Assume that we have 4 usual axioms for the inner product (denoted as $(x, y)$ for elements $x, y \in V$ and some scalar $c$) in the real Euclidean space $V$:
- $(x, y) = (y, x)$
- $(x, y + z) = (x, y) + (x, z)$
- $c(x, y) = (cx, y)$
- $(x, x) > 0, x \neq O$
Now, assume that the Axiom 4 is removed and replaced with the statement: $(x, x) = O$ if and only if $x = O$. I need to show, that either $(x, x) > 0$ for all $x \neq O$, or else $(x, x) < 0$ for all $x \neq O$!
Apostol gives the following hint: "Assume $(x, x) > 0$, and $(y, y) < 0$ for some elements $x, y$, s.t. $x, y \neq O$. In the space spanned by $\{x, y\}$, find an element $z \neq O$, s.t. $(z, z) = 0$"
Can someone show how to prove this? I tried quite many combinations, like $(x + y, x - y), $(x - x, y - y), $(x + ax, y + ay), ...$. Nothing worked.
In the end, I somehow managed to prove this partially with Cauchy-schwarz ineqaulity. But since the Cauchy-Schwarz itself depends on positivity axiom, I had to reprove it from the start, and I have to also assume that $(x, y)x - (x, x)y$ has a positive inner product with itself, which I CANNOT assume from the the Apostol hint (can be $(x, x) > 0, (y, y) < 0$, but no guarantee there exists $((x, y)x - (x, x)y, (x, y)x - (x, x)y) > 0$).
How to show this?
NB. Apostol 15.12, calculus 1.
You may stick to $(x + t y, x + t y)$, where $t$ is a real number that you can choose.
You want to choose $t$ such that $(x + ty , x+ ty) = 0$. By the other three axioms, the left hand side is equal to $$(x, x) + 2(x, y)t + (y, y)t^2.$$
This now becomes the following problem:
I assume that you have enough knowledge to solve the above problem. This then gives you the choice of $t$.
The final step is to show that $x + ty$ is not the zero vector. This is easy: assume that $x + ty = 0$, we will arrive at $(x, x) = (-ty, -ty) = t^2(y, y)$, contradicting the assumptions $(x, x) > 0$ and $(y, y) < 0$.