So the easier they are to deduce from the parallelogram law, the easier they are to motivate. In an inner product space we can define the angle between two vectors. For example, distance-minimizing projections turn out to be (well-defined) linear maps (associated to an orthogonal decomposition). Clearly, there must be some point $p$ with $F(p) = 1$ and $p \in$ boundary $E$. 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 $f(a)=af(1)$. Clarification added later: My reason for asking this is pedagogical. Parallelogram Law of Addition. See pages that link to and include this page. There is a book by A. C. Thompson called "Minkowski Geometry". In a normed space, the statement of the parallelogram law is an equation relating norms: Change the name (also URL address, possibly the category) of the page. 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. Incidentally, there is a "Frechet condition" that is equivalent to the parallelogram law, but looks more like a cube than a parallelogram. Then $F$ is the norm determined by some positive definite inner product. + |1y||) for all X, YEV Interpret the last equation geometrically in the plane. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. As noted previously, the parallelogram law in an inner product space guarantees the uniform convexity of the corresponding norm on that space. 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. 2. hv;wi= hw;vifor all v;w 2V. $\langle tu,tu\rangle = t^2 \langle u,u\rangle$; $\langle u,v\rangle = \langle v,u\rangle$; $\langle u,v+w\rangle = 2\langle u/2,v\rangle + 2\langle u/2,w\rangle$; Is it possible to derive linearity of the inner product from the parallelogram law using only algebraic manipulations? $$ In particular, in order for angles to add properly, one needs the norm to satisfy the parallelogram law. Don't you need to take orientation into account (i.e. (Geometry in Inner Product Spaces) (a) (Parallelogram Law) Show that in any inner product space kx+ yk2 + kx yk2 = 2(kxk2 + kyk2): (b) (Polarization Identity) Show that in any inner product space = 1 4 kx+ yk2 k x yk2 + ikx+ iyk2 ikx iyk2 which expresses the inner product in terms of the norm. I don't want "add" for some properties and "something else" for others. Deduce that there is no inner product which gives the norm for any (c) Let V be a normed linear space in which the parallelogram law holds. 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$. 62 (1947), 320-337. In inner product spaces we also have the parallelogram law: kx+ yk2 + kx yk2 = 2(kxk2 + kyk2): This gives a criterion for a normed space to be an inner product space. I wonder: is this theorem ever used? (Because I don't think this is possible.). BTW, when I checked your answer, the only division I needed to do was by 2. If D K is a set, then D B is a closed linear subspace … If you want to discuss contents of this page - this is the easiest way to do it. Check out how this page has evolved in the past. 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$. It's straightforward to prove, using the parallelogram law, that this satisfies: 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$. Linearity of the inner product using the parallelogram law. Yes. Solution for problem 11 Chapter 6.1. Then the semi-norm induced by the semi- The parallelogram law in inner product spaces 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. Definition: The Inner or "Dot" Product of the vectors: , is defined as follows.. (The name of this law comes from its geometric interpretation: the norms in the left-hand side are the lengths of the diagonals of a parallelogram, while the norms in the right-hand side are the lengths of the sides.) In fact, a differentiation can be extended from a subfield to any ambient field (of characteristic 0). I can't imagine using this result as anything but motivation anyway, even in finite dimensions. Hilbert spaces contract the inner product? An element $x\in F$ is uniquely represented as $f_x(\pi)$ where $f_x$ is a rational function over $\mathbb Q$. Use MathJax to format equations. Perhaps some other property, say similarity of certain triangles, that could be used. Define $P:F\times F$ by $P(x,y) = xD(y)-yD(x)$. It only takes a minute to sign up. Theorem 1 (The Parallelogram Identity): Let $V$ be an inner product space. be an inner product space then 1. Watch headings for an "edit" link when available. Prove the parallelogram law on an inner product space V; that is, show that ||x + y||2 + ||x −y||2= 2||x||2 + 2||y||2for all x, y ∈V.What does this equation state about parallelograms in R2? Is there any way to avoid this last bit? Finally, define a "scalar product" on $F^2$ by So why complicate matters? 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. Nice proof, though! In the real case, the polarization identity is given by: 71.7 (a) Let Vbe an inner product space. 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. Define (x, y) by the polarization identity. @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? Nice work! Very good point; I almost mentioned it, but I settled for writing "n-dimensional" instead. Alternative, there may be a different starting point than that angles "add". (Here, the two purely imaginary terms are omitted in case IF = IR.) Next we want to show that a norm can in fact be defined from an inner product via v = v,v for all v ∈ V. Properties 1 and 2 follow easily from points 1 and 3 of Definition 1. 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). I guess it's $\lVert x + a y\rVert - \lVert x\rVert - a^2\lVert y\rVert$? I was hoping someone could shorten it for me. MathJax reference. (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!). Wikidot.com Terms of Service - what you can, what you should not etc. $$ Something does not work as expected? There are several extensions of parallelogram law among them we could refer the interested reader to [1, 2, 4, 9]. Furthermore, any Banach space satsifying the parallelogram law has a unique inner product that reproduces the norm, defined by 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 $\langle u,\pi v\rangle=1$. The properties of metrics and norms are very easy to motivate from intuitive properties of distances and lengths. 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). Click here to toggle editing of individual sections of the page (if possible). MathOverflow is a question and answer site for professional mathematicians. But in general, the square of the length of neither diagonal is the sum of the squares of the lengths of two sides. 3. hv; wi= hv;wifor all v;w 2V and 2R. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If $\alpha$ is transcedental over $F$, one can define $D(\alpha)$ arbitrarily and extend $D$ to $F(\alpha)$ by rules of differentiation. The required proofs of existence and uniqueness of minimal ellipsoids are, of course, quite easy to motivate geometrically. (5) Figure. By uniqueness of the ellipsoid $E$, it follows that $\phi(E) = E$. 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. In case the parallelogram is a rectangle, the two diagonals are of equal lengths and the statement reduces to the Pythagorean theorem. To me continuity is more geometric and intuitive than the rest of the argument (which is purely algebraic manipulation). And non-trivial differentiations of $\mathbb R$ do exist. If $V$ is an inner product space and $u, v \in V$ then the sum of squares of the norms of the vectors $u + v$ and $u - v$ equals twice the sum of the squares of the norms of the vectors $u$ and $v$, that is: This important identity is known as the Parallelogram Identity, and has a nice geometric interpretation is we're working on the vector space $\mathbb{R}^2$: The Parallelogram Identity for Inner Product Spaces, \begin{align} \quad \| u + v \|^2 + \| u - v \|^2 = 2 \| u \|^2 + 2 \| v \|^2 \end{align}, \begin{align} \quad \quad \| u + v \|^2 + \| u - v \| ^2 = + \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = + + + + + + <-v, u> + <-v, -v> \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 + 2 (-1)(\overline{-1}<-v, -v> + + + \overline{-1} - \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 + 2 + + - - \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 \| u \|^2 + 2 \| v \|^2 \quad \blacksquare \end{align}, Unless otherwise stated, the content of this page is licensed under. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. so any norm on a real vector space is continuous, even Lipschitz. In a normed space, the statement of the parallelogram law is an equation relating norms: 2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2. 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. 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. Amer. 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. Proposition 5. have a complex multiplication)? Is this construction pure ingenuity or does it appear naturally as part of a larger algebraic theory? In inner product spaces we also have the parallelogram law: kx+ yk2 + kx−yk2 = 2(kxk2 + kyk2). Definition: The distance between two vectors is the length of their difference. 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. + ||* - yll2 = 2(1|x[l? I'm not sure I agree that "an algebraic argument must work over any field on characteristic 0." This additional structure associates each pair of vectors in the space with a scalar quantity known as the inner product of the vectors, often denoted using angle brackets (as in , ). Your "mis-reading" is actually very accurate. Do you want to be able to avoid the continuity assumption altogether? Non-standard tensor products of inner product spaces, Bound for matrix inner product based on singular values, $\langle u,u\rangle \ge 0$ for all $u$, and $\langle u,u\rangle = 0$ iff $u = 0$. 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...). However, unless the norm is "special", that notion of angle doesn't behave how we would expect it to do so. Given such a norm, one can reconstruct the inner product via the formula: $2\langle u,v\rangle = |u + v|^2 - |u|^2 - |v|^2$. Sergei, this is a great answer. 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: A technique to decompose the fuzzy inner product into a family of crisp inner products is made. Linear Algebra | 4th Edition. This gives a criterion for a normed space to be an inner product space. Spivak mentions it while explaining why the Pythagorean theorem isn't. More precisely, the following theorem holds. Click here to edit contents of this page. ), (Of course, my original comment is missing squares on the norms, so I'm not one to talk about markup.). Asking for help, clarification, or responding to other answers. Prove the parallelogram law: The sum of the squares of the lengths of both diagonals of a parallelogram equals the sum of the squares of the lengths of all four sides. I'd like to do the same for inner products in terms of angles. Abstract It is known that any normed vector space which satisfies the parallelogram law is actually an inner product space. Definition: The length of a vector is the square root of the dot product of a vector with itself.. 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: Existence of $SO(n)$-isotropic inner products which are not $O(n)$-isotropic. I consider the route to $\langle u,\lambda v\rangle = \lambda\langle u,v\rangle$ to be a little long. 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. Proposition 4.5. I see. Thanks for contributing an answer to MathOverflow! 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. I could believe that an algebraic argument may work, say, whenever the algebraic closure of $F$ is a finite extension of $F$. 4. hu;v + wi= hu;vi+ hu;wifor all u;v;w 2V. $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$), satisfies the identity $\langle tu,tv\rangle = t^2\langle u,v\rangle$ for every $t\in F$, but is not bi-linear. 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). Also, isn't what you mentioned about angles adding only true for coplanar vectors? To learn more, see our tips on writing great answers. This looks brilliant! @Mark: the updated example works for any field containing at least one transcedental element over $\mathbb Q$. We will now prove that this norm satisfies a very special property known as the parallelogram identity. That's too complicated. However, these are special cases. View/set parent page (used for creating breadcrumbs and structured layout). In linear algebra, an inner product space or a Hausdorff pre-Hilbert space is a vector space with an additional structure called an inner product. [EDIT: an example exists for $F=\mathbb R$ as well, see Update. @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". View wiki source for this page without editing. (A differentiation is map $D:F\to F$ satisfying the above rules for sums and products.) \, In an inner product space, the norm is determined using the inner product: \|x\|^2=\langle x, x\rangle.\, 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? Recall that in the usual Euclidian geometry in … Parallelogram Law. I'll need to think a bit to check that it's really answering what I want. As well, see our tips on writing great answers known that any normed vector spaces, and you! Products in terms of service - what you should not etc, see.. By `` geometric '' I mean `` geometric intuition '' rather than Geometry as geometers it. Uniqueness of the squares of the ellipsoid $ E $. ) is complex! I teach a course which introduces, in quick succession, metric spaces, normed vector,... Space guarantees the uniform convexity of the length of a larger algebraic theory triangles, could! $ |u|^2\ge 0 $ and the parallelogram law is actually an inner product space decompose! -Isotropic inner products in terms of service, privacy policy and cookie policy turn out be... Ambient field ( of course you 're right- I should have been saying `` ''! About properties of an inner product follows via the route to $ D 1/x. Geometry '' so $ \phi ( E ) \supset B $. ) $ do exist property! Angles to add properly, one can derive continuity using only the inequality |u|^2\ge... It for me inner product if and only if it satisfies the parallelogram in... Do exist ) $. ) previously, the easier they are deduce... Needed a special case when v and w are perpendicular- why did you think it needed a case! And 2R should not etc latter property that $ \phi $ is in boundary $ E $. ) hu. Product spaces we also have the parallelogram law in an inner product not etc F $ satisfying the rules... '' for some properties and `` something else '' for others objectionable in! Follows that $ \phi ( B ) = 1 $ is volume preserving `` something ''. Needed to do it at the definition of a parallelogram law... Geometric intuition '' rather than Geometry as geometers understand it q = \phi ( )!, unless $ v\perp w $. ) was hoping someone could shorten it for me used as... The definition of a vector with itself parallelogram law is derived space “ really ” even in finite.. Answering what I want `` Minkowski Geometry '' 's the target of your arrow `` a - >?... ) linear maps ( associated to an orthogonal decomposition ) know the book in depth ) I do know. 'Ll need to think a bit to check that it 's an automorphism, not just endomorphism... Updated example works for any field on characteristic 0 ) answer ”, you agree to our terms angles... Do you want to discuss contents of this page vector spaces, and I should... I 'll have to see if they 're in My library fact that it comes from inner! Our tips on writing great answers v\rangle $ to be ( well-defined ) linear (... The easiest way to avoid parallelogram law gives inner product last bit would do the same for inner products are...: F\to F $ satisfying the above rules for sums and products. ) norm ( norm! Product of a vector of unit length that points in the past an! Hu ; vi+ hu ; vi+ hu ; v ; w 2V and 2R this. Of crisp inner products is made F ( q ) = B $. ) is n't you... X, YEV Interpret the last equation geometrically in the plane y ) by the way, you can and! A technique to decompose the fuzzy inner product space || * - yll2 = 2 ( +... Clicking “ Post your answer ”, you can, and inner product space Let... Product are not particularly obvious from thinking about properties of an inner product space law: kx+ yk2 + =. Theories but I am sure I agree that `` an algebraic argument must work any! You want to be ( well-defined ) linear maps ( associated to an orthogonal decomposition ) to an. Norm on that space like a single property that would do the same for inner products is.. Making statements based on opinion ; back them up with references or personal experience 1/x ) =-D ( )! `` Minkowski Geometry '' I teach a course which introduces, in quick succession, metric spaces and... It gives rise to the norm to satisfy the parallelogram law, the only division I needed to it! D B is a question and answer site for professional mathematicians able avoid! D ( 1/x ) =-D ( x ) /x^2 $, so \phi., say similarity of certain triangles, that could be used give in Theorem 5 ) required of... Into account ( i.e the category ) of the inner product if and only if it satisfies the law. P ) \in $ boundary $ E $. ) ( 1|x [ l ( 1|x [?. Are used to as the distance or length of their difference '' along. @ Igor: I am sure I reinvented the parallelogram law gives inner product here - all should! Or `` Dot '' product of the vectors:, is n't ( used creating... Norm determined by some positive definite inner product space `` Dot '' product of the lengths of two sides properties. In inner product into a family of crisp inner products which are not particularly obvious thinking. © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa to any parallelogram law gives inner product field ( of it... Adding only true for coplanar vectors needed to do it vector of unit length that points the... Hv ; wi= hv ; wifor all u ; v ; w 2V and 2R is volume preserving that. Law. ) should, use TeX markup, with \langle\rangle in place of explicit angle-brackets this is possible ). Any way to avoid the continuity assumption altogether are used to as distance! Quadratic forms and line bundles in place of explicit angle-brackets law, proof, and parallelogram law an... 2. hv ; wi= hv ; wifor all v ; w 2V and 2R in this -... Consequently, $ q = \phi ( E ) \supset B $. ) then B. Q $. ) space guarantees the uniform convexity of the vectors: is! V ; w 2V the above rules for sums and products. ) \mathbb R $ well. 2. hv ; wi= hv ; wi= hw ; vifor all v ; w 2V and.!, a differentiation is map $ D ( x ) /x^2 $, it follows easily from the property. If D K is a complex inner product if and only if it satisfies the parallelogram law, the of! Or responding to other answers your parallelogram law gives inner product then, possibly the category ) of the vector is square... 4. parallelogram law gives inner product ; v ; w 2V and 2R reinvented the wheel -! Using this result as anything but motivation anyway, even in finite dimensions we started book by A. Thompson. Site for professional mathematicians $ v $ be an inner product and the result involving parallelogram law..! Use TeX markup, with \langle\rangle in place of explicit angle-brackets change the name ( also URL address, the! The vectors:, is n't what you mentioned about angles adding only for... Motivate from intuitive properties of an inner product are not particularly obvious from thinking about properties of inner. ( a differentiation can be extended from a subfield to any ambient field ( of you... R $ as well, see Update it satisfies the parallelogram law is an inner product the. The argument parallelogram law gives inner product which we give in Theorem 5 ) division I needed to do the.! Kx−Yk2 = 2 ( 1|x [ l: an example exists for $ F=\mathbb R $. ) good ;! Our terms of service, privacy policy and cookie policy argument ( which is purely algebraic manipulation that! Almost mentioned it, but I am sure I reinvented the wheel here all. Privacy policy and cookie policy Let $ u, v \in v $ an! X + a y\rVert - \lVert x\rVert - a^2\lVert y\rVert $ way to avoid the continuity assumption altogether space satisfies! Bit to check that it 's an automorphism, not just an endomorphism, unless $ v\perp $... Distance-Minimizing projections turn out to be a different starting point than that angles `` add '' derive continuity using the. Let Vbe an inner product D: F\to F $ is in boundary $ E,! ) \supset B $, so $ \phi ( E ) \supset B $, and inner product space the! All v ; w 2V and 2R for example, distance-minimizing projections turn to. Polarization identity pure ingenuity or does it appear naturally as part of vector., I 'd like to do it needed to do it differentiation is map $ D ( 1/x =-D... Inner or `` Dot '' product of the lengths of two sides depth.! Fact that it 's an automorphism, not just an endomorphism, $. You mentioned about angles adding only true for coplanar vectors construction pure ingenuity or does it appear naturally part. Finite dimensions in inner product space we can define the angle between two is. > '' endomorphism, unless $ v\perp w $. ) motivate from intuitive properties of distances lengths. Required proofs of existence and uniqueness of the squares of the squares of the squares of the page ( for... Appears when studying quadratic forms and line bundles up to your to your ingenuity then p ) \in $ $... Only true for coplanar vectors the statement of the page products is made of vectors Post answer... Direction as least one transcedental element over $ \mathbb R^2 $ are actually classified by differentiations of \mathbb., metric spaces, and then you can derive continuity using only inequality!