Cauchy-Schwarz

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for all vectors $v$ and $v$ of an inner product space, we have
$\mid \langle u, v \rangle \mid ^2 \le \langle u, u \rangle \dot \langle v, v \rangle$

In context of Euclidean norm:

\mid x^T y \mid \le \|x\|_2 \|y\|_2

proof

using Pythagorean theorem

special case of $v=0$ . Then $\|u\|\|v\| =0$ ,

⇒ if $u$ and $v$ are linearly dependent., then q.e.d

Assume that $v \neq 0$ . Let $z \coloneqq u - \frac{\langle u, v \rangle}{\langle v, v \rangle} v$

It follows from linearity of inner product that

\langle z,v \rangle = \langle u - \frac{\langle u,v \rangle}{\langle v, v \rangle} v,v \rangle = \langle u,v \rangle - \frac{\langle u,v \rangle}{\langle v,v \rangle}\langle v,v \rangle = 0

Therefore $z$ is orthogonal to $v$ (or $z$ is the projection onto the plane orthogonal to $v$ ). We can then apply Pythagorean theorem for the following:

u = \frac{\langle u,v \rangle}{\langle v,v \rangle} v + z

which gives

\begin{aligned} \|u\|^{2} &= \mid \frac{\langle u,v \rangle}{\langle v,v \rangle} \mid^{2} \|v\|^{2} + \|z\|^2 \\ &=\frac{\mid \langle u,v \rangle \mid^{2}}{(\|v\|^2)^{2}} \|v\|^{2} + \|z\|^2 \\ &= \frac{\mid \langle u, v \rangle\mid^2}{\|v\|^{2} } + \|z\|^2 \ge \frac{\mid \langle u,v \rangle \mid^2}{\|v\|^{2} }\\ \end{aligned}

Follows $\|z\|^{2}=0 \implies z=0$ , which establishes linear dependences between $u$ and $v$ .

q.e.d

Cauchy-Schwarz

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