ページ1:

CH8 Matrix
PAI
1. Basic Concepts:
1) Matrix = [aij ]mxn =
…
5075
A21 azzazz
Ami amz
·Amn⋅
all a 12 ain
row matrix
一有方向性的向量不一樣意思,
(ii) A man matrix can be viewed as m Ixn vectors or n mal vectors
Column matrix
Example
* min 14-1/31,
(就畫一槓!
01=03563
92 = [145]
(i)転置
93=[213]
94-18923
14x3
Mikiki
→ ýjun§ ØX 40 1×3 㣆‡ or 33 4×1 Ref
=
||
a₁
02
93
A-Lajmin Example
=
A = [αij Jnxm
2. Matrix mulitplication
A
mxn =
07
Az
am
'
ay
↑
Yow matrix set
1 Bmxk = [ b₁ bz ... bk ]
Çmxk = [Cij ] = [ai⋅bj]
3. Special types of matrix:
Diagonal matrix (AME)
[aij] man is diagonal if αij =0, V₁ #j
Example
ummmm
:000
0200
0030
2000.
0400
L0001-4x4 20050-3x4
123
456 2x3
[b, bz b3]
↑
column matrix
T
A mxn x Buck = Cmsk
H...
A column & Brow #k
Example = []
(i) Unit matrix (Identity matrix)
AxB=
B = [
D
47-3×2
24 20
A*D=C = [141 61 ] 2×2
1,i=1
[az] ] nxm is a unit matrix it aus = {t, it
arg=
Execele I=
0 0
000hxn
0,
(iii) Aman
6368
13x2
[312]
[]
-15+1+6=24
[312]
=6+0+14=20
方陣
is called a "square matrix" if m=n
(iv) A matrix is called a "column matrix (vector)"
if there is only one column in the matrix.
Example A=
A = [ } ]
(V) A matrix is called a "row matrix (vector)"
if there is only one row in the matrix.
Example A-[123]
→上半部不全。
(V) Upper triangular matrix (±Âži)
[ij] such that Vaij=0 it i > j
12345
02678
00542 23x5
Example
Dicj
①对綿查下来(丁))
000
14x4
Fi>jibo
全為oj
i=j

ページ2:

(vii) Lower triangular matrix
[aij] such that Vaij=0 if i < j
Example
-400
200→j,上半部都是。
i
4010
·6327-4x4
V.
4. Linear Algebraic System 性状系統(联立方程式)
Given that X+22+6x3 = 11
3x1+4X2+x3=12←←
42x1+3x2-x3=7
Linear Algebraic System
126
BALALA
2
41
3 13x3
3x1
⇒ A⋅ x = B, Find x = ¦
-
·linear equation
- linear equation Find x1, x2, x3 = 1
-linear equation
3x1
(i) Linear Algebraic System To find Xnx! (ii) Elementary row operations
An Xxl = Bmx1
x
equation number
I where Aman is called coefficient matrix
Xnxl is called variable matrix
Bmxl is called constant matrix
I row interchange
Scalar multiplication
rows addiction
(i) Gaussian elimination. → to solve X
To apply elementary row operation
to an augmented coefficient matrix
such that the matrix can be converted
into a upper friangular matrix.
增廣矩陣
-x1+X2+2X3=2
Example 34-12+x3=6
-x1+3×2+4x3=4
-1122
112
3
x x2
=
- 34.
3x3
X3
3×1
3-16 Augmented coefficient matrix
Step1. definition:
-1344
coefficient↓
constant
增廣矩陣
3x1
Step 2. Gaussian elimination (to convert to an upper triangular matrix)
x3
3
122 X1-1) -122.
L-1 3 4.
=
122
7122
027 12 =
0 7123
44.
02 22
1-101
Coefficient constant
|-
1-x1+x2+2x3=2
282+783=12
55733-10
Solve X3 X3=2
-x1+(-1)+4=-2
272+14=12
X2=-1
x=1

ページ3:

Summary: To solve for the linear algebaric equation:
(1) elementary row operation ⇒ upper triangular matrix
(2) back-substitution
5. Rank of a matrix 12,
linear dependence of matrix
Definition: Let Vi, Vz ... Vk be k nxl matrices
α1, az
•ak bek constants, then
線性組合
V = a₁₁ + a2V2 + ... + Akvk is called a linear combination
of V, V2.... VK
If V=0, Va, az.... ak 710, then V₁, V₂,
Cons
Vk are "linear dependent"
Example V₁=['}] V2= [?] a₁₁ + A₂V₂ = 0
⇒ a₁ = -2α2, 11 V1, Vz are linearly dep.
© It V = 0, Va₁, a 2 ... ako, then V₁, Vz,... Vk are "linear independent"
Example v₁= [ 1 ] V₂ = [ 1 ] a₁√ ₁ + A ₂ V ₂ = 0
(i) Rank of a matrix
⇒ α₁ = a₂ = 0, ₁₁ V₁, Vz are linearly independent
(Vector)
(vector)
A mxn matrix can be viewed as m lxn row matrices and n mx1 column matrices
Example A
=
35 6
5
=
213
-892-4x3
4 row matrix: [356], [145], [213],[892]
-0.0.0
3 column matrix:
Definition: The maximum number of linearly independent row matrices is
called the "rank" of the matrix
Example A
517
:
413x3
-324√3×3
(column)
3 row matrix: [312], [517], [634]
rank(A)=3
3 row matrix: [111], [213], [324]
rank (B) = 2

ページ4:

Theory: The number of linearly indepent row matrix is equivalent to the
number of linearly independent column matrix.
In other words, A and AT have the same rank!
Example
10066].
3299
3x4
In terms of row matrix, rank = 2
AQQA
獨立
無關係
=
4 column matrix
4=-
42=6
[j] = «[}] +
+9199
C3=-5
44=12
In terms of column matrix, rank = 2
6. Linear Systems:
One has m equations, unknowns, then the general expression
of the corresponding linear system is given by :
Amen
Xnxi
=
Bmx1
equation o↑
{3x+5y + bz = 2
Example
-7x-3y-2=10
356
增廣矩陣
and A= [AB] is called augmented matrix
+4;9[}}]}[M]
562
(1) A linear system has solutions if only if
|rank (A) = rank (Ã)
rank (A) + rank (Ã)
{x+y=1
1x+y=2
⇒無解
(ii) If rank (A)-rank(A)=n, then
the linear system has precisely
one solution
By using Gaussian elimination, we can
obtain the solution.

ページ5:

8. Inverse of a matrix k鿽½18`~
(i) definition:
Let A be nxn matrix. If there exists an nxn
matrix B such that AxB=Bx
可有倒拔
=
= =
where I is the nxn identity, then A is
said to be invertible, and B is said to be the
"inverse" of A, denoted as A
(ii) Properties:
nonsingular essence #
(#)
+4213x
500
unit
51
© (A")" = A
rank (A) = n
A nxn is invertible if and only it A has full rank (nunsingular) (A, B) = A
*A' Let Anxn Xnx1 = Bnx1
Xnxi is unique if t rank (A)= n
A'. A · X = A¹· B ⇒ X = A + B
=
A exists since x exists
(iii) Finding the inverse
© Gaussian elimination
[AI] elementary
Yow
operations
transforming A into I
Simultaneously transforming I into A'
for real number:
[51]→[17
Example Find the inverse of A= 2
wwwm
-30
YA is nonsingular, ii A " exists
+5
Applying the Gauss elimination of the augmented [A11]:
其它变成口
13|100
[A11]=
2
2 010
X(-3)
*留下对角線,
0 001
|
131100
=
·B
X(-1)
0142
0-3-81-30
1×(3) =
113|100
0-1-4 -210x-1
0 -3 -81-30 1
10-1-110
2-10
0
4
X(-1)
00
--3 P.

ページ6:

有解
(iii) It rank (A) = Yank (Ã) = r <
then one has infinately many solutions
and one can arbitarily assign values to
n-r unknowns to obtain the values of other
r variables.
Example {x+y+2=3
(奇異(發散)
(iv) singular/nonsingular
•Amxn is called singular it rank (A)=r<n
It rank (A)=n, then we called the matrix
has full rank (non-singular)
42x+y+2=5
!= [[ } }]
rn
rank (A) = rank (A) = 2 <3 ct *x+x+x / Let z=1 { x+y=z
x=2
n-r=1
C2x+y=4
y=0
→[]=[]
1 independent variable (2)
Z
2 dependent variable.
→X, Y Let z=2
[x+4=1
x=2
42x+y=3
y=-1
[44]
7. Determinants of matrix =
(i) 2x2 matrix
The determinant of Auz
= Laz azz↓ 1
denoted as det (A) all α12 ||
= |
= 011922-012021
Az azz
(ii) 3x3 matrix
all 12 13
+
t
all 12 13
The determinant of x= au azz az is det (A) = | A21 Azz Azz
-A31 A32 A33.
|a31 32 33
= all
922 923
✗
A21 A23
Az
a12
+ α13
1932 0331
2012(降階过程)
= all · C11 + α12 · C12 +α13. C 13°
where C11, C12, C13 are called the "cofactors"
corresponded to a11, a12, a13 respectively

ページ7:

9. Cramer's Rule (8.1)
10. Homogenous Linear Equations
A system of linear equations with n
unknowns of X1, X2
TATE
Xn is given by:
all a12 ain
A nxn Xnx1 = Bnx1, where A = az azzazn
=
1✗=
ani anz
...ann.
Xn
we know that Xnxi is unique ift A is nonsingular A' exists!
⇒ A+ A•X = A+B ⇒ x= A₁₁ B
I
We could use the inverse to solve the system.
A homogeneous linear system is such that, B-0,
and be expressed by: Anxn Xnx₁ = 0
(i) it § is nonsingular (rank (1) r=n)
"
⇒ exists! = 10=0← trivial solution
(ii) if A is singular (rank (1) ren)
There must exist an infinite number of solutions, which are
nontrivial solution
非零解
In
• arbitarily assign n-r unknowns
•obtaining other variables
conclusion, for a homogeneous system of linear equations
with n unknowns ( Anxn · Xnx₁ = 0): it and only it
only the trivial solution exists A is nonsingular (det (9)70, rank (^)=n)
a nontrivial solution may exists ift A is singular. (det (A)=0, rank (^)=r<n)

ページ8:

1 0 -1 | -1
010-12-1
00413-31
0
010
001
2
1
=
|
0
I 0-12-1
4 00
= [1 | A']
Using the determinant & cofactors
3,31
q
4
伴隨矩陣
definition: Let A be an nxn matrix. The adjont matrix
is the transpose of the matrix of cofactors
Corresponding to the elements of A
Cin T
C11 C12
C11 C 21
Chi
adj (A) = Cu Czz
Czn
C12 C22
Chz
LCn₁ Ch₂-
Cin Can
Chn-
If A is invertible, then A==
I. adj (1)
(i) 2x2 matrix:
ail 912
Given that A=
A21 Azz.
t
det (4)
⇒ C₁ = a22, C12=- A21, C21 = -α12, C22=011
then adj (9) = [Cu Cu] = [al - 24 ] = [au Au]
(12]
-C21 C22
=
a22-12
detla) L-az all-
+
-
all 12 13
(iii) 3x3 matrix: Given that A=aa
all
-921 all
azz azz
a2z a2z3
021 023
C₁₁ =
,
1 C ₁₂ = -|
C 13 =
032 033
031 033
/
921 azz
931 932
012 013
all 013
all 012
C21=-
Cz₂ =
1
023=-
032 033
931 933
| 031 032 |
Pazz
C31=
012 013
det (A)
0231
·[
(32=
C12 C22 C32
913023033
all as
A21 A231
all 012
1033=
azı azz

ページ9:

(iii) Cofactors and mirror Determinants (15-2££}$)
In general, the cofactors of Aij is given by: A 3x3 matrix has 9 cofactors:
Cij = (-1). Mij, where Mij is the determinant
of the submatrix obtained by deleting i th
row and jth column of A. Mij is thus called
"minor determinant"
Cl = MC12=-M12, C13-M13
C21=-M21, C22-M22, C23=-M23
C31 = M31 C32=-M32 C33= M33
The sign factor can be obtained by this pattern:
t
t
(#)
t
3x3 matrix
)
In conclusion, det (A) can be obtained by expanding the cofactors
along one of the rows (columns). The same approach can be
utilized for non matrix.
(iv) Properties (Chap. 8.5)
det (Amxn) = det(Amxn)
√ If any two rows (columns) of Anxn are the same, then
det (A) = 0
If all the entries in a row (column) of Anxn are zero, then
det (A)=0
√ Proportionals of a row (column):
Scalar product to a row (column):
Example det
312
613
= 0
456
189
x2
624
⇒ It A is singular, then det (A)=0
23
456
89
=
then det (b) =②x det (!)
Note that det (kA) = "det (A)
⑥det (AxB) = det(A)x det (b)

ページ10:

0λ=-1
(A - λ) k=
-4
=0
by Gauss elimination:
20
x
-420
[^\² 0 ] 2×) → [* ² 0 ] ⇒ - 4 k tzk₂ = 0
2
000
(iii) Properties
If Ann has distinct ev, then their
Ⓡ It
corresponding nevec. are linearly independent
may
nik = [ ]
not be possible to find n linearly indep
evec for an Anxn when some of the eiv, are
repeated (tk)
-22-3
Example Find the ev. and the corresponding eve of A= 21-6
-2-7
⇒ det (A-1) =
Z
2-3
+λ-6=0
-1-2-λ
-1-20
(2-2) (+2) (-2)+12+12-3(+2)+42-12 (-2-1)=0
(-2-1)(12)(-2)+24+92-3+122+24=0
(-2-a)(1a)(-a)†19a+45=0
+2+9
3
27-2x+x-x²+192+45=0 1-5 | X³3+27=-217-45
-2-77212+45=0
x³-3x²
x²+1-212-45=0
67-212
673307
(1-5)(x²+67+9)=0
92-45
(3-5)(2+3)=0
01=5:
k=
(A-AE)K = 2-4 +
-72-3
(ⅱ入=5,-3,-3(重根)
by Gauss elimination
-72-30-
k2 =0
2-4 -60
-1-2-5
k3.
1-2-50-
1-2-50
2-4-6 0
1-2-50
-> 0-8-160
--72 (x) 2016 0
9-2-50
0 -160
00
0
0
·-1-2-5.
20-816
(BOQ)PN) -
assign l-indep variable⇒k² = 1
x²)
5+1-242-543=0
-8k2-16k3=0
get other 2-dep variables → k2=21+1=-1
PH

ページ11:

1. Eigen Value Problem $1
(i) definition:
入
eigen vector
Let Anan, a scalars, knxi (column/matrix)
If Athen λ is an ev of A
I is an e.vec. of A corresponding
(ii) solution method: (to find λ and k)
"AK = 2K
純量,要完成向量,有辦法相加
to λ
⇒ AB -λE =0 (A-λ1) = 0
homogeneous
保抆矩陣
nxn
nxI
→持微方程式
To seek for a non-trivial solution for £, we must 7
we
have det (A-1)=0, which is a characteristic equation of
A
(the poly hormial equation of 1")
(11-21) is singular)
⇒ The roots of the characteristic epnation of A are the eigen value of A
And, by solving the system of equation (A-1) = 0, we can find
the solution of k, which is the eigen factor of A corresponding to λ.
Example, Find the e.v. and the corresponding even of A = [1 2 3]
To find ev. of, we have det (A-1)=0
(A-λ)
-5-22-0
det (^-^=) = | 3-1
[^{}]
2-0-2-2
= (-5-2) (2-2)-4=0
FORTR
ein of Ath
= (a+b)(2+1)=0, λ=-1or-6
To find the corresponding evec,, we solve for
|(A-1)=0
07=-6
•入)=1
40
(₤1+2 /2=0)
2/21+4/2 = 0 )
by Gauss elimination
20
2X2
2
= 0
NEXGCH)
000.
⇒ ki+zk2=0 ik [1] [2] []
or

ページ12:

• (lower triangular)
Example 4 Find the Giv, of A = ['
+4-[]
Find the characteristic equation
From the characteristic equation
det (A-1) =
3-70
02-20=0
01-2
D
2
-40
(3-2)(2-A)(1-2) = 0
入=31211
in Triangular matrix (or Diagonal matrix) by eigen
恰等於其對角線上的元素
12. Diagonalization 2-1 (eivi application 1)
(i) definition: An is called "diagonalizable" if it exists a
=
·A·P = D, where
ip nonsingular nxn such that PA. P = D
needs to
(ii) to find P:
Dis a diagonal matrix.
=
線性獨立
If Anxn has linearly indep. evec (K1, K2 1111 kn)
Let nxn [k|kz|| kn], then ·A· = nxn
where
is the ev. corresponding to ki
=
D
Az is "I
"
"12
Lain is
"
"
"kn
Enxn can diagonalize Anxn
1 proof: nxn = [kilk21 ··· | kn] nxn
=
⇒ A.kA. Ckilkz|| kn] = [Ak₁ | Ak₂ | Akn]
[1|zk|ankn]
= [ki|kz|
kn].20
0 A2
00 ....
An
]
111
multiplying on the above equation:
F
k1. K⋅ D =
= =

ページ13:

-
© λ=-3 (§*k)→ try to find two linearly indep e,vec,
[
12-3
(A-AI)K=
(A-λ1) K = 24-6
1-23
{
k2=0k1+2k2-3k3= 0 {assign 2-indep variables
get 1-dep variable
⇒(1) let k²-1, k3=1 ⇒ k₁ = 1 # k= []
(ii) let k2=1,k3-2 ⇒ k₁ = 47 kε [j]
[ ]*[}]
112=-3 (24k), two e,vec
Example 3 Find the eiv. and evec of A = [ } } ]
From the characteristic equation
1^1^1^1 | = (1-5)²= 0 24+ 10 +2 + 4 = 0
-172
det (A-1)=
3-74
入=5,5(重根)
24-10入++4=0
23-102+25=0
|λ=53 try to find two linearly indep e,vec
(9-2)-[42] []=0
=
⇒ -11 +22=0 719=2k2
Let k₂ = 1, k₁ = 2 ⇒ k = [ 1
].
not linearly
Let k₂ = 2, k₁ = 4 7 k = [ 1 ] ← indep (the same)
Let k₂
K=0
We only have 1. e,vec corresponding
to入=5(重根)

ページ14:

Example Diagonalize A = [11]
(17-05-15 = 0
入=34入+289-225=0
0=3入+64=6
⇒ det (A-1)= | 1713 15|=0 x=-342+64=0
(A-1)=
-15 51=
-15-15
A = 32 = (^-^3) E = [ 15 - 15 ] [ " ]] = 0
--15-15.
=0
k1+k2=0,k1=-12
let k₁ =1, k2=-1
15+5
λ=2 = (A-^=) · K = [115 15.
(iii) Property:
· [12] = [4]
=0
11-12-0741-2
let k₁=1,k2=1
[[][]
Anxn is diagonalizable if and only if Anxn
has n linearly indep evec, even though
some repeated e.v. exist.
(2-2313-321=0
{^-32 = k₁ = [√]
(x=2 =⇒ k²= [1]
=
131
Orthogonal
Matrix 正交矩陣
垂直
(i) definition:
The inner product of two vector x = (x1, x2,... Xn)
and y = (y," yn ) is given by : 7. j = X₁₁ +x y z + X nyn
Using the same approach, we define that the inner product
for two column matrix: Xnx=
following way : x⋅y = x². y
#x.yt
= 3nxn
=[xxxzxn]*
nx=2, in the
Lynnx1
= [141+2+x] 1x
We say is orthogonal to y, if x.y = [x₁y, +xzy₂+m+Xnyr] | x₁ =0
* The length (norm) of a column matrix is given
←