The following is a problem from Chapter 1 "Linear Spaces" from Apostol's Calculus (section 1.13)
- Suppose we retain the first three axioms for a real inner product (symmetry, linearity, and homogeneity) but replace the fourth axiom by a new axiom: $(x,x)=0 \iff x=O$. Prove that either $(x,x)>0$ for all $x\neq O$ or else $(x,x)<0$ for all $x\neq O$.
[Hint: Assume $(x,x)>0$ for some $x\neq O$ and $(y,y)<0$ for some $y\neq O$. In the space spanned by $\{x,y\}$ find an element $z\neq O$ with $(z,z)=0$.]
My question is if the solution below is correct.
Here is my attempt at a solution
Assume $(x,x)>0$ and $(y,y)>0$ for some elements $x$ and $y$ neither equal to the zero element, $O$.
Let $z=c_1x+c_2y$ be any element in the space spanned by $x$ and $y$.
Assume $c_2=1$. Then
$$(z,z)=(x,x)c_1^2+2(x,y)c_1+(y,y)=0$$
If we try to solve this for $c_1$ we find
$$\Delta=4[(x,y)^2-(x,x)(y,y)]$$
Now, if we were dealing with the usual axioms for a real inner product we could say that $\Delta\leq 0$ by the Cauchy-Schwarz inequality. But it seems that this inequality does not follow given our new fourth axiom.
Thus we can solve for $c_1$ obtaining
$$c_1=\frac{-(x,y)\pm\sqrt{(x,y)-(x,x)(y,y)}}{(x,x)}$$
In addition, $z$ is not the zero element because if it were then we'd have
$$x+z=(1+c_1)x+y=x$$
$$y=-c_1x$$
and $(y,y)=(-c_1x,-c_1x)=c_1^2(x,x)\geq 0$
which contradicts our assumption that $(y,y)<0$.
Thus, $z$ is not the zero element.
That is, given any nonzero $x$ and $y$ with $(x,x)>0$ and $(y,y)<0$ we have found a $z=c_1x+y$ such that $(z,z)=0$ and $z\neq O$. But this contradicts the new axiom from this problem.