Suppose $x,y \in V$ are vectors such that $\lVert x\rVert=\lVert y\rVert=1$ and $\langle x,y\rangle=1$. Show $x=y$

Let V be a vector space over $$\mathbb{R}$$. Let $$(x,y) \mapsto \langle x,y\rangle$$ be an inner product on $$V$$ with induced norm $$\lVert x\rVert=\sqrt{\langle x,y\rangle}$$. Suppose that $$x$$ and $$y$$ are two vectors in $$V$$ such that $$\lVert x\rVert=\lVert y\rVert=1$$ and $$\langle x,y\rangle=1$$. Show that $$x=y$$.

This looks really obvious, but I tried using $$(\lVert x\rVert-\lVert y\rVert)^2$$ and Cauchy inequality to approach it, and still didn't get it. I am running out of ideas now. Any help is appreciated.

• Please use proper delimiters (\langle, \rangle, \lVert, \rVert) for the symbols. Jul 16 '20 at 19:57

Notice that $$\|x-y\|^2 = \langle x-y,x-y \rangle = \langle x,x \rangle - 2 \langle x,y \rangle + \langle y,y \rangle = 1-2+1 = 0,$$ so $$\|x-y\| = 0$$ and hence $$x=y$$.

Here's a bit more geometric an argument. Consider the projection of $$x$$ onto $$y$$. The usual formula gives $$\text{proj}_y x = \frac{\langle x,y\rangle}{\|y\|^2} y = \frac11 y = y.$$ Now the Pythagorean Theorem says that $$x=y$$. If we have $$x=y+z$$ with $$z$$ orthogonal to $$y$$, then $$\|x\|^2 = \|y\|^2+\|z\|^2\ge \|y\|^2$$; thus, $$\|x\| = \|y\|$$ if and only if $$z=0$$.

• Beautiful proof! Jul 16 '20 at 20:06
• Thanks, @BrianBritosSimmari. I usually try to emphasize things geometric in linear algebra. But this exercise was a new one for me. Jul 16 '20 at 21:52

You have

$$\Vert x-y \Vert^2 = \Vert x\Vert^2 -2\langle x,y \rangle + \Vert y \Vert^2=0$$

Hence $$x=y$$.

By Cauchy-Schwarz inequality, \begin{align*} 1 = |\langle x, y \rangle| \leq \|x\|\|y\| = 1 \times 1 = 1. \end{align*} Hence the equality holds. Check the equality condition for Cauchy-Schwarz inequality: we know that the equality holds if and only if $$y = \lambda x$$ for some $$\lambda$$. Taking norms on both sides yields $$\lambda = \pm 1$$. But $$\langle x, y \rangle = 1$$ rules out the case $$\lambda = -1$$.