Cauchy-Schwarz inequality - Revision history

Rongzhou: /* Mathematical statement */

2026-10-01T08:20:20Z

Mathematical statement

← Older revision		Revision as of 08:20, 1 October 2026
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	== Mathematical statement ==		== Mathematical statement ==

	Let $V$ be a [[inner product space]] over $\mathbb{F}$.		Let $V$ be an [[inner product space]] over $\mathbb{F}$.

	$$\forall u,v\in V,\; \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.$$		$$\forall u,v\in V,\; \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.$$

2026-10-01T08:18:51Z

Mathematical statement

← Older revision		Revision as of 08:18, 1 October 2026
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	== Mathematical statement ==		== Mathematical statement ==

	Let $V$ be a [[~~vector~~ space]] over $\mathbb{F}$.		Let $V$ be a [[inner product space]] over $\mathbb{F}$.

	$$\forall u,v\in V,\; \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.$$		$$\forall u,v\in V,\; \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.$$

2026-10-01T08:01:54Z

Mathematical statement

← Older revision		Revision as of 08:01, 1 October 2026
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	$$\|\langle u,v\rangle\|=\\|u\\|\\|v\\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$		$$\|\langle u,v\rangle\|=\\|u\\|\\|v\\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$
			<ref>{{#cite:Q1921}}</ref>

	== Proof ==		== Proof ==

2026-09-30T20:55:31Z

← Older revision	Revision as of 20:55, 30 September 2026
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		Cauchy-Schwarz inequality states that the [[inner product]] of two vectors is always less or equal to the product of the norms of the two vector in any [[inner product space]].

		== Mathematical statement ==

		Let $V$ be a [[vector space]] over $\mathbb{F}$.

		$$\forall u,v\in V,\; \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.$$

		Furthermore,

		$$\|\langle u,v\rangle\|=\\|u\\|\\|v\\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$

		== Proof ==

		If $v=0$,

		$$
		\|\langle u,v\rangle\| = 0, \\|u\\|\\|v\\| = 0 \\
		\therefore \|\langle u,v\rangle\| = \\|u\\|\\|v\\|
		$$

		If $v\neq 0$,

		Consider the [[Orthogonality#Orthogonal_decomposition\|orthogonal decomposition]]

		$$u = \frac{\langle u, v \rangle}{\\| v \\|^2} v + u_2$$

		where $u_2\bot v$.

		By the [[Orthogonality#Generalised_Pythagorean_theorem\|generalised Pythagorean theorem]],

		$$\\| u \\|^2 = \left\\| \frac{\langle u, v \rangle}{\\| v \\|^2}v \right\\|^2 + \\| u_2 \\|^2
		= \frac{\|\langle u, v \rangle\|^2}{\\| v \\|^2} + \\| u_2 \\|^2 $$

		Since $v\neq 0$:

		$$
		\frac{\|\langle u, v \rangle\|^2}{\\| v \\|^2} + \\| u_2 \\|^2 \ge \frac{\|\langle u, v \rangle\|^2}{\\| v \\|^2} \\
		\Rightarrow \\| u \\|^2 \ge \frac{\|\langle u, v \rangle\|^2}{\\| v \\|^2} \\
		\Rightarrow \|\langle u,v\rangle\|\le\\|u\\|\\|v\\|.
		$$

2026-09-30T20:48:22Z

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