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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?

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    $\begingroup$ 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. $\endgroup$
    – user754697
    Mar 4, 2020 at 5:09
  • $\begingroup$ @user754697 thx, that seems helpful to my doubts $\endgroup$
    – xyz12354
    Mar 4, 2020 at 5:18

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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 $z=0$.

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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.

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