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If $\vec{X}$ and $\vec{Y}$ are vectors in $\Re^n$ , then $\lVert \vec{X} \cdot \vec{Y} \rVert \leq \lVert \vec{X} \rVert \lVert \vec{Y} \rVert$

ProofEdit

If either $\vec{X}$ or $\vec{Y}$ are the zero vector, the statement holds trivially, so assume that both $\vec{X}$ and $\vec{Y}$ are nonzero. Let $r$ be a scalar and $\vec{Z}$ be defined as $r\vec{X} + \vec{Y}$ . Since, for any nonzero vector $\vec{V}$ , $\vec{V} \cdot \vec{V} = \lVert \vec{X} \rVert^2 \geq 0$ (NOTE: merits own proof)

$\displaystyle 0 \leq (r\vec{X} + \vec{Y}) \cdot (r\vec{X} + \vec{Y})$

$\displaystyle {} \leq r^2(\vec{X} \cdot \vec{X}) + 2r(\vec{X} \cdot \vec{Y}) + (\vec{Y} \cdot \vec{Y})$

$\displaystyle {} \leq ar^2 + 2br + c$

where $a = \vec{X} \cdot \vec{X}$ , $b = \vec{X} \cdot \vec{Y}$ and $c = \vec{Y} \cdot \vec{Y}$ . It can be seen clearly that $p(r) = ar^2 + 2br + c$ is a quadratic polynomial that is non-negative for any $r$ . Consequently, the polynomial has two complex roots, or has a single distinct real root.^[1]

Remember that the roots of $p(r)$ are given by the quadratic formula

$\displaystyle \frac{-2b \pm \sqrt{4b^2 - 4ac}}{2a}$

In particular, the term $4b^2 - 4ac$ must either be negative, yielding two complex roots, or zero, yielding a single real root. Thus

$4b^2 - 4ac \leq 0$

$4b^2 \leq 4ac$

$b^2 \leq ac$

$b \leq \sqrt{ac}$

Substituting the values of $a$ , $b$ and $c$ into the last of these inequalities, it can be seen that

$\sqrt{(\vec{X} \cdot \vec{Y})^2} \leq \sqrt{\vec{X} \cdot \vec{X}}\ \sqrt{\vec{Y} \cdot \vec{Y}}$

Failed to parse (unknown function\lvert): \lvert \vec{X} \cdot \vec{Y} \rvert \leq \lVert \vec{X} \rVert \lVert \vec{Y} \rVert

which is equal to the original statement.

QED

NotesEdit

↑ Intuitively, the graph of $p(r)$ is either 'floating above' the horizontal axis, if it has two complex roots, or tangent if it has one real root. Since $p(r)$ is non-negative for every $r$ , it can't have two real roots because the graph of the function would have to 'pass under' the horizontal axis.

[0] Intuitively, the graph of $p(r)$ is either 'floating above' the horizontal axis, if it has two complex roots, or tangent if it has one real root. Since $p(r)$ is non-negative for every $r$ , it can't have two real roots because the graph of the function would have to 'pass under' the horizontal axis.

[1]

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Proof of the Cauchy-Schwarz inequality

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