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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?


enter image description here


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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, $e_1 = (1,0,\dots,0)^T$.

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