壓縮感知
壓縮感知是2006年纔開始興起的研究方向,它主要是藉助自然信號的規律性,從而可大大減少觀測次數。這在很多領域都有很好的應用前景。對於自然信號的規律性,用數學語言可以做很多種描述,比較流行的一種就是自然信號在一組基底表示下是稀疏的。
壓縮感知說的是對於方程組A x = b Ax=b A x = b ,A A A 是m × N m\times N m × N ,這裏m < N m<N m < N 的矩陣,什麼條件下,能保證x x x
是唯一可解的。爲什麼叫壓縮感知呢?舉個簡單的例子。一幅256 × 256 256\times 256 2 5 6 × 2 5 6 的圖像,可以把它拉成一條N = 25 6 2 N=256^2 N = 2 5 6 2 長度的列向量x x x ,通過A A A 的"壓縮",可以變成一個m m m 長的列向量b b b 進行存儲,就節省了存儲空間。另一方面,我們有了b b b ,我們希望將x x x 唯一地"感知"和恢復出來。因爲這是一個欠定的方程組,解不唯一,所以我們必須要加一些約束條件,使得能從b b b 向量,唯一地恢復處x x x ,這就是壓縮感知要乾的事情。
不做冗長的贅述和無用的借鑑參考和複製黏貼,下面我寫一些精華的內容。純自證手打,內容不多,但都是心血。
1
定理:
假設A ∈ R m × N A\in \mathbb{R^{m\times N}} A ∈ R m × N ,那麼爲任x 0 ∈ R N x_0\in \mathbb{R^N} x 0 ∈ R N ,且∥ x 0 ∥ 0 ≤ s \left\|x_{0}\right\|_0\leq s ∥ x 0 ∥ 0 ≤ s ,那麼
arg min x ∈ R n { ∥ x ∥ 0 : A x = A x 0 } = x 0 \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{0}: Ax=A x_{0}\right\}=x_{0} x ∈ R n arg min { ∥ x ∥ 0 : A x = A x 0 } = x 0
當且僅當A A A 中任2 s 2s 2 s 列線性獨立。
證明:
⇐ \Leftarrow ⇐ 假設A A A 中任2 s 2s 2 s 列線性獨立,用反證。
假設存在z 0 ∈ R N z_0 \in \mathbb{R}^{N} z 0 ∈ R N ,∥ z 0 ∥ 0 ≤ s ( z 0 ≠ x 0 ) \left \|z_{0} \right\|_0 \leq s (z_0 \neq x_0) ∥ z 0 ∥ 0 ≤ s ( z 0 = x 0 ) ,A z 0 = A x 0 Az_0=Ax_0 A z 0 = A x 0 → \rightarrow → A ( z 0 − x 0 ) = 0 A(z_0-x_0)=0 A ( z 0 − x 0 ) = 0 。
∥ z 0 ∥ 0 ≤ s \|z_{0}\|_0\leq s ∥ z 0 ∥ 0 ≤ s ,∥ x 0 ∥ 0 ≤ s \|x_{0}\|_0\leq s ∥ x 0 ∥ 0 ≤ s ,則
∥ z 0 − x 0 ∥ 0 ≤ 2 s \|z_0 - x_{0}\|_0\leq 2s ∥ z 0 − x 0 ∥ 0 ≤ 2 s 。
A ( z 0 − x 0 ) = 0 A(z_0-x_0)=0 A ( z 0 − x 0 ) = 0 → \rightarrow → 存在k k k 列線性相關且k ≤ 2 s k\leq 2s k ≤ 2 s ,矛盾。
⇒ \Rightarrow ⇒
假設arg min x ∈ R n { ∥ x ∥ 0 : A x = A x 0 } = x 0 , ∀ x 0 , ∥ x 0 ∥ 0 ≤ s \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{0}: Ax=A x_{0}\right\}=x_{0},\forall x_0,\|x_0\|_0\leq s x ∈ R n arg min { ∥ x ∥ 0 : A x = A x 0 } = x 0 , ∀ x 0 , ∥ x 0 ∥ 0 ≤ s
用反證。
假設A = ( a 1 , a 2 , ⋯ , a n ) ∈ R N A = (a_1,a_2,\cdots,a_n)\in \mathbb{R}^N A = ( a 1 , a 2 , ⋯ , a n ) ∈ R N ,其中{ a 1 , a 2 , ⋯ , a 2 s } \{a_1,a_2,\cdots,a_{2s}\} { a 1 , a 2 , ⋯ , a 2 s } 線性相關。
x 1 a 1 + ⋯ + x 2 s a 2 s = 0 x_1a_1+\cdots+x_{2s}a_{2s}=0 x 1 a 1 + ⋯ + x 2 s a 2 s = 0
那麼存在
x 0 = ( ⋯ , 0 , ⋯ , x 1 , x 2 , ⋯ , x s , ⋯ , 0 , ⋯ ) ∈ R N x_0=(\cdots,0,\cdots,x_1,x_2,\cdots,x_s,\cdots,0,\cdots)\in \mathbb{R}^N x 0 = ( ⋯ , 0 , ⋯ , x 1 , x 2 , ⋯ , x s , ⋯ , 0 , ⋯ ) ∈ R N
x ~ 0 = ( ⋯ , 0 , ⋯ , x s + 1 , x s + 2 , ⋯ , x 2 s , ⋯ , 0 , ⋯ ) ∈ R N \tilde{x}_0=(\cdots,0,\cdots,x_{s+1},x_{s+2},\cdots,x_{2s},\cdots,0,\cdots)\in \mathbb{R}^N x ~ 0 = ( ⋯ , 0 , ⋯ , x s + 1 , x s + 2 , ⋯ , x 2 s , ⋯ , 0 , ⋯ ) ∈ R N
A x ~ 0 = A x 0 A\tilde x_0=Ax_0 A x ~ 0 = A x 0 ,矛盾。
2
定義:
s p a r k ( A ) = n \mathrm {spark}(A) = n s p a r k ( A ) = n :A A A 中任n n n 列線性獨立,存在n + 1 n+1 n + 1 列線性相關。s p a r k ( A ) ≤ r a n k ( A ) \mathrm {spark}(A) \leq \mathrm{rank}(A) s p a r k ( A ) ≤ r a n k ( A ) 。
3
考慮 m i n ∥ x 1 ∥ 1 s . t . A x = A x 0 \begin{aligned}
% \nonumber to remove numbering (before each equation)
\mathrm{min} & \|x_1\|_1 \\
\mathrm{s.t.} & Ax=Ax_0\end{aligned} m i n s . t . ∥ x 1 ∥ 1 A x = A x 0 在什麼樣的條件下,
arg min x ∈ R n { ∥ x ∥ 1 : A x = A x 0 } = x 0 \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A x_{0}\right\}=x_{0} x ∈ R n arg min { ∥ x ∥ 1 : A x = A x 0 } = x 0
對任意的∥ x 0 ∥ 0 ≤ s \left \|x_{0} \right\|_0 \leq s ∥ x 0 ∥ 0 ≤ s 。
4
定義:
稱矩陣A A A 滿足k-階零空間性質,若∀ η ∈ N ( A ) \ 0 , ∀ T ⊆ { 1 , 2 , ⋯ , N } , ∣ T ∣ ≤ k \forall \eta \in \mathcal N(A)\backslash {0},\forall T\subseteq\{1,2,\cdots,N\},|T|\leq k ∀ η ∈ N ( A ) \ 0 , ∀ T ⊆ { 1 , 2 , ⋯ , N } , ∣ T ∣ ≤ k ,有∥ η T ∥ 0 < ∥ η T c ∥ 0 \|\eta_T\|_0<\|\eta_{T^c}\|_0 ∥ η T ∥ 0 < ∥ η T c ∥ 0 。這裏的N ( A ) \mathcal{N}(A) N ( A ) 表示A A A 的零空間。η T \eta_T η T 表示η \eta η 在T T T 這些位置上取值,別的地方補零,長度不變。T c T^c T c 表示T T T 在{ 1 , 2 , ⋯ , N } \{1,2,\cdots,N\} { 1 , 2 , ⋯ , N } 中的補集。
5
定理:
對任x 0 ∈ R N x0\in \mathbb{R}^N x 0 ∈ R N ,∥ x 0 ∥ ≤ k \|x_0\|\leq k ∥ x 0 ∥ ≤ k ,
arg min x ∈ R n { ∥ x ∥ 1 : A x = A x 0 } = x 0 \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A x_{0}\right\}=x_{0} x ∈ R n arg min { ∥ x ∥ 1 : A x = A x 0 } = x 0
當且僅當A滿足k-階零空間性質。
證明:
⇐ \Leftarrow ⇐ 假設A A A 滿足k-階零空間性質,用反證。
假設
arg min x ∈ R n { ∥ x ∥ 1 : A x = A x 0 } = x ♯ ≠ x 0 \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A x_{0}\right\}=x^{\sharp}\neq x_0 x ∈ R n arg min { ∥ x ∥ 1 : A x = A x 0 } = x ♯ = x 0
那麼 η = x ♯ − x 0 ∈ N ( A ) \eta = x^\sharp -x_0 \in \mathcal N(A) η = x ♯ − x 0 ∈ N ( A )
令T = supp ( x 0 ) T=\text{supp} (x_0) T = supp ( x 0 ) ,只要證 ∥ η T ∥ 1 ≥ ∥ η T c ∥ 1 \|\eta_T\|_1\geq \|\eta_{T^c}\|_1 ∥ η T ∥ 1 ≥ ∥ η T c ∥ 1
由假設,有∥ x ♯ ∥ 1 ≤ ∥ x 0 ∥ 1 \|x^\sharp\|_1\leq\|x_0\|_1 ∥ x ♯ ∥ 1 ≤ ∥ x 0 ∥ 1 ,進而
∥ x T ♯ ∥ 1 + ∥ x T c ♯ ∥ 1 = ∥ x T ♯ ∥ 1 ≤ ∥ x 0 ∥ 1 \|x_T^\sharp\|_1+\|x_{T^c}^\sharp\|_1=\|x_T^\sharp\|_1\leq\|x_0\|_1 ∥ x T ♯ ∥ 1 + ∥ x T c ♯ ∥ 1 = ∥ x T ♯ ∥ 1 ≤ ∥ x 0 ∥ 1
可得,
∥ η T c ∥ 1 = ∥ x T c ♯ ∥ 1 ≤ ∥ x 0 ∥ 1 − ∥ x T ♯ ∥ 1 ≤ ∥ ( x 0 − x ♯ ) T ∥ 1 = ∥ η T ∥ 1 \|\eta_{T^c}\|_1=\|x^\sharp_{T^c}\|_1\leq \|x_0\|_1-\|x_T^\sharp\|_1\leq\|(x_0-x^\sharp)_T\|_1=\|\eta_T\|_1 ∥ η T c ∥ 1 = ∥ x T c ♯ ∥ 1 ≤ ∥ x 0 ∥ 1 − ∥ x T ♯ ∥ 1 ≤ ∥ ( x 0 − x ♯ ) T ∥ 1 = ∥ η T ∥ 1
矛盾。
⇒ \Rightarrow ⇒
假設arg min x ∈ R n { ∥ x ∥ 1 : A x = A x 0 } = x 0 , ∀ x 0 , ∥ x 0 ∥ 0 ≤ k \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A x_{0}\right\}=x_{0},\forall x_0,\|x_0\|_0\leq k x ∈ R n arg min { ∥ x ∥ 1 : A x = A x 0 } = x 0 , ∀ x 0 , ∥ x 0 ∥ 0 ≤ k
用反證。
假設∃ η ≠ 0 \exists \eta \neq 0 ∃ η = 0 ,∃ T , ∣ T ∣ ≤ k \exists T,|T|\leq k ∃ T , ∣ T ∣ ≤ k ,∥ η T ∥ 1 ≥ ∥ η T c ∥ 1 \|\eta_T\|_1\geq \|\eta_{T^c}\|_1 ∥ η T ∥ 1 ≥ ∥ η T c ∥ 1 ,那麼
A η = A ( η T + η T c ) = 0 A\eta = A(\eta_T+\eta_{T^c})=0 A η = A ( η T + η T c ) = 0 ⇓ \Downarrow ⇓
A ( η T ) = A ( − η T c ) A(\eta_T) = A(-\eta_{T^c}) A ( η T ) = A ( − η T c )
由假設,因∥ n T ∥ 0 ≤ k \|n_T\|_0\leq k ∥ n T ∥ 0 ≤ k ,有
arg min x ∈ R n { ∥ x ∥ 1 : A x = A η T } = η T \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A \eta_{T}\right\}=\eta_{T} x ∈ R n arg min { ∥ x ∥ 1 : A x = A η T } = η T
故而η T c = η T \eta_{T^c}=\eta_T η T c = η T ,可知η = 0 \eta=0 η = 0 ,矛盾。
6
定義:
矩陣A ∈ R m × N A\in \mathbb{R}^{m\times N} A ∈ R m × N 滿足k-階RIP條件,如果
( 1 − δ k ) ∥ x ∥ 2 2 ≤ ∥ A x ∥ 2 2 ≤ ( 1 + δ k ) ∥ x ∥ 2 2 (1-\delta_k)\|x\|^2_2\leq\|Ax\|^2_2\leq(1+\delta_k)\|x\|_2^2 ( 1 − δ k ) ∥ x ∥ 2 2 ≤ ∥ A x ∥ 2 2 ≤ ( 1 + δ k ) ∥ x ∥ 2 2
對任意x ∈ R N , ∥ x ∥ 0 ≤ k x\in \mathbb{R}^N,\|x\|_0\leq k x ∈ R N , ∥ x ∥ 0 ≤ k ,成立。這裏δ k ∈ ( 0 , 1 ) \delta_k \in (0,1) δ k ∈ ( 0 , 1 ) 。
7
定理:
如果矩陣A ∈ R m × N A\in \mathbb{R}^{m\times N} A ∈ R m × N 滿足2k階RIP條件,δ 2 k < 2 − 1 \delta_{2k}<\sqrt 2-1 δ 2 k < 2 − 1 ,那麼對任意x ∈ R N , ∥ x ∥ 0 ≤ k x\in \mathbb{R}^N,\|x\|_0\leq k x ∈ R N , ∥ x ∥ 0 ≤ k ,有
arg min x ∈ R n { ∥ x ∥ 1 : A x = A x 0 } = x 0 \underset{x \in \mathbb{R}^{n}}{\arg \min }\left\{\|x\|_{1}: Ax=A x_{0}\right\}=x_{0} x ∈ R n arg min { ∥ x ∥ 1 : A x = A x 0 } = x 0
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