Proof for 6A.3

[CLARKBENHAM]

https://github.com/celiopassos/linear-algebra-done-right-solutions/blob/4d934545705a679cf534d6d41a8f0a2e3a78f49f/Chapter%2006%20-%20Inner%20Product%20Spaces/6A%20-%20Inner%20Products%20and%20Norms.md?plain=1#L59C54-L59C130

\sqrt{(4\langle u, v \rangle^2 - 4\langle u, u \rangle\langle v, v \rangle)}
I believe this line is incorrect as <u,v>^2 <= <u,u>*<v,v> by the Cauchy Swartz inequality. The value under the root is only defined when <u,v>^2 = <u,u>*<v,v> for u=v; giving λ=-1, making the initial case <0,0> = 0 without contradiction. Yes by, definition <u,u> <0 and <v,v> >0 which implies -<u,u>*<v,v> >0.

An equivalent proof to your is here https://latexonline.cc/compile?git=https://github.com/jubnoske08/linear_algebra&target=chapter_6.tex&command=pdflatex&force=true. It first shows that for any u,v s.t. <v,v> >0 and <u,u> < 0 then there is some α not 0 nor 1 such that for w=αu + (1 − α)v, <w, w> = 0. It is defining λ=α-1/α after showing that α cannot be 0 or 1, so λ is well defined. This proof does not compute the zero.

So the real contradiction in your proof is that a degree 2 polynomial must have roots by Bolzano’s theorem (not yet shown), but actually has no roots from the Cauchy Swartz inequality as applied to the quadratic formula.