I try to understand the proof of the theorem in the 1.9 section "Existence and Uniqueness" page 5 https://ocw.mit.edu/courses/mathematics/18-335j-introduction-to-numerical-methods-fall-2004/lecture-notes/lecture3.pdf
but I find it really hard to understand the line (1.29).
Why the following equation holds
$ u_{1}^{*}Av_{1}=\begin{bmatrix}
\sigma_{1} & w^{*} \\
0&B
\end{bmatrix}=S$
i.e why the multiplication of $U_{1}^{*}AV_{1}=S$
where $U_{1}^{*}=[u_{1}|...]$ ,$V_{1}=[v_{1}|...]$ are orthonormal matrices and $u_{1},v_{1}$ are orthonormal vectors
Have a look at https://s3.amazonaws.com/ulaff/LAFF-NLA.pdf starting at page 67. It walks you through this.
(My notation is slightly different from yours, but I think you will get the drift.)
The key is that $ \sigma_1 = \| A \|_2 $. This means it equals $ \sigma_1 = \max_{\| x \|_2 = 1} \| A x \|_2 $. $ v_1 $ is then the unit vector that maximizes this, and $ u_1 = A v_1/\sigma_1 $ (and is of unit length). You then create $ U = \left( \begin{array}{c | c} u_1 & U_2 \end{array} \right) $ where $ U_2^T U_2 = I $ and $ U_2^T u_1 = 0 $. In other words, $ U_2 $ fill out an orthogonal matrix. Same for $ V $: $ V = \left( \begin{array}{c | c} v_1 & V_2 \end{array} \right) $ where $ V_2^T V_2 = I $ and $ V_2^T v_1 = 0 $. Now, $$ \left( \begin{array}{c | c} u_1 & U_2 \end{array} \right)^T A \left( \begin{array}{c | c} v_1 & V_2 \end{array} \right) = \left( \begin{array}{c | c} u_1^T A v_1 & u_1^T A V_2 \\ \hline U_2^T A v_1 & U_2^T A V_2 \end{array} \right) = \left( \begin{array}{c | c} \sigma_1 & w^T \\ \hline z & B \end{array} \right) $$ where $ w^T = u_1^T A V_2 $ and $ z = U_2^T A v_1$. Now, since $ u_1 = A v_1 / \sigma_1 $ (or, equivalently, $ \sigma_1 u_1 = A v_1 $) we notice that $ z = U_2^T ( \sigma_1 u_1 ) = \sigma_1 U_2^T u_1 = 0 $.
And there you are!