odcinaknr

2022-10-05

The problem is to show that, given $‖y{‖}_{2}={\lambda }^{T}y,‖\lambda {‖}_{2}\le 1$ and $y\ne 0$, we have $\lambda =\frac{y}{‖y{‖}_{2}}$
My approach is, $‖y{‖}_{2}=|{\lambda }^{T}y|\le ‖y{‖}_{2}‖\lambda {‖}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}‖\lambda {‖}_{2}\ge 1$ which combined with $‖\lambda {‖}_{2}\le 1$ gives that $‖\lambda {‖}_{2}=1$. So $\lambda$ and y are not oppositely aligned, since $‖y{‖}_{2}\ne 0$
Also, $‖y{‖}_{2}={\lambda }^{T}y\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\left(\frac{y}{‖y{‖}_{2}}-\lambda \right)}^{T}y=0$. But since we showed that $\lambda$ and y are not oppositely aligned, this should mean that the only possibility is $\frac{y}{‖y{‖}_{2}}-\lambda =0$ 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

You are right, that $||\lambda |{|}_{2}=1$. With this information it is easy to see that
$||\frac{y}{‖y{‖}_{2}}-\lambda |{|}_{2}^{2}=0.$
To this end use: $||a|{|}_{2}^{2}=\left(a|a\right)$, where $\left(\cdot |\cdot \right)$ denotes the usual inner product.

samuelaplc

A different approach, don't know if it's more straightforward, but maybe a bit more intuitive:
Take $\frac{\lambda }{‖\lambda ‖}$ and complete it to an orthonormal basis $\left\{\frac{\lambda }{‖\lambda ‖},{e}_{2},...,{e}_{n}\right\}.$ Then
$y=⟨\lambda ,y⟩\frac{\lambda }{‖\lambda {‖}^{2}}+\sum _{i}⟨{e}_{i},y⟩{e}_{i}=\frac{‖y‖}{‖\lambda {‖}^{2}}\lambda +\sum _{i}⟨{e}_{i},y⟩{e}_{i}.$
Taking the norm of y and using $‖\lambda ‖<1,$ we see $⟨{e}_{i},y⟩=0$ for $i=2,...,n;$ and also $‖\lambda ‖=1.$ This yields
$y=‖y‖\lambda .$

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