Eckart-Young-Mirsky theorem

定理：

(The Eckhart-Young Theorem) If k<r=rank(A)k<r = rank(A)k<r=rank(A) and
Ak=∑i=1kσiμiνi⊤A_k=\sum_{i=1}^k \sigma_i\mu_i\nu_i^{\top}Ak=i=1∑kσiμiνi⊤
then
min⁡rank(B)=k∣∣A−B∣∣2=∣∣A−Ak∣∣2=σk+1\min_{rank(B)=k} ||A-B||_2=||A-A_k||_2=\sigma_{k+1}rank(B)=kmin∣∣A−B∣∣2=∣∣A−Ak∣∣2=σk+1

证明：
Since U⊤AkV=diag(σ1,⋯,σk,0,⋯,0)U^{\top}A_kV=diag(\sigma_1,\cdots,\sigma_k,0,\cdots,0)U⊤AkV=diag(σ1,⋯,σk,0,⋯,0) it follows that AkA_kAk is rank k. Moreover, U⊤(A−Ak)V=diag(0,⋯,0,σk+1,⋯,σp)U^{\top}(A-A_k)V=diag(0,\cdots,0,\sigma_{k+1},\cdots,\sigma_p)U⊤(A−Ak)V=diag(0,⋯,0,σk+1,⋯,σp) and so ∣∣A−Ak∣∣2=σk+1||A-A_k||_2=\sigma_{k+1}∣∣A−Ak∣∣2=σk+1.
Now suppose rank(B)=krank(B) = krank(B)=k for some B∈Rm×nB\in \mathbb{R}^{m\times n}B∈Rm×n. It follows that we can find orthonormal vectors x1,...,xn−kx_1 , ... , x_{n-k}x1,...,xn−k so null(B)=span{x1,...,xn−k}null (B) = span\{x_1 , ... , x_{n-k}\}null(B)=span{x1,...,xn−k}· A dimension argument shows that
span{x1,⋯,xn−k}∩span{v1,⋯,vk+1}≠{0}.\text{span}\{x_1,\cdots,x_{n-k}\}\cap\text{span}\{v_1,\cdots,v_{k+1}\}\neq\{0\}.span{x1,⋯,xn−k}∩span{v1,⋯,vk+1}={0}.
Let z be a unit 2-norm vector in this intersection. Since Bz=0Bz = 0Bz=0 and
Az=∑i=1k+1σi(vi⊤z)ui,Az=\sum_{i=1}^{k+1}\sigma_i(v_i^{\top}z)u_i,Az=i=1∑k+1σi(vi⊤z)ui,
we have
∣∣A−B∣∣22≥∣∣(A−B)z∣∣22=∣∣Az∣∣22=∑i=1k+1σi2(vi⊤z)2≥σk+12||A-B||_2^2\geq||(A-B)z||_2^2=||Az||_2^2=\sum_{i=1}^{k+1}\sigma_i^2(v_i^{\top}z)^2\geq\sigma_{k+1}^2∣∣A−B∣∣22≥∣∣(A−B)z∣∣22=∣∣Az∣∣22=i=1∑k+1σi2(vi⊤z)2≥σk+12
completing the proof of the theorem.

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