SVD and Null Space: Exploring the Relationship between A and ΣV^T

(a) Let x be in N(A). Then we have Ax = 0. Multiplying both sides by VΣ^-1 gives UΣV^T VΣ^-1x = UΣx = 0. Since Σ is diagonal with positive entries on the diagonal, this implies that Σx = 0. Therefore, V^T VΣ^-1x = V^T UΣx = 0. But V is orthogonal, so V^T V = I, and hence V^T VΣ^-1x = Σ^-1x. Therefore, Σ^-1x = 0, which implies that x is in N(ΣV^T).

Conversely, let x be in N(ΣV^T). Then Σx = 0, which implies that UΣx = 0. Multiplying both sides by V^T gives V^T UΣx = 0, which implies that Ax = 0. Therefore, x is in N(A).

(b) The rank of ΣV^T is also equal to r. This is because the rank of Σ is equal to r, and multiplying Σ by an invertible matrix (V^T in this case) does not change its rank. Therefore, the rank of ΣV^T is also equal to r.