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I tried to prove a special case of Cauchy-Schwarz:

$$x,y \text{ are linearly depending vectors} \Leftrightarrow |\langle x,y\rangle|=||x|| \cdot ||y||$$

$\Rightarrow$ is simple: \begin{eqnarray*} x= \lambda y \Leftrightarrow (x_1,\ldots,x_n)&=&(\lambda x_1,\ldots,\lambda x_n),\\ |\langle x,y\rangle|=||x|| \cdot ||y|| &\Leftrightarrow& |\langle x,\lambda x\rangle|=||x|| \cdot ||\lambda x|| \\ & \Leftrightarrow& |x_1 \lambda x_1+\ldots +x_n \lambda x_n|=\sqrt{x_1^2+\ldots+x_n^2} \cdot \sqrt{(\lambda x_1)^2+\ldots+(\lambda x_n)^2}\\ &\Leftrightarrow& |\lambda x_1^2+\ldots+\lambda x_n^2|= \sqrt{x_1^2+\ldots+x_n^2} \cdot \sqrt{\lambda^2(x_1^2+\ldots+x_n^2}\\ &\Leftrightarrow& |\lambda| \cdot |x_1^2+\ldots+x_n^2|=|\lambda| \cdot \sqrt{x_1^2+\ldots+x_n^2}^2\\ &\Leftrightarrow& |\lambda| \cdot |x_1^2+\ldots+x_n^2|=|\lambda| \cdot |x_1^2+\ldots+x_n^2| \end{eqnarray*}

Now I have absolutly no idea how to prove $\Leftarrow$. Anyone got in idea?

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  • $\begingroup$ @CarstenSchultz thanks, I made an edit $\endgroup$
    – ulead86
    Jan 31, 2014 at 16:48

2 Answers 2

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Yours is correct, but you don't need to use coordinates for this proof: just the definition of the norm $||x||$ as $\sqrt{\langle x,x \rangle}$ and the bilinear properties of the dot product.

For instance, if $y = \lambda x$, then, on the one hand,

$$ |\langle x, y \rangle| = |\langle x, \lambda x \rangle| =|\lambda \cdot \langle x, x \rangle| = |\lambda |\cdot || x||^2 \ . $$

On the other hand,

$$ ||x||\cdot ||y|| =||x||\cdot ||\lambda x|| = |\lambda | \cdot ||x||^2 $$

too.

The other implication doesn't need coordinates either: if $y \neq \lambda x$, then

$$ y - \lambda x \neq 0 \qquad \Longleftrightarrow \qquad ||y - \lambda x|| \neq 0 \ . $$

Which is the same as

$$ \ 0 <||y - \lambda x||^2 = \langle y-\lambda x , y - \lambda x\rangle = || y ||^2 -2\lambda \langle x, y\rangle + \lambda^2 ||x||^2 \ . $$

Now, if you consider this last expression as a polynomial in $\lambda$, you just have proved that

$$ P(\lambda ) = || y ||^2 -2\lambda \langle x, y\rangle + \lambda^2 ||x||^2 > 0 \qquad \text{for all} \quad \lambda \ . $$

Hence, this polynomial has no roots. So its discriminant must be negative:

$$ 4\langle x, y\rangle^2 - 4 ||y||^2||x||^2 < 0 \ . $$

In particular,

$$ |\langle x, y \rangle | \neq ||x||\cdot||y|| \ . $$

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  • $\begingroup$ Very nice answer, thank you very much. $\endgroup$
    – ulead86
    Jan 31, 2014 at 17:19
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    $\begingroup$ You're welcome. The advantage of this proof "coordinate-free", or not using the particular formula for the standard dot product in $\mathbb{R}^n$, is that it's available for no matter what Euclidian vector space. $\endgroup$ Jan 31, 2014 at 17:28
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Suppose $\lambda y-x\neq0$ for all $\lambda\in\mathbb R$. so $$0<|\lambda y-x|^2=\sum_{i=1}^n(\lambda y_i-x_i)^2=\lambda^2\sum_{i=1}^ny_i^2-2\lambda\sum_{i=1}^n x_iy_i+\sum_{i=1}^n x_i^2$$ Therefore the right side is a quadratic equation in $\lambda$ with no real solution, and its discriminant must be negative. Thus $$4\left(\sum_{i=1}^n x_iy_i\right)^2-4\left(\sum_{i=1}^nx_i^2\right)\left(\sum_{i=1}^ny_i^2\right)<0.$$

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