Does the singular value decomposition (SVD) require a complete field? SVD clearly can't work on $\mathbb Q$, since we need square roots. But can it work on $\mathbb E$, the smallest Euclidean field containing $\mathbb Q$?
I ask because, to my surprise, my synthetic geometry proof of polar decomposition required completeness, and would not, as far as I can tell, work for $\mathbb E$. (This is startling, because, it is generally accepted, as Dedekind himself wrote when introducing his cuts, that Euclidean geometry does not require, and would not be changed, by completeness, as long as it includes all constructible numbers.)
Additionally, when proving SVD, Trefethen & Bau make use of compactness to show the existence of a maximum, which of course requires a complete field.
And, from Harvard's fabled 55a, another use of completeness:
Compactness... gives us another way to prove the spectral theorem: we can find an eigenvector for $T$ by seeing where the function $w \mapsto \langle Tw,w \rangle$ achieves its maximum on the unit sphere.
Thus, can the SVD be done in a field that admits square roots but is not complete? And, if yes: What aspect of complete fields finds its way into all three of these proofs? Is SVD somehow different or "stronger" in complete fields?