Proof about inner product space $\vert \langle x,y\rangle \vert=\Vert x \Vert\Vert y \Vert$

Prove that if $$V$$ is an inner product space, then $$|\langle x,y\rangle| = \Vert x \Vert\Vert y\Vert$$ if and only if one of the vectors $$x$$ or $$y$$ is a multiple of the other.

I am given a hint to let $$a= \frac{\langle x,y\rangle}{\|y\|^2}$$, and let $$z = x - ay$$, then prove $$y$$ and $$z$$ are orthogonal and $$|a|= \frac{\|x\|}{\|y\|}$$. I don't know how they come up with this, what was the original idea?

• Check out the top answer in this question: math.stackexchange.com/questions/1357968/… and look at what happens when you get equality rather than inequality. Mar 4, 2020 at 5:09
• @user754697 thx, that seems helpful to my doubts Mar 4, 2020 at 5:18

I don't know how they come up with this, what was the original idea?

This is orthogonal projection of $$x$$ onto $$y$$. Assume that $$y\neq 0$$.

The goal is to decompose $$x$$ into a sum $$x=w+z$$ of two vectors, $$w$$ being parallel to $$y$$ and $$z$$ perpendicular to $$y$$. Since $$w$$ is parallel to $$y$$, we have $$w=a y$$ for some real number $$a$$. And note that, once $$w$$ (that is $$a$$) is known, we can compute $$z$$ by the formula

$$z=x-ay$$

We only need to compute $$a$$. But, if $$x=w+z$$, then $$\langle x,z\rangle = a\langle y,y\rangle+\langle z,y\rangle = a\left\| y\right\|^2$$ so

$$a = \frac{\langle x,y\rangle}{\left\|y \right\|^2}$$

Finally, $$x$$ is parallel to $$y$$ if and only if $$x=w$$, which is equivalent to $$z=0$$.

Let $$\lambda \in \mathbb{K}$$ and $$x, y \in V$$. Note that

$$0 \leq \langle x + \lambda y, x + \lambda y \rangle \tag{1}$$

is always true and that equality in (1) holds iff equality in

$$\lvert \langle x, y \rangle \rvert^2 = \langle x, x \rangle \langle y, y \rangle$$

holds. This proves the claim.