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In Page 185 here it says

... $M^2 y=\sigma^2y$. Since $M$ is symmetric, it follows that $y$ is an eigenvector of $M$ with eigenvalue $\pm \sigma$.

It seems to contradict the example here. What am I missing?

Milten
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Dan Feldman
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  • I don’t know a lot about the context or theory here, but I see from your link that $y$ is defined as the solution to an optimisation problem. Maybe it follows from that that $y$ must be an eigenvector, and then your paragraph gives the eigenvalue? (In that case it wouldn’t matter that $M$ is symmetric, so I might be wrong) – Milten Jun 19 '23 at 19:36
  • We indeed look for the optimal $y$, but all we know is that $y$ is a unit vector. The proof uses mainly Lagrange multipliers. BTW, the context is proving convergence of self-concordant function in Newton's method. It is mentioned e.g. in the book "Convex Optimization" of Boyd et al. However, even in the books of the great inventor (Nesterov) he gives it without proof ("We accept this statement without a proof since it needs some special facts from the theory of three-linear symmetric forms"). – Dan Feldman Jun 19 '23 at 20:02
  • An observation, given the original context. We can write $y=u+v$, where $u$ and $v$ are orthogonal, $M_y u=\sigma u$ and $M_y v = -\sigma v$. It follows that $x=u-v$, and we want to show that either $u$ or $v$ is zero. I can't see a way to continue, but maybe we can consider the optimization problem from this point again... – Milten Jun 20 '23 at 17:51
  • Update: one of the author answered me that it is indeed seems like a bug. Any other source for the proof will also be appreciated. – Dan Feldman Jun 21 '23 at 18:29
  • @Milten, I could not see how your idea work but I suspect that instead of multiplying by $M_y$ to get $M^2y$ we should multiply by $M'_y$ whose eigenvectors are the same as $M_y$ but all the eigenvalues are positive, i.e., $M'_y=\sqrt(M_y^2)$, to remember the sign of the eigenvalues. We might also need to flip sign for some entries of the eigenvector, but I am not sure. – Dan Feldman Jun 25 '23 at 13:17

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