Orthogonality - Revision history

Rongzhou: Undid revision 8543 by Rongzhou (talk)

2026-10-01T10:40:02Z

Undid revision 8543 by Rongzhou (talk)

2026-10-01T10:38:46Z

2026-10-01T10:16:27Z

Mathematical definition

← Older revision		Revision as of 10:16, 1 October 2026
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	== Mathematical definition ==		== Mathematical definition ==

	Let ~~[[vector space]]~~ $V$ be an [[inner product space]].		Let $V$ be an [[inner product space]].

	~~Two vectors~~ $u,v\in V$ ~~are '''orthogonal''' if~~ $\langle u, v \rangle=0$.		If for $u,v\in V$

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

			Then $u,v$ are '''orthogonal''' to each other.

	$u\bot v$ denotes this orthogonality.		$u\bot v$ denotes this orthogonality.

2026-10-01T10:00:30Z

Properties

← Older revision		Revision as of 10:00, 1 October 2026
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	$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$		$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$

			== Orthogonal vector list ==

			A list of '''nonzero''' vectors $(e_1,\ldots,e_m) \in V$ is '''orthogonal''' if

			$$
			\langle e_i, e_j \rangle = 0 \Leftarrow 1\le i\neq j \le m
			$$

2026-09-30T21:33:50Z

$\vec 0$ is orthogonal to every vector

← Older revision		Revision as of 21:33, 30 September 2026
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	'''Proof''':		'''Proof''':


			Recall [[Vector_space#Mathematical_definition\|the properties of a vector space by definition]]. By additive identity,

	$$		$$
	~~Recall [[Vector_space#Mathematical_definition\|the properties of a vector space by definition]]. By additive identity, $~~\vec 0 +\vec 0 = \vec 0$		\vec 0 +\vec 0 = \vec 0
			$$

	Therefore,		Therefore,

2026-09-30T21:07:27Z

Triangle inequality

← Older revision		Revision as of 21:07, 30 September 2026
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	$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$		$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$

	~~=== Triangle inequality ===~~

	~~$$\forall u,v\in V,\; \left\\| u+v \right\\| \le \left\\| u \right\\| + \left\\| v \right\\|.$$~~

	~~'''Proof''':~~

	$$
	~~\begin{aligned}~~
	~~\left\\| u+v \right\\|^2~~
	~~&=\langle u+v,u+v\rangle~~
	~~=\langle u,u\rangle+\langle v,v\rangle+\langle u,v\rangle+\langle v,u\rangle\\~~
	~~&=\langle u,u\rangle+\langle v,v\rangle+\langle u,v\rangle+\overline{\langle u,v\rangle}~~
	~~=\left\\| u \right\\|^2+\left\\| v \right\\|^2+2\operatorname{Re}\langle u,v\rangle,\\~~
	~~\text{Note }\operatorname{Re}\langle u,v\rangle~~
	~~&\le\|\langle u,v\rangle\|~~
	~~\stackrel{\text{Cauchy-Schwarz}}{\le}\left\\| u \right\\|\left\\| v \right\\|,\\~~
	~~\left\\| u \right\\|^2+\left\\| v \right\\|^2+2\operatorname{Re}\langle u,v\rangle~~
	~~&\le\left\\| u \right\\|^2+\left\\| v \right\\|^2+2\left\\| u \right\\|\left\\| v \right\\|~~
	~~=(\left\\| u \right\\|+\left\\| v \right\\|)^2,\\~~
	~~\Rightarrow \left\\| u+v \right\\|^2 \leq (\left\\| u \right\\|+\left\\| v \right\\|)^2,\\~~
	~~\because \left\\| . \right\\| \geq 0 \\~~
	~~\therefore \left\\| u+v \right\\|&\le\left\\| u \right\\|+\left\\| v \right\\|~~
	~~\end{aligned}~~
	$$

2026-09-30T20:44:30Z

Moved Cauchy-Schwarz inequality and triangle inequality into their own pages

← Older revision		Revision as of 20:44, 30 September 2026
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	$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$		$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$

	~~=== Cauchy-Schwarz inequality ===~~

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

	~~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 orthogonal decomposition~~

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

	~~where $u_2\bot v$.~~

	~~By the 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 $v\neq 0$:~~

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

	=== Triangle inequality ===		=== Triangle inequality ===

2026-09-30T19:55:29Z

Orthogonal decomposition

← Older revision		Revision as of 19:55, 30 September 2026
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	$u\bot v$ denotes this orthogonality.		$u\bot v$ denotes this orthogonality.

			== Application ==

			=== Orthogonal decomposition ===

			Given vectors $u,v\in V$ and $v\neq 0$, $u$ can be uniquely decomposed into two pieces: one parallel to $v$ and one orthogonal to $v$. This is called an '''orthogonal decomposition'''.

			In mathmetical terms:

			$$u=u_1+u_2,$$

			Where $u_1=a v, a \in \mathbb{F}; u_2\bot v$

			'''Derivation:'''

			$$
			\because u = u_1 +u_2 \\
			\therefore u_2=u-u_1=u-av \\
			\because u_2 \bot v \\
			\therefore \langle u_2, v \rangle = \langle u- av, v \rangle = \langle u, v \rangle - a \left\\| v \right\\|^2 = \vec 0 \\
			\text{Solve for }a,a = \frac{\langle u, v \rangle}{\left\\| v \right\\|^2} \\
			\text{Substitute back into }u_1=av, u_2=u-av, \\
			u=\frac{\langle u, v \rangle}{\\| v \\|^2} v + \left( u-\frac{\langle u, v \rangle}{\\| v \\|^2} v \right)
			$$

	== Properties ==		== Properties ==

2026-09-30T19:47:11Z

Triangle inequality

@@ Line 132: / Line 132: @@
 \Rightarrow  \| u \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\
 \Rightarrow |\langle u,v\rangle|\le\|u\|\|v\|.
 $$

2026-09-30T19:22:11Z

Cauchy-Schwarz inequality

← Older revision		Revision as of 19:22, 30 September 2026
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	$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$		$$\\| u+v \\|^2 = \\| u \\|^2 + \\| v \\|^2.$$

			=== Cauchy-Schwarz inequality ===

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

			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 orthogonal decomposition

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

			where $u_2\bot v$.

			By the 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 $v\neq 0$:

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