Low-Rank Matrix Recovery from Row-and-Column Affine Measurements

7. Appendix 7.1. Proofs for Noiseless GRC Case Proof of Lemma 1 Proof. First, rank(X2A ) = rank(X1A ) = r and rank(AX2) = rank(A X1) = r. Since span(X1A ), span(X2A ) are subspaces of span(X1), span(X2) respectively, and dim(span(X2)) ≤ r we get span(X2) = span(X2A ) = span(X1A ) = span(X1), and we define U ∈ On1×r a basis for this subspace. For X1, X2 there are Y1, Y2 ∈ Rr×n2 such that X1 = UY1, X2 = UY2. Therefore AUY1 = AUY2. Since rank(AUY1) = r and U ∈ On1×r we get rank(AU) = r, hence the matrix UA T AU is invertible, which gives Y1 = Y2, and therefore X1 = UY1 = UY2 = X2.