odcinaknr

2022-10-05

The problem is to show that, given $\Vert y{\Vert}_{2}={\lambda}^{T}y,\Vert \lambda {\Vert}_{2}\le 1$ and $y\ne 0$, we have $\lambda =\frac{y}{\Vert y{\Vert}_{2}}$

My approach is, $\Vert y{\Vert}_{2}=|{\lambda}^{T}y|\le \Vert y{\Vert}_{2}\Vert \lambda {\Vert}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\Vert \lambda {\Vert}_{2}\ge 1$ which combined with $\Vert \lambda {\Vert}_{2}\le 1$ gives that $\Vert \lambda {\Vert}_{2}=1$. So $\lambda $ and y are not oppositely aligned, since $\Vert y{\Vert}_{2}\ne 0$

Also, $\Vert y{\Vert}_{2}={\lambda}^{T}y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{(\frac{y}{\Vert y{\Vert}_{2}}-\lambda )}^{T}y=0$. But since we showed that $\lambda $ and y are not oppositely aligned, this should mean that the only possibility is $\frac{y}{\Vert y{\Vert}_{2}}-\lambda =0$ which gives the result.

I feel that there should be a much more straightforward way of seeing the result but can't seem to get there at the moment. Can someone help out?

My approach is, $\Vert y{\Vert}_{2}=|{\lambda}^{T}y|\le \Vert y{\Vert}_{2}\Vert \lambda {\Vert}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\Vert \lambda {\Vert}_{2}\ge 1$ which combined with $\Vert \lambda {\Vert}_{2}\le 1$ gives that $\Vert \lambda {\Vert}_{2}=1$. So $\lambda $ and y are not oppositely aligned, since $\Vert y{\Vert}_{2}\ne 0$

Also, $\Vert y{\Vert}_{2}={\lambda}^{T}y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{(\frac{y}{\Vert y{\Vert}_{2}}-\lambda )}^{T}y=0$. But since we showed that $\lambda $ and y are not oppositely aligned, this should mean that the only possibility is $\frac{y}{\Vert y{\Vert}_{2}}-\lambda =0$ which gives the result.

I feel that there should be a much more straightforward way of seeing the result but can't seem to get there at the moment. Can someone help out?

Conor Daniel

Beginner2022-10-06Added 11 answers

You are right, that $||\lambda |{|}_{2}=1$. With this information it is easy to see that

$||\frac{y}{\Vert y{\Vert}_{2}}-\lambda |{|}_{2}^{2}=0.$

To this end use: $||a|{|}_{2}^{2}=(a|a)$, where $(\cdot |\cdot )$ denotes the usual inner product.

$||\frac{y}{\Vert y{\Vert}_{2}}-\lambda |{|}_{2}^{2}=0.$

To this end use: $||a|{|}_{2}^{2}=(a|a)$, where $(\cdot |\cdot )$ denotes the usual inner product.

samuelaplc

Beginner2022-10-07Added 2 answers

A different approach, don't know if it's more straightforward, but maybe a bit more intuitive:

Take $\frac{\lambda}{\Vert \lambda \Vert}$ and complete it to an orthonormal basis $\{\frac{\lambda}{\Vert \lambda \Vert},{e}_{2},...,{e}_{n}\}.$ Then

$y=\u27e8\lambda ,y\u27e9\frac{\lambda}{\Vert \lambda {\Vert}^{2}}+\sum _{i}\u27e8{e}_{i},y\u27e9{e}_{i}=\frac{\Vert y\Vert}{\Vert \lambda {\Vert}^{2}}\lambda +\sum _{i}\u27e8{e}_{i},y\u27e9{e}_{i}.$

Taking the norm of y and using $\Vert \lambda \Vert <1,$ we see $\u27e8{e}_{i},y\u27e9=0$ for $i=2,...,n;$ and also $\Vert \lambda \Vert =1.$ This yields

$y=\Vert y\Vert \lambda .$

Take $\frac{\lambda}{\Vert \lambda \Vert}$ and complete it to an orthonormal basis $\{\frac{\lambda}{\Vert \lambda \Vert},{e}_{2},...,{e}_{n}\}.$ Then

$y=\u27e8\lambda ,y\u27e9\frac{\lambda}{\Vert \lambda {\Vert}^{2}}+\sum _{i}\u27e8{e}_{i},y\u27e9{e}_{i}=\frac{\Vert y\Vert}{\Vert \lambda {\Vert}^{2}}\lambda +\sum _{i}\u27e8{e}_{i},y\u27e9{e}_{i}.$

Taking the norm of y and using $\Vert \lambda \Vert <1,$ we see $\u27e8{e}_{i},y\u27e9=0$ for $i=2,...,n;$ and also $\Vert \lambda \Vert =1.$ This yields

$y=\Vert y\Vert \lambda .$

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