In Trefethen and Bau's proof of the SVD (see image below), they start by defining the following:
$$
U_1^* A V_1 = \begin{bmatrix} \sigma_1 & w^* \\ 0 & B \end{bmatrix}
$$
I understand the setup until this point, and I think I understand (broadly) how the proof proceeds: show that $w^* = 0$, then by the induction hypothesis, show that $B$ has the same structure, and so on.
What I don't understand is the block matrix above: why is the bottom-left element known to be zero, but the top right element isn't? Why don't we need to prove that the bottom left element is also 0?
Best Answer
We have constructed the bases so that $Av_1 = \sigma_1 u_1$. It follows that the first column of $U_1^*AV_1$ is $$ (U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (\sigma_1 u_1) = \sigma_1 e_1 $$ where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,\dots,0)^T$.