Cauchy-Schwarz inequality: Difference between revisions
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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 an [[inner product space]] over $\mathbb{F}$. | |||
$$\forall u,v\in V,\; |\langle u,v\rangle|\le\|u\|\|v\|.$$ | |||
Furthermore, | |||
$$|\langle u,v\rangle|=\|u\|\|v\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$ | |||
<ref>{{#cite:Q1921}}</ref> | |||
== 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 $v\neq 0$: | |||
$$ | |||
\frac{|\langle u, v \rangle|^2}{\| v \|^2} + \| u_2 \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ | |||
\Rightarrow \| u \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ | |||
\Rightarrow |\langle u,v\rangle|\le\|u\|\|v\|. | |||
$$ | |||
Latest revision as of 08:20, 1 October 2026
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 an inner product space over $\mathbb{F}$.
$$\forall u,v\in V,\; |\langle u,v\rangle|\le\|u\|\|v\|.$$
Furthermore,
$$|\langle u,v\rangle|=\|u\|\|v\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$ [1]
Proof
If $v=0$,
$$\begin{gathered} |\langle u,v\rangle| = 0, \|u\|\|v\| = 0 \\ \therefore |\langle u,v\rangle| = \|u\|\|v\| \end{gathered}$$
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 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 $v\neq 0$:
$$\begin{gathered} \frac{|\langle u, v \rangle|^2}{\| v \|^2} + \| u_2 \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ \Rightarrow \| u \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ \Rightarrow |\langle u,v\rangle|\le\|u\|\|v\|. \end{gathered}$$