35: Cauchy-Schwarz Inequality for Real or Complex Inner-Producted Vectors Space

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description/proof of the Cauchy-Schwarz inequality for real or complex inner-producted vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of inner product on real or complex vectors space.

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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.

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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.

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Main Body

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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, $⟨\bullet ,\bullet ⟩$
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Statements:
$\mathrm{\forall }{v}_1,v_2\in V(|⟨v_1,v_2⟩|\le \sqrt{⟨v_1,v_1⟩}\sqrt{⟨v_2,v_2⟩})$.
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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, $⟨\bullet ,\bullet ⟩$, for each $v_1,v_2\in V$, $|⟨v_1,v_2⟩|\le \sqrt{⟨v_1,v_1⟩}\sqrt{⟨v_2,v_2⟩}$.

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3: Proof

For any $r\in F$, $0\le ⟨r{v}_1+v_2,r{v}_1+v_2⟩=r⟨v_1,r{v}_1+v_2⟩+⟨v_2,r{v}_1+v_2⟩=r\overline{⟨r{v}_1+v_2,v_1⟩}+\overline{⟨r{v}_1+v_2,v_2⟩}=r\overline{r⟨v_1,v_1⟩+⟨v_2,v_1⟩}+\overline{r⟨v_1,v_2⟩+⟨v_2,v_2⟩}=|r{|}^2⟨v_1,v_1⟩+r⟨v_1,v_2⟩+\bar{r}\overline{⟨v_1,v_2⟩}+⟨v_2,v_2⟩$.

Let us suppose that $v_1\ne 0$.

$|r\sqrt{⟨v_1,v_1⟩}+\overline{⟨v_1,v_2⟩}/\sqrt{⟨v_1,v_1⟩}{|}^2-|⟨v_1,v_2⟩{|}^2/⟨v_1,v_1⟩+⟨v_2,v_2⟩=|r{|}^2⟨v_1,v_1⟩+|⟨v_1,v_2⟩{|}^2/⟨v_1,v_1⟩+r\sqrt{⟨v_1,v_1⟩}⟨v_1,v_2⟩/\sqrt{⟨v_1,v_1⟩}+\bar{r}\sqrt{⟨v_1,v_1⟩}\overline{⟨v_1,v_2⟩}/\sqrt{⟨v_1,v_1⟩}-|⟨v_1,v_2⟩{|}^2/⟨v_1,v_1⟩+⟨v_2,v_2⟩=|r{|}^2⟨v_1,v_1⟩+r⟨v_1,v_2⟩+\bar{r}\overline{⟨v_1,v_2⟩}+⟨v_2,v_2⟩$.

So, $0\le ⟨r{v}_1+v_2,r{v}_1+v_2⟩=|r\sqrt{⟨v_1,v_1⟩}+\overline{⟨v_1,v_2⟩}/\sqrt{⟨v_1,v_1⟩}{|}^2-|⟨v_1,v_2⟩{|}^2/⟨v_1,v_1⟩+⟨v_2,v_2⟩$, and as it holds for each $r$, $0\le -|⟨v_1,v_2⟩{|}^2/⟨v_1,v_1⟩+⟨v_2,v_2⟩$, which implies that $|⟨v_1,v_2⟩{|}^2\le ⟨v_1,v_1⟩⟨v_2,v_2⟩$ and $|⟨v_1,v_2⟩|\le \sqrt{⟨v_1,v_1⟩}\sqrt{⟨v_2,v_2⟩}$.

Let us suppose that $v_1=0$. Then, $|⟨v_1,v_2⟩|=|⟨0,v_2⟩|=0\le 0=\sqrt{⟨0,0⟩}\sqrt{⟨v_2,v_2⟩}=\sqrt{⟨v_1,v_1⟩}\sqrt{⟨v_2,v_2⟩}$ ($⟨0,v_2⟩=⟨r0,v_2⟩=r⟨0,v_2⟩$, which implies that $⟨0,v_2⟩=0$).

So, the inequality holds anyway.

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References

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