TITLE: A proof of the uniqueness of svd?
QUESTION [2 upvotes]: I understand the geometric intuition, but the proof by induction in Trefethen book confuses me : it seems to me that a 1*1 complex matrix has infinitely many left and right singular vector pairs? 
The actual statement in the book is :
"If A is square and the singular values are distinct, the left and right singular vectors are uniquely determined up to complex signs (i.e., complex scalar factors of absolute value 1)."

REPLY [2 votes]: What is unique in the SVD are the singular values. The unitary matrices are not unique.
This can be seen as follows: Assume $A=U\Sigma V^*$ is a SVD, and $W$ is a diagonal unitary matrix. Since all diagonal matrices commute, we have
$$U\Sigma V^* = U\Sigma WW^*V^* = UW\Sigma W^*V^* = (UW)\Sigma(VW)^*$$
Obviously in the $1\times 1$ case all matrices are diagonal.
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