When proving that every complex matrix has a singular value decomposition why does it follow that (U1* A V1) is an upper triangular block matrix?

Question

Let $ \sigma_1 $ be the 2-norm of $\mathbf A$; there exist unit length vectors $\mathbf x_1 \in \mathbb{C}^m,\space \mathbf x_1^*\mathbf x_1 = 1 $ and $ \mathbf y_1 \in \mathbb{C}^n,\space \mathbf y_1^*\mathbf y_1 = 1$ , such that $ \mathbf A \mathbf x_1 = \sigma_1 \mathbf y_1. $ Define the unitary matrices $ \mathbf V_1, \mathbf U_1 $ so that their first column is $ \mathbf x_1, \mathbf y_1 $, respectively:$ \mathbf V_1 = [\mathbf x_1\space \hat{\mathbf V}_1],\space \mathbf U_1 = [\mathbf y_1\space \hat{\mathbf U}_1] $

Why does it follow that $\mathbf U_1^* \mathbf A \mathbf V_1 = \pmatrix{\sigma_1 & \mathbf w^* \\ 0 & \mathbf B}$ is an upper triangular block matrix? Please may you explain where the zero entry comes from.

Many thanks,

Tri

Answer 1

$\mathbf U_1 = [ \mathbf y_1 \space \hat{\mathbf U}_1], \space \mathbf U_1^* \mathbf U_1 = 1 \implies \hat{\mathbf U}_1 \sigma_1 \mathbf y_1 = 0 $