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:

  1. $\langle u,u\rangle \ge 0$ for all $u$, and $\langle u,u\rangle = 0$ iff $u = 0$
  2. $\langle tu,tu\rangle = t^2 \langle u,u\rangle$;
  3. $\langle u,v\rangle = \langle v,u\rangle$;
  4. $\langle u,v+w\rangle = 2\langle u/2,v\rangle + 2\langle u/2,w\rangle$;

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.

  • $\begingroup$ Let me make sure I understand your question. Do you want to be able to avoid the continuity assumption altogether? (Because I don't think this is possible.) $\endgroup$ Dec 4, 2009 at 21:36
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    $\begingroup$ Yes. I agree with your instinct, but also have a sneaking suspicion that there's a cunning parallelogram somewhere that one could draw that would give the result without continuity. $\endgroup$ Dec 4, 2009 at 22:01
  • $\begingroup$ To give another point of view about this issue: the bilinearity of the inner product is responsible for geometry the way we're used to. For example, distance-minimizing projections turn out to be (well-defined) linear maps (associated to an orthogonal decomposition). If you take two points $v_1, v_2$, then the set of all points equidistant from the two is some hyperplane through the midpoint. Also, isn't what you mentioned about angles adding only true for coplanar vectors? Don't you need to take orientation into account (i.e. have a complex multiplication)? $\endgroup$
    – Phil Isett
    Oct 2, 2011 at 3:11
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    $\begingroup$ I wonder: is this theorem ever used? That is, are there spaces in which at first sight there is an obvious norm/length, appearing naturally by some geometric considerations, but there is no obvious scalar product (or concept of an angle), and only a posteriori one notices that the norm satisfies the parallelogram law, hence there is a scalar product after all? $\endgroup$
    – Goldstern
    Nov 10, 2019 at 22:27

5 Answers 5


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:

  • Is it possible to derive linearity of the inner product from the parallelogram law using only algebraic manipulations?

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.

  • $\begingroup$ This looks brilliant! I'll need to think a bit to check that it's really answering what I want. Your "mis-reading" is actually very accurate. (Though to counter your "geometric/algebraic" split, I would say that whilst I agree with your classification, I think that only someone advanced in geometry would split it there and for students it isn't obviously the right place to put the break - if that makes sense!) $\endgroup$ Apr 17, 2011 at 18:17
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    $\begingroup$ I'm not sure I agree that "an algebraic argument must work over any field on characteristic 0." I could believe that an algebraic argument may work, say, whenever the algebraic closure of $F$ is a finite extension of $F$. $\endgroup$ Apr 17, 2011 at 23:59
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    $\begingroup$ Sergei, this is a great answer. The same question as Andrew's had been sitting in the back of my mind for quite some time, but I never thought of comparing the failure of $\langle\cdot,\cdot\rangle$-bilinearity to a derivation. Is this construction pure ingenuity or does it appear naturally as part of a larger algebraic theory? $\endgroup$ Apr 18, 2011 at 2:46
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    $\begingroup$ @Mark: the updated example works for any field containing at least one transcedental element over $\mathbb Q$. @Igor: I am not aware of relevant theories but I am far from algebra. Just write down additivity in the first argument as an equation of that function (using quadratic homogenuity to move $t$ from one argument to the other). It soon leads to $D(1/x)=-D(x)/x^2$, and then you can derive the product rule. $\endgroup$ Apr 18, 2011 at 7:57
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    $\begingroup$ I see. I'll chalk it up to your to your ingenuity then. Nice work! BTW, when I checked your answer, the only division I needed to do was by 2. So, I think your construction generalizes straightforwardly to the case of modules over a ring $R$ where 2 is an invertible element. The "mock inner products", as you've defined them, which are merely $R\supset R'$-bilinear modulo exactly $R$-bilinear ones are then in one-to-one correspondence with antisymmetric, $R$-bilinear, $\mathrm{Der}_{R'}(R)$-valued forms, where $\mathrm{Der}_{R'}(R)$ is the $R$-module of derivations of $R$ that annihilate $R'$. $\endgroup$ Apr 18, 2011 at 9:39

The easiest way to see how things "break" in the case of a more general norm is to look at the shape of its unit "sphere" — unless it's an ellipsoid, no linear transformation exists taking it to a Euclidean sphere, and it follows from the principal axis theorem that each ellipsoid is associated with a unique inner product, and conversely.

Quoting Spivak's Comprehensive Introduction to Differential Geometry, Vol. 2, p. 210 (he defines a Minkowski metric to be a map $F: v \to \mathbb{R}$ such that $F(v) > 0$ for all $v \neq 0$ and $F(\lambda v) = | \lambda | F(v)$, so this holds a fortiori under the stronger hypothesis of a norm):

THEOREM. Let $F: V \to \mathbb{R}$ be a continuous Minkowski metric on an $n$-dimensional vector space $V$. Suppose that, for all $p$ and $q$ in the unit sphere $ \{ v \in V: F(x) = 1 \} $, there is a linear transformation $\phi: V \to V$ such that $\phi(p) = q$ and $F(\phi(v)) = F(v)$ for all $v \in V$. Then $F$ is the norm determined by some positive definite inner product.

PROOF: Let $B = \{ v : F(v) \leq 1 \} $, and let $E$ be the unique ellipsoid containing $B$ of smallest volume. Clearly, there must be some point $p$ with $F(p) = 1$ and $p \in$ boundary $E$. Let $q$ be any other point with $F(q) = 1$, and $\phi: V \to V$ a linear transformation with $\phi(p) = q$ such that $F(\phi(v)) = F(v)$ for all $v \in V$. It follows easily from the latter property that $\phi(E) \supset B$. Moreover, $\phi(B) = B$, so $\phi$ is volume preserving. By uniqueness of the ellipsoid $E$, it follows that $\phi(E) = E$. Consequently, $q = \phi(p) \in$ boundary $E$. In other words, every point $q$ with $F(q) = 1$ is in boundary $E$. This means that $E = B$.

The required proofs of existence and uniqueness of minimal ellipsoids are, of course, quite easy to motivate geometrically. It's also easy to see why this settles the matter: if isometries don't act transitively on the unit sphere, it's hard to define the "angle between two vectors" in a sensible way.

With all that said, a simpler way of looking at things, for me, at least, is this: given any inner product on an $n$-dimensional real vector space, an orthonormal basis exists, in terms of which things are "computationally indistinguishable" from $\mathbb{R}^n$ with the usual inner product — the coefficients don't care what the basis vectors look like. This reduces things to the case of the usual inner product, where "geometric intuition" has been axiomatized: however well-motivated it may be, algebraically, the law of cosines is essentially true by definition.

In other words, we always have the option of writing our vectors in a way that makes "ordinary" intuition apply; we even have the option of thinking about things in the usual terms, even when working in a "skew basis." So why complicate matters?

Finally, continuity is hardly "un-geometric" in this context: by the triangle inequality, the difference between the lengths of two sides of a triangle is never greater than the length of the third side: $$\Big| \lVert x \rVert - \lVert y \rVert \Big| \leq \lVert x - y \rVert,$$ so any norm on a real vector space is continuous, even Lipschitz.

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    $\begingroup$ One problem with this approach (depending on one's priorities) is that it only works in finite dimensions, whereas the result is also true in infinite dimensions. Another is that the proofs of existence and uniqueness of the ellipsoid of smallest volume containing a convex body may be beyond the scope of the course that motivated the question. Nice proof, though! $\endgroup$ Apr 16, 2011 at 20:17
  • $\begingroup$ Very good point; I almost mentioned it, but I settled for writing "n-dimensional" instead. I can't imagine using this result as anything but motivation anyway, even in finite dimensions. Spivak mentions it while explaining why the Pythagorean theorem isn't quite "true by definition," hence his choice of emphasis (that and it's a geometry book!). Pedantically: an ellipsoid E = {x in V: <x,x> <= 1} contains a set B iff it contains the convex hull of B. Proof: sufficiency is clear, since B < co(B). So assume B < E, then co(B) < co(E) = E, as E is already convex. (sorry, I couldn't resist...) $\endgroup$
    – jasomill
    Apr 17, 2011 at 5:57
  • $\begingroup$ @MarkMeckes, isn't the finite-dimensionality restriction not a problem, since the statement that $\langle u, t v\rangle = t\langle u, v\rangle$ is only talking about an at-worst-$2$-dimensional subspace, even if the ambient space is infinite dimensional? $\endgroup$
    – LSpice
    Apr 2, 2020 at 23:05
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    $\begingroup$ @LSpice: That depends on where you draw the line between "not a problem" and "a problem that can be circumvented with a little additional work". $\endgroup$ Apr 3, 2020 at 15:36

I'm having some difficulty understanding what you mean by "geometric" since well-behaved inner products are essentially what you need to make geometry work as expected.

The theorem under consideration (due to Jordan and von Neumann, 1935) is given two proofs on pages 114-118 in Istratescu's Inner product spaces: theory and applications (I found it on Google Books). The first is your proof, and the second involves first proving that for fixed u and v, |u + tv|^2 is a degree 2 polynomial in t (this is where continuity is used, together with arithmetic sequences). This is followed by an algebraic manipulation showing that the linear term of the polynomial is an inner product.

Incidentally, there is a "Frechet condition" that is equivalent to the parallelogram law, but looks more like a cube than a parallelogram. It appears when studying quadratic forms and line bundles.


There is a book by A. C. Thompson called "Minkowski Geometry". In this context "Minkowski Geometry" means the geometry of a vector space with a norm (not the geometry of special relativity as one might think...). Chapter 3.4 is called "Characterizations of the Euclidean Space" and it has many theorems stating that a norm comes from a inner product iff such and such (mostly geometric) conditions is satisfied. They might be helpful (as far as I can tell... I don't know the book in depth). In Thompson's book there's a reference to Dan Amir's book "Characterizations of inner product spaces" that might also be useful (but I haven't even seen this book).

  • $\begingroup$ Those sound very good references to chase. I'll have to see if they're in my library. $\endgroup$ Dec 4, 2009 at 21:34

I certainly agree that that proof is deeply dissatisfying. However, it seemed a bit better when I split things out into a lemma:

The only continuous endomorphisms of $(\mathbb R,+)$ are those of the form $f(a)=af(1)$. Similarly, any continuous endomorphism $f$ of $(\mathbb C,+)$ for which $f(i)=if(1)$ must have $f(a)=af(1)$ $\forall a$.

We've already shown $a \mapsto \langle av,w \rangle$ is an endomorphism of $(\mathbb R,+)$ or $(\mathbb C,+)$ (step 4 above), and we know it's continuous (composition of contin. fns), so linearity follows.

(Of course it's an automorphism, not just an endomorphism, unless $v\perp w$.)

This way we can see that the last part of the proof isn't just a random bit of analysis creeping unnaturally through the cracks but rather an important fact about the fields associated with our vector spaces.

  • $\begingroup$ What's the target of your arrow "a ->"? I guess it's $\lVert x + a y\rVert - \lVert x\rVert - a^2\lVert y\rVert$? (Of course, this actually is not an automorphism of $\mathbb C$.) How do you handle the case $x \perp y$? If you omit it, then it seems to me that one still needs to prove that $x \perp y \Rightarrow x \perp a y$ for all $a$ (or is that easy?). $\endgroup$
    – LSpice
    Apr 16, 2011 at 13:56
  • $\begingroup$ Sorry about that- I'd written the inner product with regular angle brackets rather than &lt; and &gt; so it was interpreted as a HTML tag. In the complex case, rather than the real parallelogram identity presented in the question we of course use the polarization identity to define the inner product, and it's once again easy to show <u+v,w>=<u,w>+<v,w> so a-> <av,w> is an automorphism of (C,+) under that definition. There's no need for a special case when v and w are perpendicular- why did you think it needed a special case? $\endgroup$
    – Prodicus
    Apr 18, 2011 at 16:26
  • $\begingroup$ Because, in the orthogonal case, $a \mapsto \langle a v, w\rangle$ should be the constant map with value $0$, which is not an automorphism. I guess you meant ‘endomorphism’. (By the way, you can, and I think should, use TeX markup, with \langle\rangle in place of explicit angle-brackets.) $\endgroup$
    – LSpice
    Apr 21, 2011 at 4:10
  • $\begingroup$ (Of course, my original comment is missing squares on the norms, so I'm not one to talk about markup.) $\endgroup$
    – LSpice
    Apr 21, 2011 at 4:11
  • $\begingroup$ Yes, of course you're right- I should have been saying "endomorphism" all along. $\endgroup$
    – Prodicus
    Apr 21, 2011 at 22:02

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