The Unapologetic Mathematician

The Cauchy-Schwarz Inequality

Today I want to present a deceptively simple fact about spaces equipped with inner products. The Cauchy-Schwarz inequality states that

$\displaystyle\langle v,w\rangle^2\leq\langle v,v\rangle\langle w,w\rangle$

for any vectors $v,w\in V$. The proof uses a neat little trick. We take a scalar $t$ and construct the vector $v+tw$. Now the positive-definiteness, bilinearity, and symmetry of the inner product tells us that

$\displaystyle0\leq\langle v+tw,v+tw\rangle=\langle v,v\rangle+2\langle v,w\rangle t+t^2\langle w,w\rangle$

This is a quadratic function of the real variable $t$. It can have at most one zero, if there is some value $t_0$ such that $v+t_0w$ is the zero vector, but it definitely can’t have two zeroes. That is, it’s either a perfect square or an irreducible quadratic. Thus we consider the discriminant and conclude

$\displaystyle\left(2\langle v,w\rangle\right)^2-4\langle w,w\rangle\langle v,v\rangle\leq0$

which is easily seen to be equivalent to the Cauchy-Schwarz inequality above. As a side effect, we see that we only get an equality (rather than an inequality) when $v$ and $w$ are linearly dependent.

April 16, 2009 Posted by | Algebra, Linear Algebra | 5 Comments