1 Suppose that the n n matrix A has the decomposition A UD

1. Suppose that the n × n matrix A has the decomposition A = UDV^T , where U and V are n × n matrices ewith the property that (U^T) U = I and (V^T) V = I. The matrix D is diagonal (only nonzero entries are along the diagonal of D) with positive numbers k1, ...kn on the diagonal. Show that A is invertible and find a formula for A^1 .

Solution

We are given that U^TU = I . Therefore, det(U^TU) = 1 [ as det (AB) = det(A)det (B] or, det(U^T) det(U) = 1 so that det(U^T) = det(U) = ± 1. Similarly, det(V^T) = det(V) = ± 1.Since the matrix D is a diagonal matrix with non-zero entries along the leading diagonal, therefore det(D) 0. Then det(A) = det(UDV^T) = det(U)det(D)det(V^T) . Now, we have shown that none of det(U), det(D)and det(V^T) is 0., therefore, det(A) 0. Therefore, A is invertible. Further, we know that for invertible matrices A,B, we have   (AB)^-1 = B^-1 A^-1. Also, since (U^T)U = I and (V^T)V = I. therefore U^{-1 =} U^T and V^-1 = V^T . Also, D being a diagonal matrix with non-zero entries along the leading diagonal in invertible. Then A^-1 = (UDV^T)^-1 = (V^T)^-1( UD)^-1 = (V^T)^-1 D^-1 U^-1 = VD^-1U^T