1
$\begingroup$

I am reading a book that seems to claim the following. I suspect that there may be a misprint, or some assumptions missing. For $A=(x_1,y_1,z_1),B=(x_2,y_2,z_2)\in R^3$, define $$\langle A,B\rangle=-z_1z_2+x_1x_2+y_1y_2.$$ Suppose $C=(c_1,c_2,c_3)$, $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$ are such that $\langle C,C\rangle=-1,c_3\ge 1$,$\langle u,u\rangle=1$, $\langle C,u\rangle=0$, $\langle v,v\rangle=1$ and $\langle C,v\rangle=0$. The book claims that $$|\langle u,v\rangle|\le 1,$$ which does not seem to be correct.

$\endgroup$
0
2
$\begingroup$

The book is correct. Define $$V = \{x \in \mathbb{R}^3\ |\ \langle x,C \rangle = 0\}.$$ Thus $V$ is a two-dimensional subspace of $\mathbb{R}^3$. The key point here is the following exercise :

Exercise : The restriction of $\langle \cdot, \cdot \rangle$ to $V$ is positive-definite (ie it is an actual inner product!).

This exercise implies the fact from your book via an application of the Cauchy-Schwarz inequality to your vectors $u$ and $v$.

Here's a hint for the exercise : your quadratic form on $\mathbb{R}^3$ has signature $1$. What must the signature of the restriction of it to $V$ be?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.