TITLE: Mysterious Proof about Induced Norms (was: Uniqueness of SVD)
QUESTION [7 upvotes]: In order to prove non-uniqueness of singular vectors when a repeated singular value is present, the book (Trefethen), 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.
I am pretty well convinced of partial non-uniqueness of SVD in certain situations. Other proofs are feasible, but I wish to undestand this specific algebraic step of this specific proof.
Thanks.

REPLY [2 votes]: Trefethen's argument is indeed very strange.
As I understood the above answers, everything would be fine if one knew that $Av_1$ and $Av_1$ are orthogonal. 
This can be done by observing that since $\sigma_1=\Vert A\Vert$ and $v_1$ is a unit vector such that $\Vert Av_1\Vert=\sigma_1$, then $v_1$ is an eigenvector for $A^*A$ with eigenvalue $\sigma_1^2$. 
Indeed, we have $\sigma_1^2=\Vert Av_1\Vert^2=\langle A^*Av_1,v_1\rangle$, i.e. $\langle (\sigma_1^2Id-A^*A)v_1,v_1\rangle=0$. Since the sesquilinear form $B(u,v)=\langle (\sigma_1^2Id-A^*A)u,v\rangle$ is positive semi-definite (because $\sigma_1^2=\Vert A\Vert^2=\Vert A^*A\Vert$) it follows, by Cauchy-Schwarz's inequality applied to $B$, that $(\sigma_1^2Id-A^*A)v_1=0$, i.e. $A^*Av_1=\sigma_1^2v_1$. 
The details for Cauchy-Schwarz are as follows: we have $B(v_1,v_1)=0$, hence by Cauchy-Schwarz:
$$\vert B(v_1,u)\vert\leq B(v_1,v_1)^{^1/2}B(u,u)^{1/2}=0;$$
that is, $B(v_1,u)=0$ for all vectors $u$. This means that $(\sigma_1^2Id-A^*A)v_1$ is orthogonal to everybody, and hence is equal to $0$.
Once you know that $A^*Av_1=\sigma_1^2v_1$, simply write $$\langle Av_1,Av_2\rangle=\langle A^*Av_1,v_2\rangle=\sigma_1^2\langle v_1,v_2\rangle=0\, .$$
Having said that, it seems however much simpler to prove that $\Vert Av_2\Vert=\sigma_1$ as follows. By the above reasoning, $v_1$ and $w$ are both eigenvectors of $A^*A$ with eigenvalue $\sigma_1^2$. Hence, so is $v_2$, since $v_2$ is a linear combination of $v_1$ and $w$. So we have $\Vert Av_2\Vert^ 2=\langle A^*Av_2,v_2\rangle=\langle\sigma_1^2v_2,v_2\rangle=\sigma_1^2$, as required.
Edit Actually, I didn't read Ewan's answer carefully, since he's also proving that $Av_1$ and $Av_2$ are orthogonal! His argument is in fact the one that is used in the proof of Cauchy-Schwarz's inequality.