正交补集(Orthogonal Complements) |
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正交补集
接续博客:计算矩阵的秩、行空间、列空间、零空间、左零空间 因为高维空间本人无法进行可视化,这里仅用上述博客中提到的低维度空间的例子来可视化
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A=\begin{bmatrix}2 & -1 & -3\\ -4 & 2 & 6\end{bmatrix}\quad A^T=\begin{bmatrix}2 & -4\\ -1 & 2\\ -3 & 6 \end{bmatrix}
A=[2−4−12−36]AT=⎣⎡2−1−3−426⎦⎤ 零空间(Nullspace)正交于 行空间(Row space) 零空间是行空间的正交补集、行空间是零空间的正交补集 C ( A T ) = ( N ( A ) ) ⊥ N ( A ) = ( C ( A T ) ) ⊥ C(A^T)=(N(A))^{\perp}\\ N(A)=(C(A^T))^{\perp} C(AT)=(N(A))⊥N(A)=(C(AT))⊥
左零空间是列空间的正交补集、列空间是左零空间的正交补集 C ( A ) = ( N ( A T ) ) ⊥ N ( A T ) = ( C ( A ) ) ⊥ C(A)=(N(A^T))^{\perp}\\ N(A^T)=(C(A))^{\perp} C(A)=(N(AT))⊥N(AT)=(C(A))⊥
a ⃗ ⋅ v ⃗ = 0 f o r a n y v ⃗ ∈ V , a ⃗ ∈ V ⊥ b ⃗ ⋅ v ⃗ = 0 f o r a n y v ⃗ ∈ V , b ⃗ ∈ V ⊥ ( a ⃗ + b ⃗ ) ⋅ v ⃗ = a ⃗ ⋅ v ⃗ + b ⃗ ⋅ v ⃗ = 0 + 0 = 0 c a ⃗ ⋅ v ⃗ = c ( a ⃗ ⋅ v ⃗ ) = c ⋅ 0 = 0 \vec{a}\cdot \vec{v}=0\quad for\ any\ \vec{v}\in V,\vec{a}\in V^{\perp}\\ \vec{b}\cdot \vec{v}=0\quad for\ any\ \vec{v}\in V,\vec{b}\in V^{\perp}\\ (\vec{a}+\vec{b})\cdot \vec{v}=\vec{a}\cdot \vec{v}+\vec{b}\cdot \vec{v}=0+0=0\\ c\vec{a}\cdot \vec{v}=c(\vec{a}\cdot \vec{v})=c\cdot 0=0 a ⋅v =0for any v ∈V,a ∈V⊥b ⋅v =0for any v ∈V,b ∈V⊥(a +b )⋅v =a ⋅v +b ⋅v =0+0=0ca ⋅v =c(a ⋅v )=c⋅0=0 N ( A ) i s o r t h o g o n a l c o m p l e m e n t o f C ( A T ) N ( A ) = ( C ( A T ) ) ⊥ N(A)\ is\ orthogonal\ complement\ of\ C(A^T)\\ N(A)=(C(A^T))^{\perp} N(A) is orthogonal complement of C(AT)N(A)=(C(AT))⊥ [ ⋯ r 1 ⃗ T ⋯ ⋯ r 2 ⃗ T ⋯ ⋮ ⋯ r m ⃗ T ⋯ ] m × n ⋅ x ⃗ = [ 0 0 ⋮ 0 ] \begin{bmatrix} \cdots\quad\vec{r_1}^T\quad\cdots\\ \cdots\quad\vec{r_2}^T\quad\cdots\\ \vdots\\ \cdots\quad\vec{r_m}^T\quad\cdots\\ \end{bmatrix}_{m×n}\cdot \vec{x}= \begin{bmatrix} 0\\ 0\\ \vdots\\ 0 \end{bmatrix} ⎣⎢⎢⎢⎡⋯r1 T⋯⋯r2 T⋯⋮⋯rm T⋯⎦⎥⎥⎥⎤m×n⋅x =⎣⎢⎢⎢⎡00⋮0⎦⎥⎥⎥⎤ r 1 ⃗ T ⋅ x ⃗ = 0 r 2 ⃗ T ⋅ x ⃗ = 0 ⋮ r m ⃗ T ⋅ x ⃗ = 0 \vec{r_1}^T\cdot\vec{x}=0\\ \vec{r_2}^T\cdot\vec{x}=0\\ \vdots\\ \vec{r_m}^T\cdot\vec{x}=0\\ r1 T⋅x =0r2 T⋅x =0⋮rm T⋅x =0 N ( A ) = { x ⃗ ∈ R n ∣ A x ⃗ = 0 } w ⃗ ∈ C ( A T ) v ⃗ ∈ N ( A ) w ⃗ 是 行 空 间 内 某 个 向 量 , 它 是 由 行 空 间 的 基 向 量 线 性 组 合 而 成 v ⃗ 是 零 空 间 内 某 个 向 量 w ⃗ = c 1 r 1 ⃗ + c 2 r 2 ⃗ + ⋯ + c m r m ⃗ v ⃗ ⋅ w ⃗ = c 1 ( r 1 ⃗ ⋅ v ⃗ ) + c 2 ( r 2 ⃗ ⋅ v ⃗ ) + ⋯ + c m ( r m ⃗ ⋅ v ⃗ ) = c 1 ⋅ 0 + c 2 ⋅ 0 + ⋯ + c m ⋅ 0 = 0 E v e r y m e m b e r o f n u l l s p a c e o f A i s o r t h o g o n a l t o e v e r y m e m b e r o f t h e r o w s p a c e o f A N(A)=\{\vec{x}\in \mathbb{R}^n | A\vec{x}=0\}\\ \vec{w}\in C(A^T)\\ \vec{v}\in N(A)\\ ~\\ \vec{w}是行空间内某个向量,它是由行空间的基向量线性组合而成\\ \vec{v}是零空间内某个向量 \\ \vec{w}=c_1\vec{r_1}+c_2\vec{r_2}+\cdots+c_m\vec{r_m}\\ ~\\ \vec{v}\cdot\vec{w}=c_1(\vec{r_1}\cdot\vec{v})+c_2(\vec{r_2}\cdot\vec{v})+\cdots+c_m(\vec{r_m}\cdot\vec{v})=c_1\cdot0+c_2\cdot0+\cdots+c_m\cdot0=0\\ ~\\ Every\ member\ of\ nullspace\ of\ A\ is\ orthogonal\ to\ every\ member\ of\ the\ row\ space\ of\ A N(A)={x ∈Rn∣Ax =0}w ∈C(AT)v ∈N(A) w 是行空间内某个向量,它是由行空间的基向量线性组合而成v 是零空间内某个向量w =c1r1 +c2r2 +⋯+cmrm v ⋅w =c1(r1 ⋅v )+c2(r2 ⋅v )+⋯+cm(rm ⋅v )=c1⋅0+c2⋅0+⋯+cm⋅0=0 Every member of nullspace of A is orthogonal to every member of the row space of A w ⃗ ∈ C ( A T ) u ⃗ ∈ ( C ( A T ) ) ⊥ u ⃗ ⋅ w ⃗ = 0 u ⃗ ⋅ w ⃗ = u ⃗ ⋅ r 1 ⃗ + u ⃗ ⋅ r 2 ⃗ + ⋯ + u ⃗ ⋅ r m ⃗ = 0 A u ⃗ = 0 由 于 u ⃗ ⋅ w ⃗ = 0 , v ⃗ ⋅ w ⃗ = 0 , u ⃗ ∈ ( C ( A T ) ) ⊥ , v ⃗ ∈ N ( A ) 所 以 u ⃗ 、 v ⃗ 在 同 一 子 空 间 内 , 故 u ⃗ ∈ N ( A ) ( C ( A T ) ) ⊥ ⊆ N ( A ) N ( A ) = ( C ( A T ) ) ⊥ \vec{w}\in C(A^T)\\ \vec{u}\in (C(A^T))^{\perp}\\ ~\\ \vec{u}\cdot \vec{w}=0\\ \vec{u}\cdot \vec{w}=\vec{u}\cdot\vec{r_1}+\vec{u}\cdot\vec{r_2}+\cdots+\vec{u}\cdot\vec{r_m}=0\\ A\vec{u}=0\\ ~\\ 由于 \vec{u}\cdot \vec{w}=0,\vec{v}\cdot\vec{w}=0,\vec{u}\in (C(A^T))^{\perp},\vec{v}\in N(A)\\ 所以\vec{u}、\vec{v}在同一子空间内,故 \vec{u}\in N(A)\\ (C(A^T))^{\perp}\subseteq N(A)\\ N(A)=(C(A^T))^{\perp} w ∈C(AT)u ∈(C(AT))⊥ u ⋅w =0u ⋅w =u ⋅r1 +u ⋅r2 +⋯+u ⋅rm =0Au =0 由于u ⋅w =0,v ⋅w =0,u ∈(C(AT))⊥,v ∈N(A)所以u 、v 在同一子空间内,故u ∈N(A)(C(AT))⊥⊆N(A)N(A)=(C(AT))⊥ 假 设 A = B T N ( B T ) = ( C ( B T ) T ) ⊥ N ( B T ) = C ( B ) ⊥ 换 个 字 母 N ( A T ) = C ( A ) ⊥ 假设A=B^T\\ N(B^T)=(C(B^T)^T)^{\perp}\\ N(B^T)=C(B)^{\perp}\\ 换个字母\\ N(A^T)=C(A)^{\perp} 假设A=BTN(BT)=(C(BT)T)⊥N(BT)=C(B)⊥换个字母N(AT)=C(A)⊥ |
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