Cauchy-Schwarz inequality: Difference between revisions
imported>Aleksander Stos m (→Statement of the Cauchy-Schwarz inequality: more precise?) |
imported>Aleksander Stos |
||
Line 7: | Line 7: | ||
::<math>|\langle x_1,x_2 \rangle|\leq \|x_1\|\|x_2\|,\quad (1)</math> | ::<math>|\langle x_1,x_2 \rangle|\leq \|x_1\|\|x_2\|,\quad (1)</math> | ||
where <math>\|y\|=\langle y,y \rangle^{1/2} </math> for all <math>y \in V</math>. Furthermore, the equality in (1) holds if and only if the vectors <math>x_1</math> and <math>x_2</math> are linearly dependent. | where <math>\|y\|=\langle y,y \rangle^{1/2} </math> for all <math>y \in V</math>. Furthermore, the equality in (1) holds if and only if the vectors <math>x_1</math> and <math>x_2</math> are [[linear independence|linearly dependent]] (in this case proportional one to the other). | ||
==Proof of the inequality== | ==Proof of the inequality== |
Revision as of 15:27, 3 November 2007
In mathematics, the Cauchy-Schwarz inequality is a fundamental and ubiquitously used inequality that relates the absolute value of the inner product of two elements of an inner product space with the magnitude of the two said vectors. It is named in the honor of the French mathematician Augustin-Louis Cauchy[1] and German mathematician Hermann Amandus Schwarz[2].
Statement of the Cauchy-Schwarz inequality
Let V be a complex inner product space with inner product . Then for any two elements it holds that
where for all . Furthermore, the equality in (1) holds if and only if the vectors and are linearly dependent (in this case proportional one to the other).
Proof of the inequality
A standard yet clever idea for a proof of the Cauchy-Schwarz inequality is to exploit the fact that the inner product induces a quadratic form on V. Let be some fixed pair of vectors in V and let be the argument of the complex number . Now, consider the expression for any real number t and notice that, by the properties of a complex inner product, f is a quadratic function of t. Moreover, f is non-negative definite: for all t. Expanding the expression for f gives the following:
Since f is a non-negative definite quadratic function of t, if follows that the discriminant of f is non-positive definite. That is,
from which (1) follows immediately by the substitution and .
References
- ↑ Biography at MacTutor History of Mathematics, John J. O'Connor and Edmund F. Robertson, School of Mathematics and Statistics, University of St Andrews, Scotland.
- ↑ Biography at MacTutor History of Mathematics, John J. O'Connor and Edmund F. Robertson, School of Mathematics and Statistics, University of St Andrews, Scotland.