A proof of the uniqueness of svd?

Question

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)."

score 2 · Accepted Answer · answered Sep 10 '15 at 16:14

2

If $a$ is a complex number ($a\ne0$ to avoid trivialities), the singular value is $|a|$, so we need to find $u$ and $v$ with $|u|=1$ and $|v|=1$ such that $$ a=u|a|\bar{v} $$ Since $a=|a|w$, for a unique $w$, any pair $(u,v)$ such that $u\bar{v}=w$ will do.

There is no uniqueness of $U$ and $V$ in the singular value decomposition of $A=U\Sigma V^*$, in general, see https://en.wikipedia.org/wiki/Singular_value_decomposition

answered Sep 10 '15 at 16:14

egreg

238,574

Sorry I have updated the statement of the theorem. I agree with what you said. – yong Sep 10 '15 at 17:02
@yong And my analysis reflects the statement of the theorem, doesn't it? – egreg Sep 10 '15 at 17:05

score 2 · Answer 2 · answered Sep 10 '15 at 16:39

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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.

answered Sep 10 '15 at 16:39

celtschk

43,384

I have updated the statement of the theorem. Sorry for the previous incorrect one. – yong Sep 10 '15 at 17:03

A proof of the uniqueness of svd?

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