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?