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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.

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  • $\begingroup$ Any idea why the rest of the proof is valid? I.e. up to the part of "We conclude that, if the singular vector $v_1$ is not unique, then the corresponding singular value $\sigma_1$ is not simple" - I don't understand why the fact that $||Av_2|| = \sigma_1$ leads to anything... also what does "simple" even means? $\endgroup$ Mar 24, 2020 at 11:02

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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.

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  • $\begingroup$ Very well exposed, thanks. I think it only applies to real case, however. $\endgroup$
    – MadHatter
    Apr 13, 2014 at 17:41
  • $\begingroup$ I did not look at it, but I don't immediately see what might prevent the argument for going through for the complex case... $\endgroup$
    – Suvrit
    Apr 26, 2014 at 20:07
  • $\begingroup$ the last math expression is missing a transpose on the $v_1$ $\endgroup$ Mar 24, 2020 at 10:45

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