The Unapologetic Mathematician

Mathematics for the interested outsider

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


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