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# 大學線性代數初步

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. 學 大 大 ( ) 線性

, 學 , 線性代數

. , 線性代數

, .

, . 學

. , (Question).

, 大

. , 線性代數 . ,

, ,

, . , 性

, . , .

v

(3)

(4)

(5)

## Vectors in R n

1.1.

. . 學,

.

, (a, b) , a, b∈ R ( R

P, P a (a > 0 ; a < 0

), b (b > 0 ; b < 0 ), Q.

P Q (a, b) , −→

PQ = (a, b).

( (a, b) ) , 大 :

(a, b) (c, d) ( (a, b) = (c, d)), a = c b = d; ,

, P, P, Q, Q −→

PQ =−−→

PQP P

Q Q . P = P , Q = Q. Q = Q

P = P. ( : P, P “ ” “數”, P = P

“ ” “數”). −→

PQ ,

. 性, ,

, u, v . R2

, v∈ R2, v

, a, b∈ R v = (a, b).

(scalar multiplication).

3

(6)

Deﬁnition 1.1.1. u = (a1, a2), v = (b1, b2)∈ R2 r∈ R.

u + v = (a1+ b1, a2+ b2) and ru = (ra1, ra2).

( deﬁnition) ,

P, Q, R , −→

PQ +−→

QR =−→

PR . : P, Q, R

P(x1, y1), Q(x2, y2), R(x3, y3) ( : , 代 ),

−→PQ = (x2− x1, y2− y1),−→

QR = (x3− x2, y3− y2),−→

PR = (x3− x1, y3− y1).

−→PQ +−→

QR = ((x2− x1) + (x3− x2), (y2− y1) + (y3− y2)) = (x3− x1, y3− y1) .

−→PQ +−→

QR =−→

PR. (1.1)

, u = (1, 2), v = (3, 4) r = 5, Deﬁnition 1.1.1 u + v = (1 + 3, 2 + 4) = (4, 6) ru = (5× 1,5 × 2) = (5,10). : u = (1, 2), v = (3, 4) r = 5, u + v = (1 + 3, 2 + 4) = (4, 6) ru = (5×1,5×2) = (5,10)

a1, a2, b1, b2, r 代 數 . Deﬁnition 1.1.1

rv , 初 u R2 , ru ,

rv. : , 數 ,

, , ,

, ;

(Proposition Theorem) 性 ,

Proposition 1.1.2. R2 ,:

(1) u, v∈ R2, u + v = v + u.

(2) u, v, w∈ R2, (u + v) + w = u + (v + w).

(3) O∈ R2 u∈ R2 O + u = u.

(4) u∈ R2 u∈ R2 u + u= O.

(5) r, s∈ R u∈ R2, r(su) = (rs)u.

(7)

1.1. 5

(6) r, s∈ R u∈ R2, (r + s)u = ru + su.

(7) r∈ R u, v∈ R2 r(u + v) = ru + rv.

(8) u∈ R2, 1u = u.

, . 學

. .

(1) 性.

. 學 . ,

. 大 大 數 ,

, .

(2) , 數 性 . (u + v) + w

u v w . v w

u . , (1)

, 大 .

(3) , .

. 性 .

(4) , .

u u u + u= O. u u ,

. 數學 .

(5),(6),(7) 數 性 , r(su) u s

r. 性 數 性 ,

.

(8) 1 . , 1

u =1u =1

2(2u) =1 2v.

: 性 , 大 性 .

Proof. (of Proposition 1.1.2) 性 ,

(1) u = (a1, a2), v = (b1, b2),

u + v = (a1+ b1, a2+ b2), v + u = (b1+ a1, b2+ a2).

(8)

(2) u = (a1, a2), v = (b1, b2), w = (c1, c2), u + v = (a1+ b1, a2+ b2) (u + v) + w = ((a1+ b1) + c1, (a2+ b2) + c2).

v + w = (b1+ c1, b2+ c2)

u + (v + w) = (a1+ (b1+ c1), a2+ (b2+ c2)).

(3) 性 , O .

O = (0, 0), u = (a1, a2)

O + u = (0 + a1, 0 + a2) = (a1, a2) = u.

.

(4) O = (0, 0), u = (a1, a2) u =

(−a1,−a2),

u + u= (a1+ (−a1), a2+ (−a2)) = (0, 0) = O.

(5) u = (a1, a2), su = (sa1, sa2), r(su) = (r(sa1), r(sa2)).

(rs)u = ((rs)a1, (rs)a2),

(6) u = (a1, a2),

(r + s)u = ((r + s)a1, (r + s)a2).

ru = (ra1, ra2), su = (sa1, sa2),

ru + su = (ra1+ sa1, ra2+ sa2),

(7) u = (a1, a2), v = (b1, b2),

r(u + v) = (r(a1+ b1), r(a2+ b2)).

ru = (ra1, ra2), rv = (rb1, rb2),

ru + rv = (ra1+ rb1, ra2+ rb2),

(8) u = (a1, a2), 數 a 1a = a, 1u = (1a1, 1a2) 1u = u.



Question 1.1. R2 ,:

(9)

1.1. 7

(1) O = (0, 0) R2 u∈ R2 O + u = u.

(2) u = (a, b)∈ R2, u= (−a,−b) R2 u + u= O.

. 性 ,

.

Example 1.1.3. P, Q, R, S , −→

PQ =−→

SR, 性 :

(1) −→

QR =−→ PS.

(2) TPRSQ , −→

PT =−→

T R −→

ST =−→

T Q.

P

Q R

S

T

(1) −→

PQ =−→

SR −→

QR =−→

PS. 前 (1.1) −→

PQ +−→

QR =−→

−→ PR PS +−→

SR =−→

PR.

−→PQ +−→

QR =−→ PS +−→

SR. (1.2)

Proposition 1.1.2 (4), u −→

PQ + u = O. Proposition

1.1.2 (1) −→

PQ =−→

SR ,

−→PQ + u = u +−→

PQ =−→

SR + u = u +−→ SR = O

(1.2) u

u + (−→

PQ +−→

QR) = u + (−→ PS +−→

SR) (1.3)

Proposition 1.1.2 (2), (3) u + (−→

PQ +−→

QR) = (u +−→

PQ) +−→

QR = O +−→

QR =−→

QR.

u + (−→ PS +−→

SR) = (−→ PS +−→

SR) + u =−→ PS + (−→

SR + u) =−→

PS + O =−→ PS.

(1.3) −→

QR =−→ PS.

(2) , ,

, ,

. 線 PRSQ T PR , SQ

. , PR SQ

(10)

. PR SQ , PR SQ .

PR SQ , , 線 PRSQ

PR SQ . T, T′′ PR SQ .

T= T′′, T PR SQ , T = T= T′′,

.

T, T′′ PR SQ ,

−→PT=−→

TR = 1 2

−→PR (1.4)

−−→ST′′=−−→

T′′Q = 1 2

−→SQ. (1.5)

−→PT=−−→

PT′′ T= T′′. −−→

PT′′=−→ PS +−−→

ST′′, (1.5)

−−→PT′′=−→ PS +−−→

ST′′=−→ PS +1

2

−→SQ.

(1) −→

PS =−→

QR

−−→PT′′=−→

QR +1 2

−→SQ = (1 2

−→QR +1 2

−→QR) +1 2

−→SQ = 1 2

−→QR +1 2(−→

QR +−→

SQ) =1 2(−→

QR +−→ SR)

−→PQ =−→ SR,

−−→PT′′=1 2(−→

QR +−→ SR) =1

2(−→

QR +−→

PQ) =1 2

−→PR

(1.4) −→

PT=−−→

PT′′, T= T′′ PR SQ ,

T = T= T′′. (1.4,1.5) −→

PT =−→

T R −→

ST =−→

T Q.

, Proposition 1.1.2 性 ,

1.1.2 性 , Example 1.1.3 性 . Example 1.1.3

, Proposition 1.1.2 性

, 大 .

1.2. Rn

. Rn , n∈ N

. 線性代數 ,

, , R2 性 , R2

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1.2. Rn 9

R3 , (a1, a2, a3) a1, a2, a3∈ R ,

(a1, a2, a3), (b1, b2, b3) a1= b1, a2= b2 a3= b3.

n∈ N :

Deﬁnition 1.2.1. n∈ N, Rn (a1, . . . , an), a1, . . . , an∈ R.

(a1, . . . , an) (b1, . . . , bn) a1= b1, . . . , an= bn. , (a1, . . . , an) n ,

n an . n = 4 ,R4 (a1, a2, a3, a4) . n

n = 4 a1, a2, . . . , (a1, . . . , an)

. a1, . . . , an∈ R, a1 an n

ai∈ R . : i 1 n 數, i ai

, 數 . a1= b1, . . . , an= bn

, ai= bi,∀1 ≤ i ≤ n .

R2 數 Rn (addition)

(scalar multiplication).

Deﬁnition 1.2.2. u = (a1, . . . , an), v = (b1, . . . , bn)∈ Rn r∈ R.

u + v = (a1+ b1, . . . , an+ bn) and ru = (ra1, . . . , ran).

, Rn ,

u = (1, 1, 2, 2, 3), v = (5, 4, 3, 2, 1)∈ R5

u + v = (1 + 5, 1 + 4, 2 + 3, 2 + 2, 3 + 1) = (6, 5, 5, 4, 4).

u, v Rn u + v. u R3 v

R4 u + v !

R2 , Rn 數 性 . 性 R2

( 數 性 ), .

Proposition 1.2.3. Rn ,:

(1) u, v∈ Rn, u + v = v + u.

(2) u, v, w∈ Rn, (u + v) + w = u + (v + w).

(3) O∈ Rn u∈ Rn O + u = u.

(4) u∈ Rn u∈ Rn u + u= O.

(12)

(5) r, s∈ R u∈ Rn, r(su) = (rs)u.

(6) r, s∈ R u∈ Rn, (r + s)u = ru + su.

(7) r∈ R u, v∈ Rn r(u + v) = ru + rv.

(8) u∈ Rn, 1u = u.

Proposition 1.2.3 (3) O additive identity, (a1, . . . , an), ai= 0, ∀1 ≤ i ≤ n, O . u = (a1, . . . , an)∈ Rn, Proposi- tion 1.2.3 (4) u additive inverse, (b1, . . . , bn), bi=−ai,

∀1 ≤ i ≤ n. Question 1.1 , Rn , O Rn

u∈ Rn O + u = u , u∈ Rn, u=−1u u + u= O

. , Rn , Proposition 1.2.3 (3) 性

O . O∈ Rn additive identity ( Proposition

1.2.3 (3) 性 ), O additive identity O + O= O. , O additive

identity O + O= O. O = O, 性. , u∈ Rn,

u, u′′ Proposition 1.2.3 (3) 性 , u + u= u + u′′= O, u= u+ O = u+ (u + u′′) = (u+ u) + u′′= O + u′′= u′′.

Corollary 1.2.4. Rn O u∈ Rn O + u = u. ,

u∈ Rn, u∈ Rn u + u= O.

Remark 1.2.5. 數學 , 前 ( 性 ),

Corollary . Proposition 1.2.3 (3),(4) additive identity additive

inverse 性 Corollary 1.2.4 性. 學

? Corollary , Rn

Corollary 1.2.4 ( ). 前 ,

−u u additive inverse.

Proposition 1.2.3 (3) additive identity 性 Corollary 1.2.4

additive identity 性. , Proposition 1.2.3

(3) O u∈ Rn u + O = u. 初 u∈ Rn

u + O = u, Corollary 1.2.4 additive identity 性 .

Proposition 1.2.3 性 , v

, , u u + v = u, v

. .

(13)

1.2. Rn 11

Corollary 1.2.6. v∈ Rn u∈ Rn u + v = u, v = O.

Proof. u = u + v = v + u, −u,

O = u + (−u) = (v + u) + (−u) = v + (u + (−u)) = v + O = v.



. Proposition 1.2.3 ,

Corollary 1.2.4 Corollary 1.2.6. Corollary 1.2.6 Rn

v , Rn u, v + u = u , Rn

. 性 .

Corollary 1.2.7. v∈ Rn, . (1) 0v = O.

(2) (−1)v = −v.

Proof. Rn 數 , .

Proposition 1.2.3 Corollaries .

(1) Corollary 1.2.6, u 0v + u = u . u = v,

0v + v = 0v + 1v = (0 + 1)v = 1v = v.

0v = O.

(2) (−1)v = −v, Corollary 1.2.4 additive inverse 性 (−1)v + v = O . ,

(−1)v + v = (−1)v + 1v = (−1 + 1)v = 0v, (1) (−1)v + v = O.



, Corollaries 1.2.4, 1.2.6, 1.2.7, Rn

, 性 Proposition 1.2.3 , 大 Proposition 1.2.3

“ ” , w + (−v) w− v.

u = 12(w− v).

(14)

1.3. Span of Vectors

Rn Rn “ ”. .

, , 大 Rn

(subset).

. 線. 線

, 線

, 線 P −→

PQ 線. −→

PQ

QP, Q , −−→

PQ 線 . P, Q, Q

PQ= r−→

PQ.

v 線 , 線 P, Q 數 r

−→PQ = rv. 線 L P v, Q −→

PQ = rv,

r∈ R, Q 線 L . P

v 線 L ,

L ={Q |−→

PQ = rv, r∈ R}.

, ,

. “|” , 性

. Q Q , −→

PQ = rv, r∈ R

Q −→

PQ = rv, r 數.

v 線 L 數 r, w = rv, w L

. w v parallel ( ), v∥ w . 線

, 線 .

, 線 , ,

. Rn v ,

{w ∈ Rn| w = rv, r ∈ R}

. , | “ ”,

, 前 線 L | “ ”

, 大 .

. ,

, 線 . ,

P , L1 Q r1∈ R −→

PQ = r1v1. Q L2

(15)

1.3. Span of Vectors 13

−→QR = r2v2. −→

PR =−→

PQ +−→

QR

−→PR = r1v1+ r2v2,

R r1, r2∈ R −→

PR = r1v1+ r2v2. R

−→PR = r1v1+ r2v2, R L1 L2 線 , R

. H P v1, v2 H 線 ,

H ,

H ={R | −→PR = r1v1+ r2v2, r1, r2∈ R}.

H v1, v2 , H

, , 學 線性代數 .

{w ∈ Rn| w = rv,r ∈ R} {w ∈ Rn| w = r1v1+ r2v2, r1, r2∈ R}

.

, , Rn

. Rn 線 , 線

Deﬁnition 1.3.1. v1, . . . , vm∈ Rn, r1, . . . , rm∈ R w = r1v1+··· + rmvm,

w v1, . . . , vm linear combination (線 性 ). v1, . . . , vm linear

combinations span ( ), Span(v1, . . . , vm)

,

Span(v1, . . . , vm) ={w ∈ Rn| w = r1v1+··· + rmvm, r1, . . . , rm∈ R}.

v1, . . . , vm Rn , v1, . . . , vm 線性 Rn .

Span(v1, . . . , vm) Rn . 性 , Rn

( ). 性 ?

O Span(v1, . . . , vm) ( ri 0), w∈ Span(v1, . . . , vm), w = r1v1+···+rmvm −w = (−r1)v1+···+(−rm)vm −w ∈ Span(v1, . . . , vm).

Proposition 1.3.2. v1, . . . , vm∈ Rn, u, w∈ Span(v1, . . . , vm) s,t∈ R

su + tw∈ Span(v1, . . . , vm).

Proof. u, w∈ Span(v1, . . . , vm) r1, . . . , rm∈ R r1, . . . , rm∈ R u = r1v1+··· + rmvm w = r1v1+··· + rm vm.

su + tw = (sr1+ tr1)v1+··· + (srm+ trm)vm.

su + tw v1, . . . , vm 線性 , su + tw∈ Span(v1, . . . , vm). 

(16)

Proposition 1.3.2 Span(v1, . . . , vm) 線性 Span(v1, . . . , vm) , 數學 , Span(v1, . . . , vm)

Question 1.2. v1, . . . , vm∈ Rn, w1, . . . , wk∈ Span(v1, . . . , vm), Span(w1, . . . , wk)⊆ Span(v1, . . . , vm).

span of vectors, vectors vectors

span . (1, 2, 3)∈ R3 Span((1,−1,2),(2,1,−2))?

r, s∈ R (1, 2, 3) = r(1,−1,2) + s(2,1,−2). ,



1r + 2s = 1

−1r + 1s = 2 2r − 2s = 3

.

. 大 , (

column vector), r(1,−1,2) + s(2,1,−2) = (1,2,3)

r

 1

−1 2

 + s

 2 1

−2

 =

 1 2 3

.

. vector column vector

. column vector , column vector row vector

( ) . row vector column vector.

.

column vector standard basis. R2

i = [ 1

0 ]

, j = [ 0

1 ]

, R2 vector i, j 線性 ,

[ 2 3

]

2i + 3j. R3 standard basis vectors

i =

 1 0 0

, j =

 0 1 0

, k =

 0 0 1

.

Rn standard basis vectors

e1=





 1 0 0 ... 0





 , e2=





 0 1 0 ... 0







, . . . , en=





 0 0 ... 0 1





 ,

1≤ i ≤ n, ei i 1, 0 column vector. ,

Rn vector e1, . . . , en linear combination. Rn

standard basis vectors 性 vectors, .

(17)

1.4. Dot Product 15

1.4. Dot Product

R2 R3 大 Rn.

Rn , 大 性 Rn

.

R2 R3 . R2 u = (a1, a2), v = (b1, b2), u, v u· v u· v = a1b1+ a2b2. R3 u = (a1, a2, a3), v = (b1, b2, b3), u, v

u· v u· v = a1b1+ a2b2+ a3b3. Rn

:

Deﬁnition 1.4.1. u = (a1, . . . , an), v = (b1, . . . , bn)∈ Rn. u, v dot product (inner product)

u· v = a1b1+··· + anbn=

### ∑

n i=1

aibi.

,

Proposition 1.4.2. u, v, w∈ Rn, 性 : (1) u· v = v · u.

(2) u· u ≥ 0 u· u = 0 u = O.

(3) r∈ R (ru)· v = u · (rv) = r(u · v).

(4) u· (v + w) = u · v + u · w.

Proof. 性 R2 R3 大 , Rn , n

u = (a1, . . . , an), v = (b1, . . . , bn), w = (c1, . . . , cn). ∑ (summation)

, 大 .

(1)

u· v =

### ∑

n

i=1

aibi

u· v aibi i 1 n 數. n aibi

biai ( 數 ) ,

n i=1

aibi=

n i=1

biai= v· u.

u· v = v · u.

(2)

u· u =

n

i=1

aiai=

### ∑

n i=1

a2i.

(18)

ni=1a2i = 0, a2i 0, 1≤ i ≤ n ai= 0,

u = (a1, . . . , an) = (0, . . . , 0) = O.

u = (a1, . . . , an) = O 1≤ i ≤ n ai= 0, u· u =

### ∑

n

i=1

aiai= 0.

(3) (ru)· v ru v , ru = (ra1, . . . , ran) (ru)· v =

### ∑

n

i=1

(rai)bi. 1≤ i ≤ n (rai)bi= r(aibi) ( 數 )

n i=1

(rai)bi=

### ∑

n i=1

r(aibi)

ni=1r(aibi) r , 數

n i=1

r(aibi) = r

### ∑

n i=1

aibi= r(u· v)

(ru)· v = r(u · v). u· (rv) = r(u · v),

(1) u· (rv) = (rv) · u (rv)· u = r(v · u), (1)

r(v· u) = r(u · v) u· (rv) = r(u · v).

(4) u· (v + w) u v + w ,

v + w = (b1+ c1, . . . , bn+ cn)

u· (v + w) =

### ∑

n

i=1

ai(bi+ ci).

ai(bi+ ci) aibi+ aici,

n i=1

ai(bi+ ci) =

n i=1

(aibi+ aici).

,

n i=1

(aibi+ aici) =

n i=1

aibi+

### ∑

n i=1

aici= u· v + u · w,

u· (v + w) = u · v + u · w. 

Proposition 1.4.2 (2) O , v v· v > 0,

.

Deﬁnition 1.4.3. v = (a1, . . . , an)∈ Rn, v (length)

∥v∥ =√

v· v =

a21+ a22+··· + a2n.

(19)

1.4. Dot Product 17

Proposition 1.4.2 性 , .

Lemma 1.4.4. u, v∈ Rn, ∥u + v∥2=∥u∥2+ 2u· v + ∥v∥2. Proof. ∥u + v∥2= (u + v)· (u + v), Proposition 1.4.2 (4)

(u + v)· (u + v) = (u + v) · u + (u + v) · v = u · u + v · u + u · v + v · v.

Proposition 1.4.2 (1) v· u + u · v = 2u · v . 

Question 1.3. (parallelogram relation):

, u, v∈ Rn

∥u + v∥2+∥u − v∥2= 2∥u∥2+ 2∥v∥2.

Lemma 1.4.4 性 ,

Deﬁnition 1.4.1 ( )

Proposition 1.4.2 性 , Lemma 1.4.4 性 . Lemma 1.4.4

.

Proposition 1.4.5 (Cauchy-Schwarz inequality). u, v∈ Rn, |u · v| ≤ ∥u∥∥v∥.

u, v , λ ∈ R v =λu.

Proof. u v , u· v = 0 ∥u∥∥v∥ = 0, .

u, v , u0= u/∥u∥ v0= v/∥v∥.

∥u02= u0· u0= 1

∥u∥u· 1

∥u∥u = 1

∥u∥2u· u = 1.

∥v02= 1, Lemma 1.4.4

∥u0+ v02= 2 + 2u0· v0, (1.6)

∥u0− v02= 2− 2u0· v0.

w∈ Rn ∥w∥2≥ 0, −1 ≤ u0· v0≤ 1. u, v

−∥u∥∥v∥ ≤ u · v ≤ ∥u∥∥v∥.

|u · v| ≤ ∥u∥∥v∥.

u, v , u0· v0= 1 u0· v0=

−1. (1.6) ∥u0−v02= 0 ∥u0+ v02= 0, u0= v0 u0=−v0. u, v

v =∥v∥

∥u∥u v =−∥v∥

∥u∥u.

λ ∥v∥/∥u∥ −∥v∥/∥u∥, v =λu.

v =λu, Proposition 1.4.2

|u · v| = |λ||u · u| = |λ|∥u∥2=∥u∥∥λu∥ = ∥u∥∥v∥.



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Proposition 1.4.5 , , u 1

u0= u/∥u∥. 1 , unit vector. u

unit vector, u0= u/∥u∥. unit vector

Proposition 1.4.5, .

Corollary 1.4.6 (Triangle inequality). u, v∈ Rn, ∥u + v∥ ≤ ∥u∥ + ∥v∥.

Proof. Lemma 1.4.4 Proposition 1.4.5,

∥u + v∥2=∥u∥2+ 2u· v + ∥v∥2≤ ∥u∥2+ 2∥u∥∥v∥ + ∥v∥2= (∥u∥ + ∥v∥)2.

∥u + v∥ ≤ ∥u∥ + ∥v∥. 

Question 1.4. ∥u + v∥ = ∥u∥ + ∥v∥.

. u, v θ, u· v = ∥u∥∥v∥cosθ,

. u· v = 0 u v . Rn.

n≥ 4 , “ ”Rn ( ),

R2,R3 u, v∈ Rn θ, 0≤θ ≤ π

cosθ = u· v

∥u∥∥v∥.

“well-deﬁned”.

θ , 性 . 0≤θ ≤ π ,

|cosθ| ≤ 1. θ 性 Rn u, v

u· v

∥u∥∥v∥

≤ 1.

Proposition 1.4.5 , θ 性 .

, ? 性 .

θ̸=θ cosθ = cosθ , θ 0≤θ ≤ π,

. ,

well-deﬁned.

Example 1.4.7. R4 u = (1, 1, 1, 1), v = (1, 0,−2,−2) θ, cosθ = u· v

∥u∥∥v∥= −3 2× 3=1

2, θ = 120.

.

Deﬁnition 1.4.8. u, v∈ Rn , u v orthogonal u·v = 0.

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1.4. Dot Product 19

Rn , orthogonal

perpendicular. , R2 R3

(projection) Rn.

R2 , u∈ R2, v∈ R2, u v u

, v− u ( 線 ) u , (v− u)· u = 0,

v· u = u· u.

-

*6

- u v

u

u Span(u), r∈ R u= ru, (v− u)· u = 0.

u= ru 代 v· u = ru · u = r∥u∥2, r = (v· u)/∥u∥2. , v u

, ∥u∥v·u2u ( 性). r = (v· u)/∥u∥2 (

u ), u= ru (v− u)· u = 0 ( 性).

Rn .

Proposition 1.4.9. u∈ Rn, v∈ Rn, v = u+ v,

u, v∈ Rn v· u = 0 u= ru, r∈ R. ,

r = v· u

∥u∥2.

Proof. 前 Rn , r = (v·u)/∥u∥2 數 (v−ru)·u = 0.

u= v· u

∥u∥2u.

u= ru (v−u)·u = 0. u , v v+ u= v,

v= v− u, . 

Question 1.5. r Proposition 1.4.9 性?

Proposition 1.4.9, 大 Rn u , Rn

v , Span(u) ( u) u

( v), . Span(u)

v· u

∥u∥2u v u projection ( ).

Example 1.4.10. R4 u = (1, 1, 1, 1), v = (1, 0,−2,−2). ∥u∥ = 2 v· u = −3, v u projection

3

4u =3

4(1, 1, 1, 1).

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v = (1, 0,−2,−2) = −3

4(1, 1, 1, 1) + (7 4,3

4,−5 4,−5

4),

3

4(1, 1, 1, 1)∈ Span((1,1,1,1)) and (7 4,3

4,−5 4,−5

4)· (1,1,1,1) = 0.

1.5.

R2,R3 性 . 性

, R2,R3 . 言 , 性 ,

, “ ”. 大

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## Systems of LinearEquations

. 大 (

) . ,

. ,

, 性 . 性 ,

.

2.1.

n n 數 (variable) (linear equation).

2x1+ 5x2− x3+ x4= 1 4 ( 5

). n

a1x2+··· + anxn= b,

a1, . . . , an b 數, xi 數. n

, (system of linear equations).

a11x1 + a12x2 + ··· + a1nxn = b1

a21x1 + a22x2 + ··· + a2nxn = b2 ...

am1x1 + am2x2 + ··· + amnxn = bm

m n . a11x1+ a12x2+··· + a1nxn = b1

, a21x1+ a22x2+··· + a2nxn= b2 , 1≤ i ≤ m , i ai1x1+ ai2x2+··· + ainxn= bi, ( m )

am1x1+ am2x2+··· + amnxn= bm. ai j, bi 數, 數

21

(24)

, ,

A =





a11 a12 ··· a1n

a21 a22 ··· a2n

... ... ... ... am1 am2 ··· amn



, x =



 x1 x2 ... xn



, b =



 b1 b2 ... bm



,

Ax = b . A ai j A entry.

A entry 數 數, A

column ( ). row 數 , row

row, row row, . column 數 ,

column column, column column,

. 大 A row , row

, row , . column

, A m row n column, m× n matrix.

x x, b column vector ( )

. 前大 .

3x1− 2x2+ 9x4 = 4

2x1+ 2x2− 4x4 = 6 (2.1)

[ 3 −2 0 9

2 2 0 −4

]



x1

x2

x3 x4



 = [ 4

6 ]

[ 0 0

]

column x3 數 0.

? 線性代數

. 1.3 span . u = (1,−1,2,2),v = (3,1,−1,2)

w = (1, 0, 1, 0) Span(u, v) c1, c2∈ R w = c1u + c2v.

,

(1, 0, 1, 0) = c1(1,−1,2,2) + c2(3, 1,−1,2).

x1 + 3 x2 = 1

−x1 + x2 = 0 2 x1 x2 = 1 2 x1 + 2 x2 = 0

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2.1. 23

. x = (x1, x2, x3, x4) u· x = 0 v· x = 0, x1 − x2 + 2 x3 + 2 x4 = 0

3 x1 + x2 x3 + 2 x4 = 0.

, .

, :

(1)

(2) 數

(3) 數

: 步 ?

? 步 ,

.

a11x1 + a12x2 + ··· + a1nxn = b1

a21x1 + a22x2 + ··· + a2nxn = b2

...

am1x1 + am2x2 + ··· + amnxn = bm

, augmented matrix ( )





a11 a12 ··· a1n b1

a21 a22 ··· a2n b2 ... ... . .. ... ... am1 am2 ··· amn bm





(2.1) augmented matrix

[ 3 −2 0 9 4

2 2 0 −4 6

]

augmented matrix [A| b] Ax = b.

:

(1) row

(2) row 數

(3) row 數 row.

(26)

matrix 前 步 .

augmented matrix [A| b]A elementary row operation echelon form.

echelon form. row

0 row leading entry. 數 entry

variable 數, leading entry variable xi 數,

leading entry xi . , leading entry

i column.

 1 2 1 1 4

0 0 5 0 2

0 0 1 −1 1

row leading entry 1 row 1,

row leading entry x1 , row row leading

entry 5 1 x3.

echelon form leading entry row ( row

0) , leading entry row leading entry

. 言 , row leading entry xi, row

lading entry xj, i < j. echelon form, 3 row

2 row leading entry x3, .

 1 2 −1 0

0 0 0 0

0 0 3 0

,

 0 1 1 2 0 0 2 −1

3 0 0 0

echelon form, 前 0 row , 3

row leading entry 2 row leading entry .



0 2 1 1 4

0 0 3 0 2

0 0 0 −1 1

0 0 0 0 0



echelon form. echelon form , row leading entry

pivot, pivot pivot variable.

augmented matrix [A| b] elementary row operation [A| b]

A echelon form . A . A row 0; A

row 0. .

(1) A row 0: consistent, .

.

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2.1. 25

(a) 數 (variable) xi pivot variable. pivot 數

 2 1 1 4

0 3 1 2

0 0 −1 1

echelon form pivot variable x1, x2, x3

“代 ” . 前 augmented matrix

2x1 +x2 +x3 = 4 3x2 +x3 = 2

−x3 = 1

−x3= 1 x3=−1. x3=−1 代

3x2+x3= 2, 3x2−1 = 2, x2= 1. x3=−1,x2= 1 代 2x1+x2+x3= 4, x1= 2. x1= 2, x2= 1, x3=−1.

(b) variable xi pivot variable. 數

 2 1 3 1 4

0 3 3 1 2

0 0 0 −1 1

echelon form pivot variable x1, x2, x4

x1, x2, x3, x4. .

, free variables. free variable

pivot variable variable. 前 , x3 free variable. Free

variable , free variables

. augmented matrix

2x1 +x2 +3x3 +x4 = 4 3x2 +3x3 +x4 = 2

−x4 = 1

free variable x3 數 t ( 數 t ∈ R).

−x4= 1 x4=−1. x3= t, x4=−1 代

3x2+ 3x3+ x4= 2, 3x2+ 3t− 1 = 2, x2= 1− t. x2= 1− t,x3= t, x4=−1 代 2x1+ x2+ 3x3+ x4= 4, x1= 2−t. x1= 2−t,x2=

1−t,x3= t, x4=−1, t 數. t 數,

.

(2) A row 0: , :

(28)

(a) A row 0 b row 0.

[A| b] =

 2 1 1 4 0 3 1 2 0 0 0 1

A row 0, b row 1.

inconsistent, . augmented matrix row

0x1+ 0x2+ 0x3= 1

x1, x2, x3 代 數 0x1+ 0x2+ 0x3= 1, .

(b) A 0 row, b row 0.

 2 1 4 0 3 2 0 0 0

,

 2 1 3 1 4 0 3 3 1 2 0 0 0 0 0

augmented matrices .

consistent. 0 row, 前

augmented matrices [ 2 1 4

0 3 2 ]

,

[ 2 1 3 1 4 0 3 3 1 2

]

. 前 (1) A row 0

.

, pivot 數 variables ( 數) 數 .

A echelon form , column pivot (

leading term ), pivot 數 column 數.

A column 數 variables 數, pivot 數

variables 數. row pivot, pivot 數

Question 2.1. n variables m .

(1)(a) m = n ; (1)(b) m < n

;

Ax = b 步 . augmented matrix

[A| b] elementary row operations [A| b] A echelon form .

. elementary row

operations A echelon form A, .

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2.2. Elementary Row Operations 27

2.2. Elementary Row Operations

Ax = b, augmented matrix [A| b]

elementary row operations [A| b] A echelon form .

elementary row operations echelon

form elementary row operations .

. row 數 數學 . row

echelon form, row elementary

row operations echelon form. row 3 row

elementary row operations echelon form,

4, 5, 6, . . . row . 數 row

( 10 row), ( 數 row). 數學

. k row elementary row

operations echelon form k + 1 row

elementary row operations echelon form, row

elementary row operations echelon form row

elementary row operations echelon form, 3 row ,

4 row , elementary

row operations echelon form.

, . 3 row

 0 0 1 1 1

0 1 2 1 3

0 2 2 0 −1

echelon form row leading entry ,

, row elementary row operation

 0 1 2 1 3

0 0 1 1 1

0 2 2 0 −1

row leading entry row 數 .

row −2 row elementary row operation

 0 1 2 1 3

0 0 1 1 1

0 0 −2 −2 −7

entry . row row .

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row [

0 0 1 1 1

0 0 −2 −2 −7

]

row echelon , .

, row 2 row

[ 0 0 1 1 1

0 0 0 0 −5 ]

echelon form. row

 0 1 2 1 3

0 0 1 1 1

0 0 −2 −2 −7

row 2 row

 0 1 2 1 3

0 0 1 1 1

0 0 0 0 −5

echelon form.

. row .

row echelon form. row .

echelon form row leading entry ( ) row

row ( row leading entry row) row

row operation row. row leading entry

row leading entry . row leading entry

row , row leading entry , row

echelon form. row leading entry row ,

row −b/a, a row leading entry b row leading

entry, row . row leading entry 0,

leading entry , echelon form.

3 row , 數學 ,

k row , k row elementary row

operation echelon form. k + 1 row . 前 ,

leading entry row row row operation

row. row leading entry a. leading entry

row leading entry row , row leading entry

b, row −b/a row . row leading entry

. 步 , row row leading entry

row leading entry . , row row

augmented matrix [A |I 4 ], elementary row operation A

column vector

Proposition 9.4.2, A orthogonal diagonalizable, Spectral Theorem.. Theorem 9.4.6

Theorem 8.2.6 (3) elementary column operation.. determinant elementary row

, A echelon form ( reduced echelon form) pivot column vectors.. elementary row operations column

minimal element; vector space linearly independent set

( diagonalizable), linear operator rational form Jordan form. cyclic subspace