[KOCW 선형대수] 8강. 벡터공간의 차원과 4가지 부벡터공간

제 8강. 벡터공간의 차원과 4가지 부벡터공간

If, given linearly vectors -> linear combination => unique ( 조합은 딱 1가지이다 )

basis is a maximal independent set / minimal spanning set

4 Fundamental subspaces in A

How to find an explicit basis -> a systematic procedure

(1) Column space

ℂ(A)$\mathbb{}$ -> linear combination of column vectors ⊂ℝm${\mathbb{}}^{}$

(2) Null space : N(A)$$, {x|Ax=0$\mathrm{}$} ⊂ℝn${\mathbb{}}^{}$

(3) Row space : ℂ(AT)$\mathbb{}{}^{}$ => linear combination of row vectors ⊂ℝn${\mathbb{}}^{}$

(4) Left Null space : N(AT)${}^{}$ => {y|ATy=0${}^{}\mathrm{}$}  ⊂ℝm${\mathbb{}}^{}$

• N(A)$$ and ℂ(AT)$\mathbb{}{}^{}$ are subspaces of ⊂ℝn${\mathbb{}}^{}$

• N(AT)${}^{}$ and ℂ(A)$\mathbb{}$ are subspaces of  ⊂ℝm${\mathbb{}}^{}$

ex) 

A=U=R=[1 00000]$$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\begin{array}{ccc}\mathrm{}& \mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}& \mathrm{}\end{array}$$

ℂ(A)=[1 0]⊂ℝ2$$\mathbb{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}{\mathbb{}}^{\mathrm{}}$$

ℂ(AT)=⎡⎣⎢⎢1 0 0⎤⎦⎥⎥⊂ℝ3$$\mathbb{}{}^{}\begin{array}{c}\mathrm{}\\ \mathrm{}\\ \mathrm{}\end{array}{\mathbb{}}^{\mathrm{}}$$

ℕ(A)=⎡⎣⎢⎢0 1 0⎤⎦⎥⎥+⎡⎣⎢⎢0 0 1⎤⎦⎥⎥⊂ℝ3$$\mathbb{}\begin{array}{c}\mathrm{}\\ \mathrm{}\\ \mathrm{}\end{array}\begin{array}{c}\mathrm{}\\ \mathrm{}\\ \mathrm{}\end{array}{\mathbb{}}^{\mathrm{}}$$

Free variable 위치는 -를 붙여주고 자시자신은 1

ℕ(AT)=[0 1]⊂ℝ2$$\mathbb{}{}^{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}{\mathbb{}}^{\mathrm{}}$$

Independent vector의 개수로 column space diemension을 알 수 있다.

• Subspace의 특징

차원 같은 것 끼리는 수직이다.

• N(A)$$ 와 ℂ(AT)$\mathbb{}{}^{}$ 는 수직관계

• N(AT)${}^{}$ 와 ℂ(A)$\mathbb{}$ 는 수직 관계

A=⎡⎣⎢⎢1 2 −136−3393274⎤⎦⎥⎥−>U=⎡⎣⎢⎢1 0 0300330230⎤⎦⎥⎥$$\mathbf{}\begin{array}{cccc}\mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\end{array}\mathbf{}\begin{array}{cccc}\mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}& \mathrm{}& \mathrm{}\end{array}$$

• Row space of A

-> non zero rows of U$$ are a basis, and the of pivots( or non-zero rows) is the dimension of row space of A$$ or U$$

dim(A) = dim(U) -> No change of dimension by G.E$$

• Null space of A 

-> by elimination Ax=0$\mathrm{}$ -> Ux=0$\mathrm{}$

nullspaces of A$$ and U$$ are the same. its dimension is n-r. the special solution are a basis

• Column space of A

-> The pivot columns of A(or U) are a basis

Dim(A) and Dim(ℂ(A)$\mathbb{}$ equals the rank r -> row space dim

number of independent columns equals number of independent rows

• Left nullspace of A ( AT${}^{}$ nullspace )

Dim(ℂ(A)$\mathbb{}$) + Dim(N(AT${}^{}$)) = m

( rank(r) + (m-r))

Dim(ℂ(AT)$\mathbb{}{}^{}$) + Dim(N(A)) = n

( rank (r) + (n-r) )

ex) 

A=[1326]−>G.E−>[1020]$$\mathbf{}\begin{array}{cc}\mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}\end{array}\begin{array}{cc}\mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}\end{array}$$

AT=[1236]$$\mathbf{}\mathbf{}\begin{array}{cc}\mathrm{}& \mathrm{}\\ \mathrm{}& \mathrm{}\end{array}$$

ℂ(A)=[1 3]$$\mathbb{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}$$

ℂ(AT)=[1 2]$$\mathbb{}{}^{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}$$

ℕ(A)=[−2 1]$$\mathbb{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}$$

Free variable 위치는 -를 붙여주고 자시자신은 1

ℕ(AT)=[−3 1]⊂ℝ2$$\mathbb{}{}^{}\begin{array}{c}\mathrm{}\\ \mathrm{}\end{array}{\mathbb{}}^{\mathrm{}}$$

Existence of Inverses ( Am×n${}_{}$)

(1) square case

AA−1=A−1A=I${}^{\mathrm{}}{}^{\mathrm{}}$

(2) Am×n${}_{}$ , m<n$$ ( 열의 개수 더 많음 )

오른쪽에 Inverse가 존재해서 -> Right inverse 라고 함

직사각형(가로가 긴 A$$) X 직사각형 ( 세로가 긴 A−1${}^{\mathrm{}}$) = mxm형태의 I$$

오른쪽에 붙지 않으면, 역행력이 존재하지 않음

(3) Am×n${}_{}$ , m>n$$ ( 행의 개수 더 많음 )

직사각형(가로가 긴 A−1${}^{\mathrm{}}$) X 직사각형 ( 세로가 긴 A$$) = mxm형태의 I$$

역행렬이 왼쪽에 와야한다. -> Left inverse

• Ax=b$$일 때 

만약에 left-inverser가 존재한다면

A−1Ax=A−1b$${}^{\mathrm{}}{}^{\mathrm{}}$$

x=A−1b$${}^{\mathrm{}}$$

Unique한 X를 구할 수 있음