description/proof of the Cauchy-Schwarz inequality for real or complex inner-producted vectors space
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of inner product on real or complex vectors space.
Target Context
- The reader will have a description and a proof of the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the vectors spaces over } F\}\) with any inner product, \(\langle \bullet, \bullet \rangle\)
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Statements:
\(\forall v_1, v_2 \in V (\vert \langle v_1, v_2 \rangle \vert \le \sqrt{\langle v_1, v_1 \rangle} \sqrt{\langle v_2, v_2 \rangle})\).
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2: Natural Language Description
For any field, \(F \in \{\mathbb{R}, \mathbb{C}\}\), and any vectors space, \(V\), over \(F\), with any inner product, \(\langle \bullet, \bullet \rangle\), for each \(v_1, v_2 \in V\), \(\vert \langle v_1, v_2 \rangle \vert \le \sqrt{\langle v_1, v_1 \rangle} \sqrt{\langle v_2, v_2 \rangle}\).
3: Proof
For any \(r \in F\), \(0 \le \langle r v_1 + v_2, r v_1 + v_2 \rangle = r \langle v_1, r v_1 + v_2 \rangle + \langle v_2, r v_1 + v_2 \rangle = r \overline{\langle r v_1 + v_2, v_1 \rangle} + \overline{\langle r v_1 + v_2, v_2 \rangle} = r \overline{r \langle v_1, v_1 \rangle + \langle v_2, v_1 \rangle} + \overline{r \langle v_1, v_2 \rangle + \langle v_2, v_2 \rangle} = \vert r \vert^2 \langle v_1, v_1 \rangle + r \langle v_1, v_2 \rangle + \overline{r} \overline{\langle v_1, v_2 \rangle} + \langle v_2, v_2 \rangle\).
Let us suppose that \(v_1 \neq 0\).
\(\vert r \sqrt{\langle v_1, v_1 \rangle} + \overline{\langle v_1, v_2 \rangle} / \sqrt{\langle v_1, v_1 \rangle} \vert^2 - \vert \langle v_1, v_2 \rangle \vert^2 / \langle v_1, v_1 \rangle + \langle v_2, v_2 \rangle = \vert r \vert^2 \langle v_1, v_1 \rangle + \vert \langle v_1, v_2 \rangle \vert^2 / \langle v_1, v_1 \rangle + r \sqrt{\langle v_1, v_1 \rangle} \langle v_1, v_2 \rangle / \sqrt{\langle v_1, v_1 \rangle} + \overline{r} \sqrt{\langle v_1, v_1 \rangle} \overline{\langle v_1, v_2 \rangle} / \sqrt{\langle v_1, v_1 \rangle} - \vert \langle v_1, v_2 \rangle \vert^2 / \langle v_1, v_1 \rangle + \langle v_2, v_2 \rangle = \vert r \vert^2 \langle v_1, v_1 \rangle + r \langle v_1, v_2 \rangle + \overline{r} \overline{\langle v_1, v_2 \rangle} + \langle v_2, v_2 \rangle\).
So, \(0 \le \langle r v_1 + v_2, r v_1 + v_2 \rangle = \vert r \sqrt{\langle v_1, v_1 \rangle} + \overline{\langle v_1, v_2 \rangle} / \sqrt{\langle v_1, v_1 \rangle} \vert^2 - \vert \langle v_1, v_2 \rangle \vert^2 / \langle v_1, v_1 \rangle + \langle v_2, v_2 \rangle\), and as it holds for each \(r\), \(0 \le - \vert \langle v_1, v_2 \rangle \vert^2 / \langle v_1, v_1 \rangle + \langle v_2, v_2 \rangle\), which implies that \(\vert \langle v_1, v_2 \rangle \vert^2 \le \langle v_1, v_1 \rangle \langle v_2, v_2 \rangle\) and \(\vert \langle v_1, v_2 \rangle \vert \le \sqrt{\langle v_1, v_1 \rangle} \sqrt{\langle v_2, v_2 \rangle}\).
Let us suppose that \(v_1 = 0\). Then, \(\vert \langle v_1, v_2 \rangle \vert = \vert \langle 0, v_2 \rangle \vert = 0 \le 0 = \sqrt{\langle 0, 0 \rangle} \sqrt{\langle v_2, v_2 \rangle} = \sqrt{\langle v_1, v_1 \rangle} \sqrt{\langle v_2, v_2 \rangle}\) (\(\langle 0, v_2 \rangle = \langle r 0, v_2 \rangle = r \langle 0, v_2 \rangle\), which implies that \(\langle 0, v_2 \rangle = 0\)).
So, the inequality holds anyway.
