A norm on a vector space comes from an inner product if and only if it satisfies the parallelogram law. Given such a norm, one can reconstruct the inner product via the formula:
$2\langle u,v\rangle = |u + v|^2 – |u|^2 – |v|^2$
(there are minor variations on this)
It's straightforward to prove, using the parallelogram law, that this satisfies:
- $\langle u,u\rangle \ge 0$ for all $u$, and $\langle u,u\rangle = 0$ iff $u = 0$
- $\langle tu,tu\rangle = t^2 \langle u,u\rangle$;
- $\langle u,v\rangle = \langle v,u\rangle$;
- $\langle u,v+w\rangle = 2\langle u/2,v\rangle + 2\langle u/2,w\rangle$;
From 4 with the special case $w=0$ one quickly deduces that $\langle u,v+w\rangle = \langle u,v\rangle + \langle u,w\rangle$.
The usual method of proving $\langle u,tv\rangle = t\langle u,v\rangle$ is to use 4 with induction to prove that $\langle u,nv\rangle = n\langle u,v\rangle$, then deduce $\langle u,tv\rangle = t\langle u,v\rangle$ for $t$ rational, and finally appeal to continuity to extend to the reals.
Is there any way to avoid this last bit? In particular, is there a more geometric view of why $\langle u,tv\rangle = t\langle u,v\rangle$ for all real $t$? Pictures would be great!
If not, is there a different way to express the condition that a norm comes from an inner product that does make all the conditions obviously geometrical?
Clarification added later: My reason for asking this is pedagogical. I teach a course which introduces, in quick succession, metric spaces, normed vector spaces, and inner product spaces. The properties of metrics and norms are very easy to motivate from intuitive properties of distances and lengths. I'd like to do the same for inner products in terms of angles. Thus by "geometric" I mean "geometric intuition" rather than geometry as geometers understand it. Since the inner product is introduced after the norm, I argue that using the cosine law one can define the notion of "angle" between two vectors using any norm. However, unless the norm is "special", that notion of angle doesn't behave how we would expect it to do so. In particular, in order for angles to add properly, one needs the norm to satisfy the parallelogram law. Here "add" means that (modulo a pi or two), the angle from $u$ to $v$ plus the angle from $v$ to $w$ should be the angle from $u$ to $w$. Once one has the parallelogram law then the fact that it comes from an inner product follows via the route above. However, the properties of an inner product are not particularly obvious from thinking about properties of angles. So the easier they are to deduce from the parallelogram law, the easier they are to motivate. I consider the route to $\langle u,\lambda v\rangle = \lambda\langle u,v\rangle$ to be a little long. I was hoping someone could shorten it for me.
Alternative, there may be a different starting point than that angles "add". Perhaps some other property, say similarity of certain triangles, that could be used. However, I'd like a single property that would do the lot. I don't want "add" for some properties and "something else" for others. That's too complicated.
Best Answer
To me continuity is more geometric and intuitive than the rest of the argument (which is purely algebraic manipulation). So I take the liberty to mis-read you question as follows:
By "only algebraic" I mean that you are not allowed to use inequalities. (It is triangle inequality that allows one to use continuity. In fact, one can derive continuity using only the inequality $|u|^2\ge 0$ and the parallelogram law.) Also, an algebraic argument must work over any field on characteristic 0.
The answer is that it is not possible. More precisely, the following theorem holds.
Theorem. There exists a field $F\subset\mathbb R$ and a function $\langle\cdot,\cdot\rangle: F^2\times F^2\to F$ which is symmetric, additive in each argument (i.e. $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$), satisfies the identity $\langle tu,tv\rangle = t^2\langle u,v\rangle$ for every $t\in F$, but is not bi-linear.
Note that the above assumptions imply that the "quadratic form" $Q$ defined by $Q(v)=\langle v,v\rangle$ satisfies $Q(tv)=t^2Q(v)$ and the parallelogram identity, and the "product" $\langle\cdot,\cdot\rangle$ is determined by $Q$ in the usual way. [EDIT: an example exists for $F=\mathbb R$ as well, see Update.]
Proof of the theorem. Let $F=\mathbb Q(\pi)$. An element $x\in F$ is uniquely represented as $f_x(\pi)$ where $f_x$ is a rational function over $\mathbb Q$. Define a map $D:F\to F$ by $D(x) = (f_x)'(\pi)$. This map satisfies
$D(1) = 0$;
$D(\pi)=1$;
$D(x+y) = D(x)+D(y)$;
$D(xy) = x D(y) + y D(x)$.
Define $P:F\times F$ by $P(x,y) = xD(y)-yD(x)$. From the above identities it is easy to see that $P$ is additive in each argument and satisfies $P(tx,ty)=t^2 P(x,y)$ for all $x,y,t\in F$. Finally, define a "scalar product" on $F^2$ by $$ \langle (x_1,y_1), (x_2,y_2) \rangle = P(x_1,y_2) + P(x_2,y_1) . $$ It satisfies all the desired properties but is not bilinear: if $u=(1,0)$ and $v=(0,1)$, then $\langle u,v\rangle=0$ but $\langle u,\pi v\rangle=1$.
Update. One can check that if $\langle\cdot,\cdot\rangle$ is a "mock scalar product" as in the theorem, then for any two vectors $u,v$, the map $t\mapsto \langle u,tv\rangle - t\langle u,v\rangle$ must be a differentiation of the base field. (A differentiation is map $D:F\to F$ satisfying the above rules for sums and products.) Thus mock scalar products on $\mathbb R^2$ are actually classified by differentiations of $\mathbb R$.
And non-trivial differentiations of $\mathbb R$ do exist. In fact, a differentiation can be extended from a subfield to any ambient field (of characteristic 0). Indeed, by Zorn's Lemma it suffices to extend a differentiation $D$ from a field $F$ to a one-step extension $F(\alpha)$ of $F$. If $\alpha$ is transcedental over $F$, one can define $D(\alpha)$ arbitrarily and extend $D$ to $F(\alpha)$ by rules of differentiation. And if $\alpha$ is algebraic, differentiating the identity $p(\alpha)=0$, where $p$ is a minimal polynomial for $\alpha$, yields a uniquely defined value $D(\alpha)\in F(\alpha)$, and then $D$ extends to $F(\alpha)$. The extensions are consistent because all identities involved can be realized in the field of differentiable functions on $\mathbb R$, where differentiation rules are consistent.
Thus there exists a mock scalar product on $\mathbb R^2$ such that $\langle e_1,e_2\rangle=0$ but $\langle e_1,\pi e_2\rangle=1$. And I am sure I reinvented the wheel here - all this should be well-known to algebraists.