# 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?

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 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 $$y = \lVert y \rVert \lambda.$$

Building up on Fred's answer, here's the full solution:

1. 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 $$\Vert \lambda\Vert_2=1$$.

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 $$\Vert \lambda\Vert_2=1$$.

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 $$\lambda= \frac{y}{\|y\|_2}$$.