https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/video_galleries/video-lectures/
오늘은 "4 Fundmental Subspaces"에 대해서 알아본다.A : m × n A:m\times n A : m × n
columnspace: C ( A ) ∈ R m C(A) \in \mathbb{R}^m C ( A ) ∈ R m
nullspace: N ( A ) ∈ R n N(A) \in \mathbb{R}^n N ( A ) ∈ R n
rowspace: all combinations of rows of A = all combinations of columns of A T A^T A T C ( A T ) ∈ R n C(A^T) \in \mathbb{R}^n C ( A T ) ∈ R n
nullspace of A T A^T A T A A A N ( A ) T ∈ R m N(A)^T \in \mathbb{R}^m N ( A ) T ∈ R m
이전 강의들에서 columnspace와 nullspace에 대해서는 배웠지만, 이번에는 rowspace와 nullspace of A T A^T A T
우선 각 subspace들의 basis와 dimension을 정리해보자.
C ( A ) C(A) C ( A )
dimension : r r r
bias : pivot columns
C ( A T ) C(A^T) C ( A T )
dimension : r r r
bias : ?
N ( A ) N(A) N ( A )
dimension : n − r n-r n − r
bias : special solutions
N ( A T ) N(A^T) N ( A T )
dimension : m − r m-r m − r m − r m-r m − r
bias : ?
그렇다면 C ( A T ) C(A^T) C ( A T ) N ( A T ) N(A^T) N ( A T )
A = [ 1 2 3 1 1 1 2 1 1 2 3 1 ] → [ 1 2 3 1 0 − 1 − 1 0 0 0 0 0 ] → [ 1 0 1 1 0 1 1 0 0 0 0 0 ] = R A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 2 & 3 & 1 \\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} = R A = ⎣ ⎢ ⎡ 1 1 1 2 1 2 3 2 3 1 1 1 ⎦ ⎥ ⎤ → ⎣ ⎢ ⎡ 1 0 0 2 − 1 0 3 − 1 0 1 0 0 ⎦ ⎥ ⎤ → ⎣ ⎢ ⎡ 1 0 0 0 1 0 1 1 0 1 0 0 ⎦ ⎥ ⎤ = R 그리고 위 예시에서 R의 칼럼 공간이 A의 칼럼 공간과 같다고 생각할 수 있는데, 이는 잘못된 생각이다.
C ( R ) ≠ C ( A ) b u t , “same rowspace" C(R) \ne C(A) \ but, \ \text{``same rowspace"} C ( R ) = C ( A ) b u t , “same rowspace" 그리고 위에서 basis는 [ 1 0 1 1 ] \begin{bmatrix} 1 & 0 & 1 & 1 \\ \end{bmatrix} [ 1 0 1 1 ] [ 0 1 1 0 ] \begin{bmatrix} 0 & 1 & 1 & 0 \\ \end{bmatrix} [ 0 1 1 0 ]
basis of C ( A T ) = first r rows of R \text{basis of $C(A^T)$ = first $r$ rows of $R$} basis of C ( A T ) = first r rows of R 다음으로 N ( A T ) N(A^T) N ( A T )
lecture3 내용)
[ A m × n I m × m ] → [ R m × n E m × m ] [A_{m\times n}I_{m\times m}] \rightarrow [R_{m\times n}E_{m\times m}] [ A m × n I m × m ] → [ R m × n E m × m ] [ 1 2 3 1 1 0 0 1 1 2 1 0 1 0 1 2 3 1 0 0 1 ] → [ 1 2 3 1 1 0 0 0 − 1 − 1 0 − 1 1 0 0 0 0 0 − 1 0 1 ] → [ 1 0 1 1 − 1 2 0 0 1 1 0 1 − 1 0 0 0 0 0 − 1 0 1 ] \begin{bmatrix} \begin{array}{c c c c | c c c} 1 & 2 & 3 & 1 & 1 & 0 & 0 \\ 1 & 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 1 & 0 & 0 & 1 \\ \end{array} \end{bmatrix} \rightarrow \begin{bmatrix} \begin{array}{c c c c | c c c} 1 & 2 & 3 & 1 & 1 & 0 & 0 \\ 0 & -1 & -1 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \end{array} \end{bmatrix} \\ \rightarrow \begin{bmatrix} \begin{array}{c c c c | c c c} 1 & 0 & 1 & 1 & -1 & 2 & 0 \\ 0 & 1 & 1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \end{array} \end{bmatrix} ⎣ ⎢ ⎡ 1 1 1 2 1 2 3 2 3 1 1 1 1 0 0 0 1 0 0 0 1 ⎦ ⎥ ⎤ → ⎣ ⎢ ⎡ 1 0 0 2 − 1 0 3 − 1 0 1 0 0 1 − 1 − 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ → ⎣ ⎢ ⎡ 1 0 0 0 1 0 1 1 0 1 0 0 − 1 1 − 1 2 − 1 0 0 0 1 ⎦ ⎥ ⎤ 따라서 E A = R EA = R E A = R
[ − 1 2 0 1 − 1 0 − 1 0 1 ] ⏟ E [ 1 2 3 1 1 1 2 1 1 2 3 1 ] ⏟ A = [ 1 0 1 1 0 1 1 0 0 0 0 0 ] ⏟ R \underbrace{ \begin{bmatrix} -1 & 2 & 0 \\ 1 & -1 & 0 \\ -1 & 0 & 1 \\ \end{bmatrix} }_{E} \underbrace{ \begin{bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 2 & 3 & 1 \\ \end{bmatrix} }_{A} = \underbrace{ \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} }_{R} E ⎣ ⎢ ⎡ − 1 1 − 1 2 − 1 0 0 0 1 ⎦ ⎥ ⎤ A ⎣ ⎢ ⎡ 1 1 1 2 1 2 3 2 3 1 1 1 ⎦ ⎥ ⎤ = R ⎣ ⎢ ⎡ 1 0 0 0 1 0 1 1 0 1 0 0 ⎦ ⎥ ⎤ 그리고 위에서 N ( A T ) N(A^T) N ( A T ) N ( A T ) N(A^T) N ( A T ) E E E n − r = 3 − 2 = 1 n-r=3-2=1 n − r = 3 − 2 = 1 R R R r o w 3 row_3 r o w 3 N ( A T ) N(A^T) N ( A T ) E E E r o w 3 row_3 r o w 3 [ − 1 0 1 ] [-1 \ 0 \ 1] [ − 1 0 1 ]
