In order to prove non-uniqueness of singular vectors when a repeated singular value is present, the book (Trefethen-Bau, considered the most authotitative book on the subject), argues as follows: Let $\sigma$ be the first singular value of A, and $v\_{1}$ the corresponding singular vector. Let $w$ be another linearly independent vector such that $||Aw||=\sigma$, and construct a third vector $v\_{2}$ belonging to span of $v\_{1}$ and $w$, and orthogonal to $v\_{1}$. All three vectors are unitary, so $w=av\_{1}+bv\_{2}$ with $|a|^2+|b|^2=1$, and $v\_{2}$ is constructed (Gram-Schmidt style) as follows:


$$ {v}\_{2}= \dfrac{{w}-({v}\_{1}^{T} w ){v}\_{1}}{|| {w}\_{1}-({v}\_{1}^{T} {w} ){v}\_{1} ||\_{2}}$$



> 
> Now, Trefethen says, $||A||=\sigma$, so $||Av\_{2}||\le \sigma$ **but
>  this must be an equality** (and so $v\_{2}$ is another singular vector
>  relative to $\sigma$), **since otherwise we would have
>  $||Aw||<\sigma$**, in contrast with the hypothesis.
> 
> 
> 


How that? I cannot see any elementary application of triangle inequality or Schwarz inequality to prove this claim.


Thanks.



About partial uniqueness of SVD
We have $\|Av\_1\|=\sigma$. Assume, $\|Av\_2\| < \sigma$. Now, $w= av\_1+bv\_2$, so \begin{equation}
\|Aw\|^2=|a|^2 \|Av\_1\|^2 + |b|^2 \|Av\_2\|^2 + 2ab\langle Av\_1,Av\_2\rangle < \sigma^2 + 2ab\langle Av\_1,Av\_2\rangle.\quad (\*)
\end{equation}


Let $u\_1$ be the first left-singular vector of $A$, so that $Av\_1 = \sigma u\_1$. Now, $b Av\_2 = Aw - v\_1^Tw \sigma u\_1$, whereby $2a\langle\sigma u\_1, Aw- a\sigma u\_1\rangle = 2a\sigma u\_1^TAw - 2a^2\sigma^2=0$, due to $u\_1^TA=\sigma v\_1$. Thus, the last term in $(\*)$ disappears, which completes the claim.


