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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

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    But did you understand how they got to the penultimate inequality? – 2017-01-14
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    @J.M.isn'tamathematician Thank you for your quick reply. Yes, I understand how they get from A1 to the penultimate inequality. I also understand how they arrive at the top left entry for A1. My issue is with the bottom left entry. How does (U1* A V1) create a zero entry. I have probably missed somthing very basic here. – 2017-01-14
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    The $\mathbf U_1$ and $\mathbf V_1$ are partitioned matrices. Have you tried conformally partitioning $\mathbf A$ and performing the required block multiplications? – 2017-01-14
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    @J.M.isn'tamathematician I see. That helps alot, thank you. I think I should be able to take it from here. How would you partition $$ \mathbf A = \sigma_1 \mathbf y_1 \mathbf x_1^* .$$ A worked example would let me know im on the right track or not. – 2017-01-14
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    @J.M.isn'tamathematician Thank you again for your help so far. I am still struggling to get a zero entry even after patritioning A. Could you please explain how you would partition A and how you get a zero entry in the bottom left quater through multiplication of the partitioned matrices. Ta! – 2017-01-15

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$\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 $

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    I realize you are answering your own Question here, but what you wrote is terse to the point of being inscrutable. The difficulty may well be caused by the Question's reliance on a link to set up the notation and context of your problem, but if you want to provide information helpful to future Readers, you should add more explanation. – 2017-01-15
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    Very true. I have added to it a bit. Please expand. – 2017-01-16
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    And have swapped the image in the question with code. I'm using LATEX in a browser for the first time. Works well though. :) What is missing from the current answer? – 2017-01-16