Given $\Vert y \Vert_2=\lambda^Ty, \Vert \lambda\Vert_2\leq1$ and $y\neq0$, show that $\lambda=\frac{y}{\Vert y \Vert_2}$ The problem is to show that, given $\Vert y \Vert_2=\lambda^Ty, \Vert \lambda\Vert_2\leq1$ and $y\neq0$, we have $\lambda=\frac{y}{\Vert y \Vert_2}$.
My approach is, $\Vert y \Vert_2=\vert \lambda^Ty \vert\leq \Vert y \Vert_2\Vert \lambda \Vert_2 \implies \Vert \lambda\Vert_2\geq1$ which combined with $\Vert \lambda\Vert_2\leq1$ gives that $\Vert \lambda\Vert_2=1$. So $\lambda$ and $y$ are not oppositely aligned, since $\Vert y \Vert_2\neq0$.
Also, $\Vert y \Vert_2=\lambda^Ty \implies \left(\frac{y}{\Vert y \Vert_2}-\lambda\right)^Ty=0$. 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?
 A: 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.
A: A different approach, don't know if it's more straightforward, but maybe a bit more intuitive:
Take $\frac{\lambda}{\lVert \lambda \rVert}$ and complete it to an orthonormal basis $\{ \frac{\lambda}{\lVert \lambda \rVert}, e_2, ..., e_n \}.$ Then
$$y = \langle \lambda, y \rangle \frac{\lambda}{\lVert \lambda \rVert^2} +  \sum_i \langle e_i, y \rangle e_i = \frac{\lVert y \rVert}{\lVert \lambda \rVert^2} \lambda + \sum_i \langle e_i, y \rangle e_i. $$
Taking the norm of $y$ and using $\lVert \lambda \rVert < 1,$ we see $\langle e_i, y \rangle = 0$ for $i=2,...,n;$ and also $\lVert \lambda \rVert = 1.$ This yields
$$ y = \lVert y \rVert \lambda. $$
A: Building up on Fred's answer, here's the full solution: 


*

*Establish that $\Vert \lambda\Vert_2=1$
Cauchy-Schwartz $\implies \Vert y \Vert_2=\vert \lambda^Ty \vert\leq \Vert y \Vert_2\Vert \lambda \Vert_2 \implies \Vert \lambda\Vert_2\geq1$ which combined with $\Vert \lambda\Vert_2\leq1$ gives that $\Vert \lambda\Vert_2=1$.


*Rewrite $\Vert y \Vert_2 = \lambda^Ty~$ using $\Vert \lambda\Vert_2=1$
$\Vert y \Vert_2 = \lambda^Ty \implies 2\lambda^Ty=\frac{y^Ty}{\Vert y \Vert_2^2}+\Vert \lambda\Vert_2^2 \implies \Vert\frac{y}{\Vert y \Vert_2}-\lambda\Vert_2^2=0 \implies \lambda=\frac{y}{\Vert y \Vert_2}$
where the second equality holds because $\frac{y^Ty}{\Vert y \Vert_2^2}=1$ and $\Vert \lambda\Vert_2=1$.
A: We have
$$\|y\|_2 = |\lambda^Ty| \le \|\lambda\|_2\|y\|_2 \le \|y\|_2$$
so in fact $|\lambda^Ty| = \|y\|_2$. The equality clause in Cauchy-Schwarz inequality gives that vectors $y$ and $\lambda$ are proportional, i.e. $\lambda = \alpha y$ for some scalar $\alpha$.
Now we get
$$\|y\|_2 = \lambda^Ty = \alpha y^Ty = \alpha\|y\|_2^2 \implies \alpha = \frac1{\|y\|_2}$$
or $\lambda= \frac{y}{\|y\|_2}$.
