Let $u, v \in V$, where $V$ is a vector space with an inner product.

$|u,v| \le ||u|| * ||v||$. This inequality is an equality iff one of $u,v$ is a scalar multiple of the other.

Proof: let $u, v \in V$. If $v = 0$, then both sides of the equation equal 0 and the inequality holds. Thus we can assume that $v \not = 0$. Consider the orthogonal decomposition:

$u = \frac{<u,v>}{||v||^2}v + w$,

where $w$ is orthogonal to $v$ (here $w$ equals the second term on the right side). By the Pythagorean theorem,

(1.) $||u||^2 = ||\frac{<u,v>}{||v||^2}v||^2 + ||w||^2$

(2.) = $\frac{|<u,v>|^2}{||v||^2} + ||w||^2$

(3.) $\ge \frac{|<u,v>|^2}{||v||^2}$

How in the world do you get (2.) from (1.)?


1 Answer 1


Note that $\frac{\langle u, v\rangle}{||v||^2}$ is a scalar. Call it $k$. Then one has $|| k\ v||^2=|k|^2\ ||v||^2$. The rest follows when you put back the expression for $k$.

  • $\begingroup$ Let $k = \frac{<u,v>}{||v||^2}$ Then $||k*v||^2 = |k|^2*||v||^2 = |\frac{<u,v>}{||v||^2}|^2*||v||^2$ How does this equal $\frac{|<u,v>|^2}{||v||^2}$ ? Edit. I see now. $\endgroup$ Commented May 8, 2014 at 23:52

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