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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

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## Subspaces in R m

Rm , 線性 Rm ( 性),

Proposition 1.2.3 8 , vector space. Rm

vector space, Rm subspace. Rm subspaces

4.1. Subspace and basis

, subspace basis .

Rm V , Rm , Proposition 1.2.3 8

(3), (4) ( O∈ V v∈ V, −v ∈ V) 性 ,

O =0v∈ V −v = (−1)v ∈ V. V 線性 性

Proposition 1.2.3 8 , Rm vector space

. .

Deﬁnition 4.1.1. V Rm . V 線性

V ( v1, . . . , vn∈ V c1, . . . , cn∈ R c1v1+··· + cnvn∈ V), V

Rm subspace ( ).

/0 subspace. Rm {O} Rm

subspace, Rm Rm subspace.

, Span(v1, . . . , vn) Rm v1, . . . , vn 線性 ( Deﬁnition 1.3.1),

Span(v1, . . . , vn) ={v ∈ Rm| r1v1+··· + rnvn, for some r1, . . . , rn∈ R}.

V Rm subspace v1, . . . , vn∈ V

Span(v1, . . . , vn)⊆ V. (4.1) 69

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V subspace, V 線性

V , . , , 線性

.

Proposition 4.1.2. V Rm , V Rm subspace O∈ V

u, v∈ V, r ∈ R u + rv∈ V.

Proof. (⇒) : V , v∈ V. 0v = O, subspace

0v∈ V, O∈ V. subspace , u, v∈ V, r ∈ R

u + rv∈ V.

(⇐) : O∈ V u, v∈ V, r ∈ R u + rv∈ V. O∈ V, V Rm

. (4.1), v1, . . . , vn∈ V, Span(v1, . . . , vn)⊆ V.

v1∈ V, Span(v1) ⊆ V. Span(v1) rv1 ,

r ∈ R rv1∈ V. u = O, v = v1, rv1 = O + rv1=

u + rv. u + rv∈ V, n = 1 . n = k ,

v1, . . . , vk ∈ V Span(v1, . . . , vk)⊆ V. v1, . . . , vk, vk+1∈ V Span(v1, . . . , vk, vk+1)⊆ V. w∈ Span(v1, . . . , vk+1), c1, . . . , ck, ck+1∈ R

w = c1v1+··· + ckvk+ ck+1vk+1. u = c1v1+··· + ckvk, u∈ V.

r = ck+1∈ R v = vk+1∈ V, w = u + rv∈ V, Span(v1, . . . , vk+1)⊆ V.

subspace. 

Proposition 4.1.2, V subspace,

(1) O∈ V

(2) u, v∈ V, r ∈ R ⇒ u + rv ∈ V.

.

Example 4.1.3. a, b, c, d∈ R, S ={(x,y,z) ∈ R3 | ax + by + cz = d}.

a, b, c, d S R3 subspace. S subspace, (x, y, z) = (0, 0, 0) = O∈ S,

d = 0. , d = 0 u = (x, y, z)∈ S, v = (x, y, z)∈ S r∈ R, ax + by + cz = 0, ax+ by+ cz= 0

a(x + rx) + b(y + ry) + c(z + rz) = (ax + by + cz) + r(ax+ by+ cz) = 0,

u + rv = (x + rx, y + ry, z + rz)∈ S. d = 0 d = 0 , S R3 subspace.

Example 4.1.3 , a, b, c, d 0 S =R3, R3 subspace.

Rm subspaces, Proposition 1.3.2 v1, . . . , vn∈ Rm, Span(v1, . . . , vn) Rm subspace ( Rm subspace ).

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4.1. Subspace and basis 71

m×n matrix A. A column vectors , A column

space. a1, . . . , an∈ Rm A column vectors, Span(a1, . . . , an) A column space.

A row vectors , A row space. A m× n matrix,

column space Rm subspace, row space Rn subspace.

A subspace, homogeneous linear system Ax = O

, {x ∈ Rn| Ax = O}. x = O∈ Rn Ax = O . x = u∈ Rn,

x = v∈ Rn Ax = O , r∈ R,

A(u + rv) = Au + rAv = O,

x = u + rv Ax = O . proposition 4.1.2 {x ∈ Rn| Ax = O} Rn

subspace, A nullspace.

Rm standard basis e1, . . . , em, Rm

e1, . . . , em 線性 , . standard basis ,

Rm , Rm 性 . Rm

subspace 性 , .

Deﬁnition 4.1.4. V Rm subspace. v1, . . . , vn∈ V, v∈ V, c1, . . . , cn∈ R c1v1+··· + cnvn= v, v1, . . . , vn V basis ( ).

Deﬁnition 4.1.4 basis ,

basis , .

v1, . . . , vn∈ Rm column vectors m× n matrix A, A =

 v 1 v2 ··· v n

.

v = c1v1+··· + cnvn x1= c1, . . . , xn= cn Ax = v , v∈ V, c1, . . . , cn∈ R c1v1+··· + cnvn= v, v∈ V

Ax = v . 性 .

Proposition 4.1.5. V Rm subspace v1, . . . , vn∈ V. A v1, . . . , vn

column vectors m× n matrix. v1, . . . , vn V basis v∈ V,

Ax = v .

, m× n matrix A v∈ V, Ax = v

, A column vectors V basis. A column vectors

V . V basis vectors V .

, V Rm standard basis , standard basis

Rm subspace basis, basis !

v1, . . . , vn, Proposition 4.1.5 前學

v1, . . . , vn V basis.

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, basis.

v1, . . . , vn ∈ V, V Rm subspace, Span(v1, . . . , vn)⊆ V.

v1, . . . , vn V basis , v∈ V c1, . . . , cn∈ R v =

c1v1+··· + cnvn, v∈ V v∈ Span(v1, . . . , vn). V ⊆ Span(v1, . . . , vn),

V = Span(v1, . . . , vn). 性 v1, . . . , vn, .

Deﬁnition 4.1.6. V Rm subspace. v1, . . . , vn∈ Rm Span(v1, . . . , vn) = V , v1, . . . , vn V spanning vectors.

v1, . . . , vn V spanning vectors, v1, . . . , vn∈ V. v1, . . . , vn Span(v1, . . . , vn), V = Span(v1, . . . , vn) v1, . . . , vn∈ V.

v1, . . . , vn V basis , v∈ V c1, . . . , cn∈ R

c1v1+··· + cnvn= v, V subspace, O∈ V, v = O ,

c1, . . . , cn∈ R c1v1+···+cnvn= O. c1=··· = cn= 0 c1v1+···+cnvn= O,

c1=··· = cn= 0 c1v1+··· + cnvn= O. 性 v1, . . . , vn, .

Deﬁnition 4.1.7. v1, . . . , vn∈ Rm. v1, . . . , vn c1=··· = cn= 0 c1v1+··· + cnvn= O, v1, . . . , vn linearly independent.

A v1, . . . , vn column vectors m× n matrix, v1, . . . , vn linearly independent homogeneous linear system Ax = O nontrivial solution.

x1= c1, . . . , xn= cn Ax = O nontrivial solution, c1, . . . , cn 0 c1v1+··· + cnvn= O, v1, . . . , vn linearly independent .

spanning vectors linear independence 性 , .

, v1, . . . , vn∈ V V basis, v1, . . . , vn V

spanning vectors linearly independent. basis ,

.

Proposition 4.1.8. V Rm subspace v1, . . . , vn∈ V. v1, . . . , vn V basis v1, . . . , vn V spanning vectors linearly independent.

Proof. v1, . . . , vn V spanning vectors linearly independent

v1, . . . , vn V basis. Proposition 4.1.5, A v1, . . . , vn

column vectors m× n matrix, v∈ V, Ax = v

. v1, . . . , vn V spanning vectors, v1, . . . , vn∈ V, A column vectors V , Lemma 3.4.1 V = Span(v1, . . . , vn) v∈ V,

Ax = v . v1, . . . , vn linearly independent, homogeneous

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4.2. Spanning Vectors and Linear Dependence 73

linear system Ax = O nontrivial solution. Theorem 3.4.6

Ax = v . Proposition 4.1.5 v1, . . . , vn V basis. 

Question 4.1. V Rm subspace v1, . . . , vn∈ V. A v1, . . . , vn column

vectors m× n matrix. v1, . . . , vn V basis, A column space ? A

nullspace ?

4.2. Spanning Vectors and Linear Dependence

spanning vectors linear independence . ,

4.2.1. Spanning Vectors. Rm subspace V , v1, . . . , vn∈ Rm, V spanning vectors v1, . . . , vn∈ V. V spanning vectors,

V .

, 數 , spanning vectors . Corollary

3.4.4 , n < m ,Rm n Rm spanning vectors.

v1, . . . , vn 數 , vn+1 subspace Span(v1, . . . , vn, vn+1) subspace Span(v1, . . . , vn), 大,

subspace 大.

Lemma 4.2.1. v1, . . . , vn, vn+1∈ Rm. Span(v1, . . . , vn)⊆ Span(v1, . . . , vn, vn+1), Span(v1, . . . , vn)̸= Span(v1, . . . , vn, vn+1) vn+1̸∈ Span(v1, . . . , vn).

Proof. Span(v1, . . . , vn, vn+1) Rm subspace, v1, . . . , vn∈ Span(v1, . . . , vn, vn+1), (4.1)

Span(v1, . . . , vn)⊆ Span(v1, . . . , vn, vn+1). (4.2) vn+1∈ Span(v1, . . . , vn), Span(v1, . . . , vn) Rm subspace, v1, . . . , vn, vn+1 Span(v1, . . . , vn)

Span(v1, . . . , vn, vn+1)⊆ Span(v1, . . . , vn). (4.3) (4.2), (4.3)

Span(v1, . . . , vn) = Span(v1, . . . , vn, vn+1).

Span(v1, . . . , vn)̸= Span(v1, . . . , vn, vn+1) vn+1̸∈ Span(v1, . . . , vn).

, vn+1̸∈ Span(v1, . . . , vn), vn+1∈ Span(v1, . . . , vn, vn+1), Span(v1, . . . , vn)̸=

Span(v1, . . . , vn, vn+1). 

V Rn subspace. v1, . . . , vn ∈ V V spanning vectors , Span(v1, . . . , vn)̸= V, V vn+1 vn+1̸∈ Span(v1, . . . , vn). Lemma 4.2.1 , Span(v1, . . . , vn, vn+1) Span(v1, . . . , vn) 大,

, V spanning vectors.

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4.2.2. Linear Dependence. linearly independent ,

linearly dependent linear independent , “線

linearly dependent, ,

v1, . . . , vi−1, vi+1, . . . , vn 線性 . vi 線性

. 步 , vi= r1v1+··· + rn−1vn−1+ rn+1vn+1+··· + rnvn,

rj 數.

r1v1+··· + rn−1vn−1− vi+ rn+1vn+1+··· + rnvn= O.

0 數 c1, . . . , cn c1v1+···+cnvn= O. , 0 數 c1, . . . , cn c1v1+··· + cnvn= O. ci̸= 0,

vi=−c1

ci

v1+··· +−ci−1

ci

vi−1+−ci+1

ci

vi+1+··· +−cn

ci

vn,

vi v1, . . . , vi−1, vi+1, . . . , vn 線性 . , 0

Deﬁnition 4.2.2. v1, . . . , vn∈ Rm, 0 數 c1, . . . , cn c1v1+··· + cnvn= O,

v1, . . . , vn linearly dependent.

v1, . . . , vn∈ Rm Deﬁnition 4.2.2, 0 數 c1, . . . , cn

c1v1+··· + cnvn= O, v1, . . . , vn 線性 .

Deﬁnition 4.1.7 linearly independent . linearly

independent linear dependent . A v1, . . . , vn column vectors m×n matrix. v1, . . . , vn linearly independent homogeneous linear system Ax = O nontrivial solution, v1, . . . , vn linearly dependent

homogeneous linear system Ax = O nontrivial solution.

v1, . . . , vn linearly independent, :

c1v1+··· + cnvn= O, c1, . . . , cn 0. ,

, v1, . . . , vn linearly dependent ( 0

, 大 , .

Example 4.2.3. v1, . . . , vn∈ Rm linearly independent. v1, . . . , vn

linearly independent. , , c1v1+··· +

cn−1vn−1= O c1, . . . , cn−1 0. , 0

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4.2. Spanning Vectors and Linear Dependence 75

c1, . . . , cn

c1v1+··· + cn−1vn−1+ cnvn= c1v1+··· + cn−1vn−1= O.

v1, . . . , vn∈ Rm linearly independent , v1, . . . , vn−1 linearly independent. 大 , v1, . . . , vn−1∈ Rm linearly dependent

, vn∈ Rn , v1, . . . , vn−1, vn linearly dependent.

Example 4.2.3 , v1, . . . , vn∈ Rm linearly independent

, , linearly independent 性 .

. linearly

independent.

Lemma 4.2.4. v1, . . . , vn, vn+1∈ Rm. v1, . . . , vn linearly independent, v1, . . . , vn, vn+1 linearly independent vn+1̸∈ Span(v1, . . . , vn).

Proof. vn+1∈ Span(v1, . . . , vn), v1, . . . , vn, vn+1 線性 , linearly de- pendent. v1, . . . , vn, vn+1 linearly independent, vn+1∈ Span(v1, . . . , vn)

. vn+1̸∈ Span(v1, . . . , vn).

, vn+1̸∈ Span(v1, . . . , vn) v1, . . . , vn, vn+1 linearly independent.

, v1, . . . , vn, vn+1 linearly dependent, 0 數

c1, . . . , cn, cn+1 c1v1+··· + cnvn+ cn+1vn+1= O. cn+1 0, vn+1= −c1

cn+1

v1+··· + −cn

cn+1

vn,

vn+1∈ Span(v1, . . . , vn) . cn+1= 0, c1, . . . , cn 0 數

c1v1+··· + cnvn= c1v1+··· + cnvn+ cn+1vn+1= O,

v1, . . . , vn linearly dependent. v1, . . . , vn linearly independent

, v1, . . . , vn, vn+1 linearly independent. 

V Rn subspace, v1, . . . , vn∈ V linearly independent. v1, . . . , vn

V spanning vectors, V vn+1 vn+1̸∈ Span(v1, . . . , vn).

Lemma 4.2.1 v1, . . . , vn, vn+1 linearly independent.

, 數 , linearly independent . Corollary

3.4.8 , n > m ,Rm n linearly dependent.

.

Lemma 4.2.5. w1, . . . , wk∈ Rm. v1, . . . , vn∈ Span(w1, . . . , wk) n > k, v1, . . . , vn

linearly dependent.

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Proof. v1, . . . , vn∈ Span(w1, . . . , wk) ai j 1≤ i ≤ n, 1 ≤ j ≤ k v1 = a1 1w1 + a1 2w2 + ··· + a1 kwk

...

vn = an 1w1 + an 2w2 + ··· + an kwk. c1, . . . , cn∈ R

c1v1 = c1a1 1w1 + c1a1 2w2 + ··· + c1a1 kwk ...

cnvn = cnan 1w1 + cnan 2w2 + ··· + cnan kwk. , n× k matrix A = [ai j] c1v1+··· + cnvn= b1w1+··· + bkwk

[ c1 ··· cn

]A =[

b1 ··· bk

], AT

 c1

... cn

 =

 b1

... bk

.

AT k×n matrix n > k, Corollary 3.4.7 homogeneous linear system ATx = O nontrivial solution. c1, . . . , cn∈ R 0, AT

 c1

... cn

 =

 0

... 0

, [ c1 ··· cn

]A =[

0 ··· 0 ]

. 0 c1, . . . , ck

c1v1+··· + cnvn= 0w1+··· + 0wk= O.

v1, . . . , vn linearly dependent. 

Question 4.2. v1, . . . , vn linearly independent. w1, . . . , wk∈ Span(v1, . . . , vn) k < n, Span(w1, . . . , wk)̸= Span(v1, . . . , vn).

4.3. Dimension of subspace

Rm subspace basis, .

Rm nonzero subspace basis. subspace

basis 數 , 數 subspace dimension.

dimension 性 .

V Rm subspace. V ={O}, O V , O V

spanning vector. O linearly independent, c̸= 0

cO = O. V basis . V ̸= {O} , v1∈ V v1̸= O.

V1= Span(v1). V1= V , v1 V spanning vector, v1 linearly independent, v1 V basis. V1̸= V, v2∈ V v2̸∈ V1= Span(v1).

V2= Span(v1, v2). Lemma 4.2.1, V1⊆ V2 V1̸= V2, V2 V1 大. Lemma 4.2.4 v1, v2 linearly independent. V2= V ,

v1, v2 V spanning vectors, linearly independent, v1, v2

V basis. V2̸= V, 步 , .

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4.3. Dimension of subspace 77

Theorem 4.3.1. V Rm subspace V̸= {O}, v1, . . . , vn∈ V, n≤ m,

V basis.

Proof. 前 , V2̸= V , . , m 大,

, i vectors v1, . . . , vi linearly independent, 步

, n n≤ m Span(v1, . . . , vn) = V .

k = 1 ). Vk= Span(v1, . . . , vk). Vk= V , v1, . . . , vk V spanning vectors, linearly independent, v1, . . . , vk V basis. Vk ̸= V, Vk ⊆ V vk+1∈ V vk+1̸∈ Vk. v1, . . . , vk, vk+1, v1, . . . , vk linearly independent vK+1̸∈ Vk Lemma 4.2.4 v1, . . . , vk, vk+1 linearly independent.

, 步 n≤ m, Span(v1, . . . , vn) = V .

m , Span(v1, . . . , vm) = V , 前 數學 v1, . . . , vm linearly independent. vm+1 ∈ V vm+1̸∈ Span(v1, . . . , vm), Lemma 4.2.4 v1, . . . , vm, vm+1 linearly independent. v1, . . . , vm+1∈ V v1, . . . , vm+1∈ Rm,

Corollary 3.4.8 v1, . . . , vm+1 m + 1 linearly dependent.

v1, . . . , vm, vm+1 linearly independent . m

Span(v1, . . . , vm) = V , n≤ m v1, . . . , vn V spanning vectors. v1, . . . , vn linearly independent, v1, . . . , vn V

basis. 

Theorem 4.3.1 , v1. . . vk∈ V linearly independent,

v1, . . . , vkV basis. v1, . . . , vk V spanning vectors, 前 , vk+1, . . . , vn ∈ V v1, . . . , vk, vk+1, . . . , vn V

basis.

Theorem 4.3.1 basis 性, basis

[ 1 0

] ,

[ 0 1

] [

1 1

] ,

[ −1 1

]

R2 basis.

v1, . . . , vn Rm basis, Corollary 3.4.4 n < m, v1, . . . , vn

Rm spanning vectors, n≥ m. n > m, Corollary 3.4.8 v1, . . . , vn

linearly independent, n = m. Rm basis m

. .

Theorem 4.3.2. V Rm subspace. v1, . . . , vn w1, . . . , wk V basis, n = k.

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Proof. , n̸= k. 性, n > k. w1, . . . , wk V spanning vectors, v1, . . . , vn∈ V = Span(w1, . . . , wk). n > k Lemma 4.2.5

v1, . . . , vn linearly dependent. v1, . . . , vn V basis .

n̸= k, n = k. 

Theorem 4.3.2 V basis 數 . n

V basis, V basis n .

.

Deﬁnition 4.3.3. V Rm subspace. V basis 數 V

dimension ( ), dim(V ) .

Rm basis m , dim(Rm) = m.

Theorem 4.3.1 , V linearly independent

linearly independent spanning vectors

. , V spanning vectors, spanning vectors

linearly independent . ,

.

Proposition 4.3.4. V Rm subspace v1, . . . , vn∈ V.

(1) v1, . . . , vn V spanning vectors, dim(V )≤ n. , v1, . . . , vn linearly dependent, dim(V ) < n.

(2) v1, . . . , vn linearly independent, dim(V )≥ n. , v1, . . . , vn V spanning vectors, dim(V ) > n.

Proof. dim(V ) = k, w1, . . . , wk V basis.

(1) v1, . . . , vn V spanning vectors, Span(v1, . . . , vn) = V . w1, . . . , wk V basis, w1, . . . , wk∈ Span(v1, . . . , vn) w1, . . . , wk linearly independent.

k > n, Lemma 4.2.5 w1, . . . , wk linearly dependent. ,

dim(V ) = k≤ n. v1, . . . , vn linearly dependent, v1, . . . , vn

Lemma 4.2.1 Span(v1, . . . , vn−1) = Span(v1, . . . , vn−1, vn) = V , v1, . . . , vn−1 V spanning vectors. dim(V )≤ n − 1, dim(V ) < n.

(2) v1, . . . , vn∈ V linearly independent. w1, . . . , wk V basis, Span(w1, . . . , wk) = V , v1, . . . , vn∈ Span(w1, . . . , wk). n > k, Lemma 4.2.5

v1, . . . , vn∈ V linearly dependent . n≤ k = dim(V). v1, . . . , vn V spanning vectors, vn+1∈ V vn+1̸∈ Span(v1, . . . , vn) Lemma 4.2.4

v1, . . . , vn, vn+1 linearly independent. dim(V )≥ n + 1,

dim(V ) > n. 

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4.3. Dimension of subspace 79

v1, . . . , vn V basis, v1, . . . , vn V

spanning vectors linearly independent . dim(V )

n, spanning vectors linearly independent

.

Corollary 4.3.5. V Rm subspace v1, . . . , vn∈ V. . (1) v1, . . . , vn V basis.

(2) dim(V ) = n v1, . . . , vn V spanning vectors.

(3) dim(V ) = n v1, . . . , vn linearly independent.

Proof. (1)⇒ (2): v1, . . . , vn V basis, V basis n , dim(V ) = n. basis spanning vectors, v1, . . . , vn V spanning vectors. (1)⇒ (2).

(2)⇒ (3): v1, . . . , vn V spanning vectors, Proposition 4.3.4 (1) dim(V )≤ n. v1, . . . , vn linearly independent, Proposition 4.3.4 (1)

dim(V ) < n. dim(V ) = n , v1, . . . , vn linearly independent.

(2)⇒ (3).

(3)⇒ (1): v1, . . . , vn linearly independent, Proposition 4.3.4 (2) dim(V )≥ n. v1, . . . , vn V spanning vectors, Proposition 4.3.4 (2) dim(V ) >

n. dim(V ) = n , v1, . . . , vn V spanning vectors. v1, . . . , vn V

basis, (3)⇒ (1). 

Theorem 4.3.1 V Rm subspace, V basis 數

m, dim(V )≤ m, dim(V )≤ dim(Rm).

.

Corollary 4.3.6. V,W Rm subspaces V ⊆ W, dim(V )≤ dim(W).

dim(V ) = dim(W ) V = W .

Proof. dim(V ) = n v1, . . . , vn V basis. v1, . . . , vn∈ V V ⊆ W, v1, . . . , vn∈ W. v1, . . . , vn∈ W linearly independent Proposition 4.3.4 (2) dim(V ) = n≤ dim(W).

V = W , dimension 性 dim(V ) = dim(W ). , dim(V ) = dim(W )

V ̸= W. v1, . . . , vn∈ W linearly independent W spanning vectors, Proposition 4.3.4 (2) dim(V ) = n < dim(W ). dim(V ) = dim(W ) ,

V = W . 

dim(V ) = dim(W )V = W . R2 v̸= O, Span(v)

dimension 1 subspace. w̸= O v , Span(v)̸= Span(w),

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dimension 1 subspace. Corollary 4.3.6 V ⊆ Wdim(V ) = dim(W ) V = W .

4.4. Column Space and Nullspace

column space, row space nullspace basis.

column space row space dimension rank.

subspace basis.

, column space nullspace 數

. column space nullspace 性, :

Deﬁnition 4.4.1. A =

 a 1 a2 ··· a n

 Rm a1, . . . , an column

vectors m× n matrix.

(1) Span(a1, . . . , an) A column space, C(A) A column space.

(2) homogeneous linear system Ax = O A nullspace

N(A) A nullspace. N(A) ={x ∈ Rn| Ax = O}.

Lemma 3.4.1 Theorem 3.4.6 .

Proposition 4.4.2. A m× n matrix b∈ Rm, Ax = b.

(1) Ax = b b∈ C(A).

(2) Ax = b N(A) ={O}.

column space nullspace sub-

spaces basis. Rm subspace V basis, V

spanning vectors. spanning vectors linearly independent

. , lin-

early independent. , linearly

independent, .

Example 4.4.3.

v1=



 0 3 0 1



,v2=



 2

−5 0 0



,v3=



 0 7

−1 0



.

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4.4. Column Space and Nullspace 81

v1, v2, v3 linearly independent, c1= c2= c3= 0 , c1v1+ c2v2+ c3v3= O.

c1v1+ c2v2+ c3v3= c1



 0 3 0 1



 + c2



 2

−5 0 0



 + c3



 0 7

−1 0



 =



2c2

3c1− 5c2+ 7c3

−c3

c1



.

c1v1+ c2v2+ c3v3= O, c1v1+ c2v2+ c3v3 1-st entry 2c2, 3-rd entry

−c3 4-th entry c1 0, c1= c2= c3= 0. c1= c2= c3= 0 , c1v1+ c2v2+ c3v3= O, v1, v2, v3 linearly independent.

Example 4.4.3 , v1, . . . , vn vi entry

0, entry 0, v1, . . . , vn linearly independent. ( Example 4.4.3 v1 4-th entry 1, v2, v3 4-th entry 0; v2 1-st entry 2, v1, v3

1-st entry 0; v3 3-rd entry −1, v1, v2 3-th entry 0, ).

vi 0 entry ai, c1v1+··· + cnvn entry ciai, c1v1+··· + cnvn= O, ciai = 0, ci 0. v1, . . . , vn linearly independent.

A m×n matrix, A nullspace N(A) homogeneous linear system Ax = O

. Ax = O , nullspace

basis .

Ax = O , elementary row operations A

echelon form ( reduced echelon form) A. Ax = O Ax = O ,

A A nullspace. free variable, free variable

variables , free variable , . free

variables xi1, . . . , xik. j = 1, . . . , k, xij= 1, free variable 0 , vj. vj ij-th entry 1, vij′ ij-th entry 0, 前 v1, . . . , vk linearly independent. r1, . . . , rk∈ R, r1v1+··· + rkvk

free variables xi1, . . . , xik 代 xi1 = r1, . . . , xik= rk . 言

r1v1+··· + rkvk , v1, . . . , vk A nullspace spanning vectors. v1, . . . , vk A nullspace basis, A nullspace

dimension free variables 數, A column 數 pivot 數,

.

Proposition 4.4.4. A m× n matrix. row operations A echelon

form A , A pivotr, A nullspace dimension n− r. Ax = O

free variables xi1, . . . , xik. j = 1, . . . , k, xij= 1, free variable 0, vj. v1, . . . , vk A nullspace basis.

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nullspace echelon form , nullspace

dimension , Proposition 4.4.4 “

elementary row operations echelon form , pivot 數

”. Deﬁnition 2.3.1 rank well-deﬁned.

Example 4.4.5. A nullspace,

A =



2 1 1 0 0 0

1 0 0 1 0 0

1 1 1 0 1 2

1 2 2 −2 1 2



.

A 2-nd row −2, −1, −1 1-st, 3-rd 4-th row, 1-st, 2-nd

rows 



1 0 0 1 0 0

0 1 1 −2 0 0

0 1 1 −1 1 2

0 2 2 −3 1 2



.

2-nd row −1,−2 3-rd 4-th row



1 0 0 1 0 0

0 1 1 −2 0 0

0 0 0 1 1 2

0 0 0 1 1 2



.

3-rd row −1 4-th row, echelon form



1 0 0 1 0 0

0 1 1 −2 0 0

0 0 0 1 1 2

0 0 0 0 0 0



.

homogeneous linear system

x1 +x4 = 0

x2 +x3 −2x4 = 0

+x4 +x5 +2x6 = 0

. echelon form x1, x2, x4 pivot variable, x3, x5, x6 free variable.

x6= 1, x5= 0, x3= 0, x4=−2, x2=−4, x1= 2, x6= 0, x5= 1, x3= 0 x4=−1, x2=−2, x1= 1, x6= 0, x5= 0, x3= 1 x4= 0, x2=−1, x1= 0.

v1=







 2

−4 0

−2 0 1







, v2=







 1

−2 0

−1 1 0







, v3=







 0

−1 1 0 0 0







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4.4. Column Space and Nullspace 83

A nullspace basis. , x6, x5, x3 數 r, s,t,

x4=−2r − s, x2=−4r − 2s −t, x1= 2r + s. A nullspace







2r + s

−4r − 2s −t t

−2r − s s r







= r







 2

−4 0

−2 0 1







+ s







 1

−2 0

−1 1 0







+ t







 0

−1 1 0 0 0







= rv1+ sv2+ tv3.

v1, v2, v3 A nullspace spanning vectors, v1, v2, v3 linearly independent, v1, v2, v3 N(A) basis.

Question 4.3. Example 4.4.5 A reduced echelon form.

N(A) basis ?

matrix A column space C(A) basis.

A column space, Ax = v v .

v, A column space. A

constrain equations v. ,

, .

Example 4.4.6. Example 4.4.5 4×6 matrix A. A column vectors

basis. b A column space , Ax = b ,

b =



b1 b2

b3

b4



,

b1, b2, b3, b4 .

2x1 +x2 +x3 = b1

x1 +x4 = b2

x1 +x2 +x3 +x5 +2x6 = b3

x1 +2x2 +2x3 −2x4 +x5 +2x6 = b4

augmented matrix [A| b], Example 4.4.5 elementary row operations



2 1 1 0 0 0 b1

1 0 0 1 0 0 b2

1 1 1 0 1 2 b3

1 2 2 −2 1 2 b4



 ∼



1 0 0 1 0 0 b2

0 1 1 −2 0 0 b1− 2b2

0 1 1 −1 1 2 b3− b2

0 2 2 −3 1 2 b4− b2



 ∼



1 0 0 1 0 0 b2

0 1 1 −2 0 0 b1− 2b2

0 0 0 1 1 2 b3+ b2− b1

0 0 0 1 1 2 b4+ 3b2− 2b1



 ∼



1 0 0 1 0 0 b2

0 1 1 −2 0 0 b1− 2b2

0 0 0 1 1 2 b3+ b2− b1

0 0 0 0 0 0 b4− b3+ 2b2− b1



.

( 2.1 (2)(a)(b) ) , Ax = b

b4− b3+ 2b2− b1= 0. 言 , b1− 2b2+ b3− b4= 0 , b

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

Murphy.Woodward.Stoltzfus.. 17) The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant times the density of

minimal element; vector space linearly independent set

elementary row operations reduced echelon form,.. echelon form Gauss