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Mathematics-Online course: Linear Algebra - Basic Structures - Scalar Product and Norm

Cauchy-Schwarz Inequality

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The scalar product can be estimated by the associated norm:

$\displaystyle \vert\langle u,v \rangle\vert \le \vert u\vert\vert v\vert,\quad \vert w\vert = \sqrt{\langle w,w\rangle}\, . $

We have equality if and only if $ u\parallel v$.

(Authors: Burkhardt/Höllig/Hörner)
Equality holds for $ v=\lambda u$ as asserted.

Myltiplying $ u$ and $ v$, resp., by a scalar does not change the inequality. Thus we can assume that

$\displaystyle \langle u,u\rangle = \langle v,v\rangle = 1\, . $

If for a complex scalar product (the reasoning also includes the real case)

$\displaystyle \langle u,v\rangle = r\exp(\mathrm{i}\phi)\, ,\,r>0\,, $

then, with $ \lambda = \exp(\mathrm{i}\phi)$, it follows for non-parallel vectors that
0 $\displaystyle <$ $\displaystyle \langle u-\lambda v,u-\lambda v\rangle$  
  $\displaystyle =$ $\displaystyle 1 + \lambda\bar\lambda - \lambda \overline{r\exp(\mathrm{i}\phi)} - \overline{\lambda} r\exp(\mathrm{i}\phi)$  
  $\displaystyle =$ $\displaystyle 1 + 1 - r - r\, ,$  

hence $ r=\vert\langle u,v\rangle\vert<1$ and the assertion is proved.

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005