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In Spivak Calculus you are asked to prove that in Schwarz inequality, equality holds only when $y_1 = y_2 = 0$ or when there is a number $\lambda$ such that $x_1 = \lambda y_1$ and $x_2 = \lambda y_2$.

I can go from $x_1y_1 + x_2y_2 = \sqrt{x_1^2 + x_2^2} + \sqrt{y_1^2 + y_2^2}$ to $x_1y_2 = x_2y_1$ but then I got stuck, or is it that the implication is the other way around?

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    $\begingroup$ That's correct. Conclude that if neither of $y_1, y_2$ are 0, then $\frac {x_1}{y_1} = \frac {x_2}{y_2}$, and set it to be your favorite parameter. Then deal with the case $y_1 y_2 = 0$. $\endgroup$ – Calvin Lin Jan 15 '13 at 4:28
  • $\begingroup$ Thank you Calvin, with your $\lambda = \frac{x_1}{y_1}$ trick I was able to complete the proof. $\endgroup$ – Édgar Sánchez Gordón Jan 15 '13 at 5:23
  • $\begingroup$ $x_1,\,x_2,\,y_1,\,y_2,\lambda \in \mathbb{R}$? $\endgroup$ – scjorge Jun 7 '16 at 3:41
  • $\begingroup$ @scjorge Yep, all real numbers. $\endgroup$ – Édgar Sánchez Gordón Dec 6 '17 at 16:58
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$|\langle \mathbf{x},\mathbf{y}\rangle|=\|\mathbf{x}\|\,\|\mathbf{y}\| \iff \mathbf{x}$ and $\mathbf{y}$ are linearly dependent, i.e. if $\mathbf{x}=\lambda\mathbf{y}$ (or, trivially, if either $\mathbf{x}$ or $\mathbf{y}$ is zero; handle this first to get it out of the way).

($\implies$): Follow the normal Schwarz inequality argument, but equality will force $\mathbf{x}={\langle \mathbf{x},\mathbf{y}\rangle\over \langle\mathbf{y},\mathbf{y}\rangle}\mathbf{y}$

($\Longleftarrow$): Substitute $\mathbf{x}=\lambda\mathbf{y}$.

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Since $$ \left(x_1^2+x_2^2\right)\left(y_1^2+y_2^2\right)-(x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2 $$ we get equality if and only if $x_1y_2=x_2y_1$.

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