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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?

JohnD
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Édgar Sánchez Gordón
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2 Answers2

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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 $\mathbf{x}=\lambda\mathbf{y}$.

JohnD
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    How do you find =⟨,⟩/⟨,⟩? (i.e. which inequality argument did you follow to get that) Is it the same as x=||x||/||y||y? – james black Mar 08 '20 at 06:35
  • It's all right but to be precise, the equation that you put is in fact $x= \frac{\langle y,x\rangle}{\langle y,y\rangle} y$, the order of the braket is inverse – Euler Nov 05 '21 at 09:46
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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 $x_1y_2=x_2y_1$.

robjohn
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