ε123
= ε231 = ε312
= 1
⎪
⎩ε213
= ε132 = ε321
=−1
⇒ ( σ A)( σ B) ( σ A)( σ B )
⋅ ⋅ = ∑ ∑
i i j j
i j
= ∑∑ ABσ i j iσj i j
∑∑
= AB ( Iδ + iε
σ )
i j ij ijk k
i j
∑ i i ∑ εijk i jσk ( ∑ Aiδ ij = Ai
i
i
= I AB + i AB
= ABI ⋅ + iδ( A× B)
AB ⋅ = ∑ AB i i ≡ AB i i
i
( A× B) i = ∑∑εijk
AB j k = εijk
AB j k
j k
( A× B) 1 = A2B3− A3B2 ε AB = AB −AB
∑∑
j k
ijk j k
2 3 3 2
( , , )
x 1 2
e = 1 + x+ x + ......
2!
iσϑ
⋅ 1
2 1
3
e = I + ( iσ ⋅ ϑ) + ( iσ ⋅ ϑ) + ( iσ
⋅ ϑ)
+ ......
2! 3!
2
2
( σ ⋅ ϑ) = ( σ ⋅ϑ)( σ ⋅ ϑ) = Iϑ + iσ(
ϑ× ϑ)
( ϑ× ϑ = 0 )
1 2 1
2
= I + iσ ⋅ϑ − ϑ I − ( iσ ⋅ϑ) ⋅ ϑ I + ...... ( ϑ = ϑ ˆn )
2! 3!
2 4 3 5
ϑ ϑ ϑ ϑ
= I(1 − + − ......) + iσ ⋅nˆ( ϑ−
+ −......)
2! 4! 3! 5!
= Icosϑ + iσ ⋅nˆsinϑ
ex: σ × σ = 2iσ ≠0
( σ × σ) = σ σ − σ σ = 2σ σ = 2iσ
ex:orthogonal matrix
1 2 3 3 2 2 3 1
T T
AA= AA= I
x = ( x , x , x )
1 2 3
2 2 2 2 2
x = x1 + x2 + x3
= ∑ xi
i
T
=
X X
應用數學筆記
)

紋的筆記-應用數學
⎛x⎞ X x
1
⎜ ⎟
=
⎜ 2 ⎟
⎜x⎟ 3
X ′ = AX (A:rotational matrix)
⎝ ⎠
T T T T T
X ′ X ′= X X = ( AX ) ( AX ) = X A A X
⎛x′ ⎞ ⎛ cosϑ sinϑ⎞⎛x⎞
⎜ ⎟= ⎜ ⎟⎜ ⎟
⎝y′ ⎠ ⎝−sinϑ cosϑ⎠⎝y⎠
T
∑ ( a ) ik akj A
= δij
⇒ ∑ aa ki kj = δij
k
k
⎛ cosϑ sinϑ
⎞
ex: A = ⎜ ⎟
⎝−sinϑ cosϑ
⎠
⇒ ○1 i = 1,
j = 2
(cos ϑ,sin ϑ) ⋅− ( sin ϑ,cos ϑ)
= 0……正交
○2 i= j = 1,
2 2
cos ϑ+ ( − sin ϑ)
= 1
ex:Unitary matrix
+
UU
+
= UU= I
+
( UU) ij = δij
⇒
+
( U ) ikUkj= δij
+
( UU) ij δij
Second-order Determinant
∑ ⇒
∗
∑ UU ki kj = δij
k
k
∑ ik kj ij ⇒
∗
∑ UU ki kj = δij
k
k
+
= ⇒ U ( U ) = δ
a a
11 12
D= delA= = a11a22 − a12a21 a21 a22
⎧4x1+
3x2 = 12 ⎛4 ex: ⎨
⇒
⎩2x1+
5x2 =−8
⎜
⎝2 3⎞⎛x1⎞ ⎛12⎞ ⎟⎜ ⎟=
5 x
⎜ ⎟
⎠⎝ 2 ⎠ ⎝−8⎠ 4
D =
2
3
= 14
5
D 12 1 x1
= , D 1 =
D −8
3
= 84
5
D 4 2 x2
= , D 2 =
D 2
12
=−56
−8
Third-order determinant
,∴ x 1 = 6
,∴ x 2 = − 4
= I
……正交條件
應用數學筆記

a a a
D= a a a
a a a
11 12 13
21 22 23
31 32 33
⎧ a11x1+ a12x2+ a13x3= b1
⎪
⎨a21x1+
a22x2+ a23x3= b2
⎪
⎩a31x1+
a32x2+ a33x3= b3
x
D
D
1
2
3
1 = , x2
= , x3
=
N-order determinant
紋的筆記-應用數學
a a a a a a
= a − a + a
D
D
a a ... a
a
D= delA=
a
... a
a a ... a
11 12 1n
21 22 2n
n1 n2 nn
σ
( 1) 1 σ(1) ...... nσ( n)
σ
22 23 12 13 12 13
11
a32 a33 21
a32 a33 31
a22 a23
D
D
D= ∑ − a a
σ :permutation 相鄰交換的次數
a a a
ex: D a a a
a a a
11 12 13
= 21 22 23 = a11a22a33 + a12a23a31+ a13a21a32 −a12a21a33−a11a23a32 − a13a22a31 31 32 33
= j1 j1+ j2 j2 + j3 j3 + ...... + jn jn
j1
jk
j+ k
= ( − 1) jk
minor
D a C a C a C a C
C M
jk
C :cofactor
M :原行列式中除去第 j 個 row,第 k 個 column 所形成的子行列式
1 −1
2 3
2
ex: D= delA= A =
4
2
1
0
−1 2
− 1
1 2 3 0
2 0 2 −1 2 3 −1 2 3 −1
2 3
= 1 −1 −1−2 1 −1 − 1+ 4 2 0 2 − 2 0 2
2 3 0 2 3 0 2 3 0 1 −1
1
= 16 −2⋅ 8 + 4⋅32 − 0 = 128
ex: del( AB) = delA⋅ delB ( del( A + B) ≠ delA + delB )
應用數學筆記

del( AB) = del( ∑ Aik Bkj
)
紋的筆記-應用數學
k
σ
σ
k1 A1 k B
1 k1σ(1) k2 A2 k B
2 k2σ(2) kn
Ank B
n knσ( n)
σ k1 k2 ...
kn
σ
( 1) A1k A
1 2 k ... A
2 nk B
1 (1) 2 (2) ...
n kσ Bk σ Bknσ(
n)
k1 k2 ...
kn
A1k A
1 2 k ... A [
2 nkn σ
σ
( 1) Bk1σ(1) Bk2σ(2) ... Bk
( ) ]
nσ
n
∑ ∑ ∑ ∑
= ( −1)
( )( )......( )
∑∑∑ ∑
= −
∑∑ ∑ ∑
= −
= ( delA)( delB)
Inverse Matrix (行列式不得為零)
1
AB= BA= I ⇒ B A −
=
cofactor of A
−1
ji
( A ) ij =
delA
delA ≠ 0
delA = 0 則稱 A 為singular
pf: C ij = cofactor of ij A
T
[ AC ⋅ ] ij
T
= ( A的第 i個 row) ⋅(
C的第j個column
)
= ( ai1 ai2 ...
⎛Cj1⎞ ⎜ ⎟
C j2
ain)
⎜ ⎟
⎜ ⎟
⎜ ⎟
C ⎟
⎝ jn ⎠
= aC + aC + + aC
i1 j1 i2 j2 ... in jn
⎧delA
i = j
= ⎨
⎩ 0 i ≠ j
1 2 3
2 3 1 3 1 2
ex: 1 2 3 = 1⋅ − 2 + 3 = 0
4 2 −1 2 −1
4
−1
4 2
若 a11c11+ a12c12+ a13c13 = a21c21+ a22c12 + a23c13 則此行列式為零
−1 −1 −1 −1
ex: ( ABC) = C B A
−1 −1 −1 −1
( ABC)( ABC) = ABCC B A = I
( ) ( )
−1 −1 −1 −1
ABC ABC = C B A ABC = I
⎧x1+
3x2 + x3
=−2
⎪
ex: ⎨2x1+
x2 + x3
=−5
⎪
⎩x1+
2x2 + 3x3 = 6
應用數學筆記

⇒
紋的筆記-應用數學
⎛1 3 1⎞⎛x⎞ ⎛−2⎞ 1
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜
2 5 1
⎟⎜
x2
⎟
=
⎜
−5
⎟
⎜1 2 3⎟⎜x⎟ ⎜
3 6 ⎟
⎝
⇒ AX = B
−1
⇒ X = A B
⎠⎝ ⎠ ⎝ ⎠
⇒ 1 1 2 x = 3
Eigenvalue Problem
固有值、本徵值
x = , x 2 = − , 3
AX = λ X
λ :eigenvalue
X :eigenvector
⎛. . . ⎞⎛⎞ . ⎛⎞ .
⎜ ⎟⎜⎟ ⎜⎟
⎜
. . .
⎟⎜⎟
. = λ
⎜⎟
.
⎜. . . ⎟⎜⎟ . ⎜⎟
⎝ ⎠⎝⎠ ⎝⎠ .
⎛−5 2 ⎞
ex: A = ⎜ ⎟
⎝ 2 −2⎠
⇒ AX = λ X
⇒
⇒
⇒
⇒
⎛−2 2 −3⎞
⎜ ⎟
ex: A =
⎜
2 1 −6
⎟
⎜−1 −2
0 ⎟
⎝ ⎠
⎛−5 2 ⎞⎛x⎞
⎛x⎞ = λ
1 1
⎜ ⎟⎜
⎟ ⎜ ⎟
⎝ 2 −2⎠⎝x2⎠
⎝x2⎠ ⎛−5−λ2 ⎞⎛x⎞
= 0
1
⎜ ⎟⎜
⎟
⎝ 2 −2−λ⎠⎝x2⎠ −5−λ2 = 0
2 −2−λ 2
λ + 7λ+ 6= 0 ⇒ λ = −1, − 6
λ =− 1 ⇒
⇒
λ =− 6 ⇒
⇒
⇒
⎛−4 2 ⎞⎛x⎞
= 0
1
⎜ ⎟⎜
⎟
⎝ 2 −1⎠⎝x2⎠
⎧−
4x1+ 2x2 = 0
⎨
⇒ X
⎩ 2x1− x2
= 0
⎛1 2⎞⎛x1⎞
⎜ ⎟⎜
⎟=
0
⎝2 4⎠⎝x2⎠
⎧ x1+ 2x2 = 0
⎨
⇒ X
⎩2x1+
4x2 = 0
−2−λ2 −3
2 1−λ− 6 = 0
−1 −2 0−λ
⎛x⎞ 1 ⎛1⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ 5 ⎝2⎠ (1) 1
= =
x2
⎛x⎞ 1 ⎛ 2 ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ 5 ⎝−1⎠ (2) 1
= =
x2
應用數學筆記

紋的筆記-應用數學
⇒
⇒
3 2
λ + λ −21λ− 45 = 0
2
( λ− 5)( λ+
3) = 0
ex:stretching of an elastic membrane
2 2
circle x1 + x2
= 1
( x1, x2) → ( y1, y2)
⎧ y1 = 5x1+ 3x2
⎛5 3⎞
⎨
⇒ Y = AX , A = ⎜ ⎟
⎩y2
= 3x1+ 5x2
⎝3 5⎠
5−λ3 ⇒ = 0 ⇒
3 5−λ
λ = 8 ⇒
λ = 2 ⇒
X
X
(1)
(2)
Similarity transformation
AX = λ X ⇒
⇒
1 ⎛1⎞ = ⎜ ⎟
2 ⎝1⎠ 1 ⎛ 1 ⎞
= ⎜ ⎟
2 ⎝−1⎠ λ
−1
−1 −1
R ARR X = λ R X
−1 −1
R AX = R X
A′ Y Y
, AX = λ X
2
λ − 10λ+ 16 = 0 ⇒ λ = 8, 2
⇒
⇒ AY ′ = λY
A 和 A′有相同的eigenvalue
Z Z
8 2
2 2
1 + 2
2 = 1 2
−1
′ = , Y = R X
−1
A R AR
⎛5 ex: A = ⎜
⎝1 4⎞
⎟ ⇒
2⎠
5−λ A− λI
=
1
4
2−
λ
2
⇒ λ − 7λ+ 6= 0
⇒ λ = 1, 6
⎛5−1 λ = 1 ⇒ ⎜
⎝ 1
4 ⎞⎛x⎞ (1)
⎟⎜ ⎟=
0 ⇒ x+ y = 0 ⇒ X =
2−1⎠⎝y⎠ 1 ⎛ 1 ⎞
⎜ ⎟
2 ⎝−1⎠ ⎛−1 λ = 6 ⇒ ⎜
⎝ 1
4 ⎞⎛x⎞ (1)
⎟⎜ ⎟=
0 ⇒ x− 4y = 0 ⇒ X =
−4⎠⎝y⎠
1 ⎛4⎞ ⎜ ⎟
17 ⎝1⎠ R =
1 ⎛ 1
⎜
34 ⎝−1 4⎞
1
⎟ ⇒ R
1⎠
34 1
5 1
4
1
−
=
⎛
⎜
⎝
− ⎞
⎟
⎠
−1
1⎛1 A′ = R AR=
⎜
5⎝1 −4⎞⎛5
⎟⎜
1 ⎠⎝1 4⎞⎛ 1
⎟⎜
2⎠⎝−1 4⎞ 1⎛5
⎟= ⎜
1⎠ 5⎝0
0 ⎞ ⎛1 ⎟= ⎜
30⎠ ⎝0 0⎞
⎟
6⎠
⎛5 ex: A = ⎜
⎝1 4⎞
⎟,
2⎠
50
A = ?
⇒ f ( A) = A
− −
= R( R AR)( R AR)......( R AR) R
50 1 1 −1 −1
A′ A′ A′
R( A) R −
=
′
50 1
應用數學筆記

⎛1 A′ = ⎜
⎝0 0⎞
⎟,
R =
6⎠
1 ⎛ 1
⎜
34 ⎝−1 4⎞
⎟
1⎠
50 1 ⎛ 1
A = ⎜
5 ⎝−1 4⎞⎛1 ⎟⎜
1⎠⎝0 0 ⎞⎛1 50 ⎟⎜
6 ⎠⎝1 −4⎞
⎟
1 ⎠
1 ⎛ 1
= ⎜
5 ⎝−1 4⎞⎛ 1
⎟⎜ 50
1⎠⎝6 −4⎞
50 ⎟
6 ⎠
50
1 ⎛1+ 4× 6
= ⎜ 50
5 ⎝ − 1+ 6
50
− 4+ 4× 6 ⎞
50 ⎟
4+ 6 ⎠
紋的筆記-應用數學
, 1 34 1 4
R −
−
⎛ ⎞
= ⎜ ⎟
5 ⎝1 1 ⎠
Invariant(不變量)
⇒ TrA , delA
−1 −1
TrA′ = TrR AR = TrARR = TrA
Tr( abc) = Tr( bca) = Tr( cab)
TrA = eigenvalue的總和
′ = = =
−1 −1
delA del( R AR) ( delR )( delA)( delR) delA
del R R del I delR delR
−1 −1
⇒ delR = ( delR)
−1 −1
( ) = ( ) = 1 = ( )( )
⎛ 2
⎜
ex: H =
⎜
−1
⎜
⎝−3 −1 1
2
−3⎞
⎟
2
⎟
,(a) ∑ λ i = ? (b)
3 ⎟ i
⎠
⇒ (a) λ = 2 + 1+ 3 = 6
(b)
∑
i
i
∑
i
λ = ?
⎛ 2
2 ⎜
H =
⎜
−1 ⎜
⎝−3 −1 1
2
−3⎞⎛ 2
⎟⎜
2
⎟⎜
− 1
3 ⎟⎜
⎠⎝− 3
−1 1
2
− 3⎞ ⎛4+ 1+ 9
⎟ ⎜
2
⎟
=
⎜
3 ⎟ ⎜
⎠ ⎝
1+ 1+ 4
⎞
⎟
⎟
9+ 4+ 9⎟
⎠
2
λ = (4 + 1+ 9) + (1+ 1+ 4) + (9+ 4+ 9) = 42
∑
i
i
2
A
⎛λ1 ⎜
⎜
λ2 ⎞⎛λ1 ⎟⎜
⎟⎜
λ2 ⎞
⎟
⎟
2
λ1
2
λ2
λ3 λ3 2
λ3
⎛ ⎞
⎜ ⎟
= = ⎜ ⎟
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
⎜ ⎟
⎝ ⎠
ex:eigenvalues(有重根)(degenerate)
degenerate :一個 λ ↔ 多個eigenvector
nondegenerate :一個 λ ↔ 一個eigenvector
2
i
應用數學筆記

Conic Section
⎛0 0 0 1⎞
⎜ ⎟
0 0 1 0
A = ⎜ ⎟
⎜0 1 0 0⎟
⎜ ⎟
1 0 0 0⎟
⎝ ⎠
Quadratic form(二次項)
T
Q= X AX
紋的筆記-應用數學
3 3
⇒
= ∑∑ a X X
Transformation to principal axes
T
Q= X AX
1
( ) T −
−λ
0 0 1
A− λI
=
0
0
−λ
1
1
−λ
0
0
1 0 0 −λ
−λ
1 0 0 0 1
=−λ 1 −λ 0 −−λ
1 0
0 0 −λ 1 −λ
0
2 2
⇒ ( λ − 1) = 0 ⇒ λ = 1,1, −1, − 1
⇒ λ = 1,
(1)
X =
⎛1⎞ ⎜ ⎟
1 ⎜
0
⎟,
2 ⎜0⎟ ⎜ ⎟
1⎟
⎝ ⎠
(2)
X =
⎛0⎞ ⎜ ⎟
1 ⎜
1
⎟
2 ⎜1⎟ ⎜ ⎟
0⎟
⎝ ⎠
(2)
X 的決定是要與 (1)
(1) T (2)
X 正交比較好 ⇒ X ⋅ X =
⇒ λ =− 1,
(3)
X =
⎛ 0 ⎞
⎜ ⎟
1 ⎜
1
⎟,
2 ⎜−1⎟ ⎜ ⎟
0 ⎟
⎝ ⎠
(4)
X =
⎛ 1 ⎞
⎜ ⎟
1 ⎜
0
⎟
2 ⎜ 0 ⎟
⎜ ⎟
−1⎟
⎝ ⎠
jk j k
j= 1 k=
1
2
a11X1 2
a22 X2 2
a33X3 2a12X1X22a23X2X3 2a31X3X1
= + + + + +
=
⎛λ0⎞⎛ y ⎞
T −1 −1
T
1 1
X RR ARR
X = Y AY ′ = ( y1y2) ⎜ ⎟⎜ ⎟
T
Y A′
Y
0 λ2
y2
R = R ⇒
−1
Y R X
= ⇒
T
X X = 不變量
− 1 T
R = R
T T −1
T T
= ( ) =
Y X R X R
X → RX = X ′ X → X R = X′
T T
⇒ X ′ X′= X
T
R R X
T
= X X
− 1 T
R = R (symmetric)
I
右乘 R
⇒
T T T T
−1
T
R R R R
2 2
ex: Q= 17X − 30X X + 17X =
128
1 1 2 2
= ⇒
⎝ ⎠⎝ ⎠
T
I = RR
應用數學筆記
0

T
= X AX
= ( x1 ⎛ 17
x2)
⎜
⎝−15 −15⎞⎛x1⎞ ⎟⎜
⎟ = ( x1 17 ⎠⎝x2⎠ ⎛ 17x1−15x2 ⎞
x2)
⎜ ⎟
⎝− 15x1+ 17x2⎠
⎛ 17
A = ⎜
⎝−15 −15⎞
⎟ ⇒
17 ⎠
17
−15
−15
= 0 ⇒ λ = 2,32
17
(1)
λ = 2 ⇒ X =
1 ⎛1⎞ ⎜ ⎟
2 ⎝1⎠ (2)
λ = 32 ⇒ X =
1 ⎛−1⎞ ⎜ ⎟
2 ⎝ 1 ⎠
2 2 2 2
∵ Q= λ1Y1+ λ2Y2=
2Y1 + 32Y2 = 128 ……為一橢圓
⎛
⎜
−1
Y = R X ⇒ X = RY = ⎜
⎜
⎜
⎝
1
2
1
2
1 ⎞
−
2
⎟
⎛ y1⎞ ⎛x1⎞ ⎟⎜
⎟= ⎜ ⎟
1 ⎟⎝y2⎠ ⎝x2⎠ ⎟
2 ⎠
∴ X 1 =
Y1 −
2
Y2
, X 2 =
2
Y1 Y2
+
2 2
紋的筆記-應用數學
應用數學筆記
∞ ∞ 2 2
−( x − xy+ y )
2 2
ex: ∫ e dxdy
−∞∫ ⇒ Q= x − xy+ y = ( x
−∞
⎛
⎜ 1
y) ⎜
⎜ 1
⎜− ⎝ 2
1 ⎞
−
2
⎟⎛x⎞ T
⎟⎜ ⎟=
X AX
y
1
⎟⎝ ⎠
⎟
⎠
⎛
⎜ 1
A = ⎜
⎜ 1
⎜− ⎝ 2
1 ⎞
−
2
⎟
⎟
1 ⎟
⎠
A− λI
1−λ
=
1
−
2
1
−
2
2 1
3 1
= 0 ⇒ ( λ −1) − = 0 ⇒ λ = ,
4
2 2
1−λ
3 2 1 2
Q= S1 + S2
2 2
3 2 1 2
∞ ∞ − ( S1 + S2
)
2 2
⇒ ∫ e JdS1dS −∞∫ −∞
2 =
2π ⋅
3
2π
2π
=
3
3
(1)
λ = ⇒ X =
2
1 ⎛−1⎞ ⎜ ⎟
2 ⎝ 1 ⎠
1
(2)
λ = ⇒ X =
2
1 ⎛1⎞ ⎜ ⎟
2 ⎝1⎠

−1
R X Y
J
= ⇒
∂X ∂X
∂S ∂S
紋的筆記-應用數學
X RY
⎛ 1 1 ⎞
⎜− 2 2
⎟
⎛S⎞ ⎜ ⎟
⎜ ⎟⎝ ⎠
⎜ ⎟
⎝ 2 2 ⎠
1
= = ⎜ ⎟
1 1 S2
1 2
= = =
∂Y ∂Y
∂S ∂S
1 2
⎧x
=− 4x
+ x + x
⎪
ex: ⎨x
2 = x1+ 5x2
−x3
⎪
⎩x3
= x2 −3x3
1 1 2 3
−
1
2
1
2
1 1
2 2
⇒ X = AX
⎛−4 ⎜
∴ A =
⎜
1
⎜
⎝ 0
1
5
1
1 ⎞ ⎛x1⎞ ⎟ ⎜ ⎟
−1
⎟
, X =
⎜
x2
⎟
3 ⎟ ⎜
⎠ x ⎟
⎝ 3 ⎠
⎛ y1⎞ ⎛λ1 −1 −1
−1
⇒ R X ⎜ ⎟ ⎜
= R AR R X ⇒ 2 0
YA′ ⎜
y ⎟
=
⎜ Y ⎜ y ⎟ ⎜
⎝ 3⎠
⎝ 0
0
λ2
0
0 ⎞⎛ y1⎞
⎟⎜ ⎟
0
⎟⎜
y2⎟
λ ⎟⎜
3 y ⎟
⎠⎝ 3⎠
−4−λ1 1
⇒ A− λI = 1 5−λ − 1 = 0 ⇒ λ = −3, − 4,5
0 1 3−λ
(1)
λ =− 3 ⇒ X =
⎛1⎞ 1 ⎜ ⎟
0
2
⎜ ⎟
⎜1⎟ ⎝ ⎠
(2)
λ =− 4 ⇒ X =
⎛10 ⎞
1 ⎜ ⎟
1
102
⎜
−
⎟
⎜ 1 ⎟
⎝ ⎠
(3)
λ = 5 ⇒ X =
⎛1⎞ 1 ⎜ ⎟
8
66
⎜ ⎟
⎜1⎟ ⎝ ⎠
R =
⎛1 1 ⎜
0
2× 102× 66
⎜
⎝1 10
−1
1
1⎞
⎟
8
⎟
1⎟
⎠
−1
Y = R X ⇒ X =
RY
1
⇒
R −
1
⎛−9 −9
81⎞
1 ⎜ ⎟
= 8 0 8
72 ⎜
−
⎟
⎜ 1 9 −1⎟
⎝ ⎠
應用數學筆記

⎛ y⎞ ⎛−3 0 0⎞⎛
y ⎞
1 1
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜
y2⎟ =
⎜
0 −4
0
⎟⎜
y2⎟
⎜ y⎟ ⎜
3 0 0 5⎟⎜
y ⎟
3
⎝ ⎠ ⎝ ⎠⎝ ⎠
∴
紋的筆記-應用數學
⎛x⎞ ⎛1 10 1⎞⎛
y ⎞
1 1
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜
x2⎟ =
⎜
0 −1
8
⎟⎜
y2⎟
⎜x⎟ ⎜
3 1 1 1⎟⎜
y ⎟
3
⎝ ⎠ ⎝ ⎠⎝ ⎠
⎧ x = ce + 10ce
+ ce
⎪
⎨x
c e c e
⎪
⎩x
= ce + c e + c e
−3t −4t
5t
1 1 2 3
2 =− 2
−4t
+ 8 3
5t
−3t −4t
5t
3 1 2 3
pf: 1
2
∫ [ P ( )]
1
n x dx
−
⇒
⎧y1
=−3y1
⎪
⎨y=−4y
⎪
⎩y3
= 5y3
2 2
2
=
2n+ 1
1
2 2
⇒ 由生成函數 (1 2 ) ( ) n
−
∞
− xu+ u =∑ Pnx u
⇒
n=
0
⇒
1
(1
−1 2 xu
2 − 1
u ) dx
m n
1
m+ n
( P ( ) ( ))
1
m x Pn x u
−
n
1
2 2n
[ P ( )]
1
n x u
−
∫ ∫
− + =∑∑
⎧ y = ce
⎪
⎨y
ce
⎪
⎩y
= ce
−3t
1 1
2 = 2
−4t
5t
3 3
應用數學筆記
2
= ∑∫ ( y = 1− 2xu
+ u )
2 2
(1 − u) 2
1 dy 1 (1 + u)
1 (1 + u)
1 1+
u
⇒ ( ) ln y
2
2
∫
=
= ln = ln
(1 + u)
(1 −u)
2
−2u
y 2u
2 u (1 − u)
u 1−
u
2 3
u u
ln(1 + u) = u−
+ − ......
2 3
2 3
u u
ln(1 − u) =−u− − − ......
2 3
1 1+ u 1
⇒ ln = [ln(1 + u) −ln(1 −u)]
u 1−
u u
1 2 3 2 5
= [2 u+ u + u + ......]
u 3 5
1 2 1 4
= 2[1 + u + u + ......]
3 5
∞ 2n
u
= 2∑
n=
0 2n+ 1
∞ 2n
∞ u
1
2 2n
⇒ 2 ∑ = ∑∫ [ P ( )]
1
n x dx u
n 0 2n1 −
= + n=
0
1
2 2
⇒ ∫ [ P ( )]
1
n x dx=
……得證
−
2n+ 1

P ( x) P ( x) dx
2
2
m! n! d
[
dx
( x
d
1) ][
dx
( x 1) ] dx
m> n
m+ n
1 1
m 2 m d 2 n
= ( −1) ( x 1) ( x 1) dx
m+ n
2 ! ! 1
m+ n
mn∫ − −
− dx
m n
1 1
2 m 2 n
1
m n = − −
− m+ n −1
m n
∫ ∫ ……利用部分積分
紋的筆記-應用數學
應用數學筆記