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

Scalar Product of Complex Vectors

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Calculating the scalar product of vectors

$\displaystyle x=\left(\begin{array}{cc}1+2\mathrm{i}\\ -2-\mathrm{i}\end{array}\right) \,,\quad y=\left(\begin{array}{cc}2\\ 2\mathrm{i}\end{array}\right) $

from $ \mathbb{C}^2$ we obtain

$\displaystyle \langle x,y \rangle = (1+2\mathrm{i})\cdot 2 + (-2-\mathrm{i})\cd... ...2+4\mathrm{i}+4\mathrm{i} +2\mathrm{i}^2 = 2+8\mathrm{i} -2 = 8\mathrm{i}\,. $

Thus, the scalar product of two complex vectors is not necessarily real-valued.

Conjugation of the second vector plays an important role in forming the associated norm. Without conjugation the first vector would have norm

$\displaystyle \sqrt{(1+2\mathrm{i})^2+(-2-\mathrm{i})^2} = \sqrt{1+4\mathrm{i}-4+4+4\mathrm{i} -1} = \sqrt{8\mathrm{i}} \not\in \mathbb{R}\,, $

and the second one would have norm

$\displaystyle \sqrt{2^2+(2\mathrm{i})^2} = \sqrt{4-4} = 0 \not>0\,. $

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