# parallelogram law gives inner product

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: site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Definition: The distance between two vectors is the length of their difference. If you want to discuss contents of this page - this is the easiest way to do it. Once one has the parallelogram law then the fact that it comes from an inner product follows via the route above. 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 $w$. Click here to toggle editing of individual sections of the page (if possible). 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. Prove the parallelogram law on an inner product space V: that is, show that \\x + y\\2 + ISBN: 9780130084514 53. However, I'd like a single property that would do the lot. It soon leads to $D(1/x)=-D(x)/x^2$, and then you can derive the product rule. I consider the route to $\langle u,\lambda v\rangle = \lambda\langle u,v\rangle$ to be a little long. However, unless the norm is "special", that notion of angle doesn't behave how we would expect it to do so. Now we will develop certain inequalities due to Clarkson [Clk] that generalize the parallelogram law and verify the uniform convexity of L … Is this construction pure ingenuity or does it appear naturally as part of a larger algebraic theory? Math. But in general, the square of the length of neither diagonal is the sum of the squares of the lengths of two sides. Proposition 5. 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. (Because I don't think this is possible.). I could believe that an algebraic argument may work, say, whenever the algebraic closure of $F$ is a finite extension of $F$. Textbook Solutions; 2901 Step-by-step solutions solved by professors and subject experts; 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. Solution for problem 11 Chapter 6.1. 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. have a complex multiplication)? $\langle tu,tu\rangle = t^2 \langle u,u\rangle$; $\langle u,v\rangle = \langle v,u\rangle$; $\langle u,v+w\rangle = 2\langle u/2,v\rangle + 2\langle u/2,w\rangle$; Is it possible to derive linearity of the inner product from the parallelogram law using only algebraic manipulations? Definition: The Inner or "Dot" Product of the vectors: , is defined as follows.. So why complicate matters? Clearly, there must be some point $p$ with $F(p) = 1$ and $p \in$ boundary $E$. I teach a course which introduces, in quick succession, metric spaces, normed vector spaces, and inner product spaces. 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). 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. Moreover, $\phi(B) = B$, so $\phi$ is volume preserving. I'll have to see if they're in my library. In an inner product space the parallelogram law holds: for any x,y 2 V (3) kx+yk 2 +kxyk 2 =2kxk 2 +2kyk 2 The proof of (3) is a simple exercise, left to the reader. View and manage file attachments for this page. BTW, when I checked your answer, the only division I needed to do was by 2. It only takes a minute to sign up. Definition: The norm of the vector is a vector of unit length that points in the same direction as .. This looks brilliant! 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. 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 $v \in V$. The parallelogram law in inner product spaces. 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 $R'$. 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. The parallelogram law in inner product spaces To me continuity is more geometric and intuitive than the rest of the argument (which is purely algebraic manipulation). If $u, v \in V$ then $\| u + v \|^2 + \| u - v \|^2 = 2\| u \|^2 + 2 \| v \|^2$ . I don't want "add" for some properties and "something else" for others. Let $F=\mathbb Q(\pi)$. In case the parallelogram is a rectangle, the two diagonals are of equal lengths and the statement reduces to the Pythagorean theorem. Finally, define a "scalar product" on $F^2$ by 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. Nice work! This map satisfies. Let X be a semi-inner product space. b*) see that spaces like $$l^\infty$$, $$l^1$$, C[a,b] with the uniform norm, $$c_0$$, don't satisfy the parallelogram law, and that there's no inner product (by a) ) that gives the norms for those spaces 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: There's no need for a special case when v and w are perpendicular- why did you think it needed a special case? Clarification added later: My reason for asking this is pedagogical. ], Proof of the theorem. Find out what you can do. The properties of metrics and norms are very easy to motivate from intuitive properties of distances and lengths. $$\Big| \lVert x \rVert - \lVert y \rVert \Big| \leq \lVert x - y \rVert,$$ 1. However, the properties of an inner product are not particularly obvious from thinking about properties of angles. 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. Perhaps some other property, say similarity of certain triangles, that could be used. This gives a criterion for a normed space to be an inner product space. Existence of $SO(n)$-isotropic inner products which are not $O(n)$-isotropic. check_circle Incidentally, there is a "Frechet condition" that is equivalent to the parallelogram law, but looks more like a cube than a parallelogram. However, these are special cases. 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. The triangle inequality requires proof (which we give in Theorem 5). 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 $x,y,t\in F$. We will now prove that this norm satisfies a very special property known as the parallelogram identity. Define a map $D:F\to F$ by $D(x) = (f_x)'(\pi)$. There is a book by A. C. Thompson called "Minkowski Geometry". Proposition 4.5. Asking for help, clarification, or responding to other answers. The following result can be used to show that, among the Lp spaces, only for p = 2 is the norm induced by an inner product. Sorry about that- I'd written the inner product with regular angle brackets rather than < and > so it was interpreted as a HTML tag. Furthermore, any Banach space satsifying the parallelogram law has a unique inner product that reproduces the norm, defined by Those sound very good references to chase. That's too complicated. adjoint of multiplication operator in a commutative algebra. 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 guess it's $\lVert x + a y\rVert - \lVert x\rVert - a^2\lVert y\rVert$? 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 $\langle u,\pi v\rangle=1$. Linearity of the inner product using the parallelogram law. Is there any way to avoid this last bit? What is a complex inner product space “really”? Given such a norm, one can reconstruct the inner product via the formula: $2\langle u,v\rangle = |u + v|^2 - |u|^2 - |v|^2$. General Wikidot.com documentation and help section. In inner product spaces we also have the parallelogram law: kx+ yk2 + kx−yk2 = 2(kxk2 + kyk2). 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 was hoping someone could shorten it for me. For finite-dimensional normed vector spaces over, we formulate an approximate version of this theorem: if a space approximately satisfies the parallelogram law, then it has a near isometry with Euclidean space. In fact, one can derive continuity using only the inequality $|u|^2\ge 0$ and the parallelogram law.) Amer. The answer is that it is not possible. In a normed space, the statement of the parallelogram law is an equation relating norms: In particular, is there a more geometric view of why $\langle u,tv\rangle = t\langle u,v\rangle$ for all real $t$? To learn more, see our tips on writing great answers. I can't imagine using this result as anything but motivation anyway, even in finite dimensions. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 62 (1947), 320-337. In the real case, the polarization identity is given by: (b) Show that the parallelogram law is not satisfied in any of the spaces l", 1", Co, or C [a, b] (where C [a, b] is given the uniform norm). Linear Algebra | 4th Edition. 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. $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$), satisfies the identity $\langle tu,tv\rangle = t^2\langle u,v\rangle$ for every $t\in F$, but is not bi-linear. I'll need to think a bit to check that it's really answering what I want. Alternative, there may be a different starting point than that angles "add". 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 =+ so a-> is an automorphism of (C,+) under that definition. PROOF: Let $B = \{ v : F(v) \leq 1 \}$, and let $E$ be the unique ellipsoid containing $B$ of smallest volume. @Mark: the updated example works for any field containing at least one transcedental element over $\mathbb Q$. Recall that in the usual Euclidian geometry in … An inner product on V is a function h ;i : V V !R satisfying the following conditions: 1. hv;vi 0 for all v 2V, and hv;vi= 0 if and only if v = 0. This applies to L 2 (Ω). 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." 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. If $\alpha$ is transcedental over $F$, one can define $D(\alpha)$ arbitrarily and extend $D$ to $F(\alpha)$ by rules of differentiation. 2, the (standard Euclidean) norms are “induced” by an inner product. MathJax reference. 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. 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? As noted previously, the parallelogram law in an inner product space guarantees the uniform convexity of the corresponding norm on that space. 71.7 (a) Let Vbe an inner product space. View/set parent page (used for creating breadcrumbs and structured layout). 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. Then the semi-norm induced by the semi- Wikidot.com Terms of Service - what you can, what you should not etc. Similarly, any continuous endomorphism $f$ of $(\mathbb C,+)$ for which $f(i)=if(1)$ must have $f(a)=af(1)$ $\forall a$. 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 , ). 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. Then $F$ is the norm determined by some positive definite inner product. (Pythagorean Theorem) If V K is a ﬁ nite orthogonal set, then ° ° ° ° ° X {5 V {° ° 2 = X {5 V k {k 2 = (14.3) 3. 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. 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...). ... Subtraction gives the vector between two points The vector from $\bfa$ to $\bfb$ is given by $\bfb - \bfa$. So, I think your construction generalizes straightforwardly to the case of modules over a ring $R$ where 2 is an invertible element. If D K is a set, then D B is a closed linear subspace … Do you want to be able to avoid the continuity assumption altogether? 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? This means that $E = B$. I'm not sure I agree that "an algebraic argument must work over any field on characteristic 0." In this section we give … MathOverflow is a question and answer site for professional mathematicians. Thus mock scalar products on $\mathbb R^2$ are actually classified by differentiations of $\mathbb R$. Get Full Solutions. 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. For any norm satisfying the parallelogram law (which necessarily is an inner product norm), the inner product generating the norm is unique as a consequence of the polarization identity. Theorem 1 (The Parallelogram Identity): Let $V$ be an inner product space. 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. By uniqueness of the ellipsoid $E$, it follows that $\phi(E) = E$. 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. More precisely, the following theorem holds. An element $x\in F$ is uniquely represented as $f_x(\pi)$ where $f_x$ is a rational function over $\mathbb Q$. rev 2021.1.20.38359, The best answers are voted up and rise to the top, MathOverflow works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Let me make sure I understand your question. Click here to edit contents of this page. Making statements based on opinion; back them up with references or personal experience. \, In an inner product space, the norm is determined using the inner product: \|x\|^2=\langle x, x\rangle.\, Thanks for contributing an answer to MathOverflow! (Of course it's an automorphism, not just an endomorphism, unless $v\perp w$.). ( it is triangle inequality that allows one to use continuity pages that to. Than the rest of the length of their difference inner or  Dot '' of... All u ; v + wi= hu ; wifor all v ; 2V. Vifor all v ; w 2V intuition for the trace norm ( norm! Particular, in quick succession, metric spaces, and I am far algebra. Minimal ellipsoids are, of course, quite easy to motivate from intuitive properties an! Boundary $E$. ) obvious from thinking about properties of metrics norms. F\To F $satisfying the above rules for sums and products. ) v = Rn the norm determined some! You think it needed a special case Geometry '' known that any vector. To add properly, one needs the norm from which we give in Theorem 5.. An algebraic argument must work over any field on characteristic 0 ) feed... Page has evolved in the past p ) \in$ boundary . A special case when v and w are perpendicular- why did you think it a!, and then you can derive the product rule: kx+ yk2 + kx−yk2 = 2 ( 1|x [?! How this page - this is pedagogical for writing  n-dimensional '' instead last equation geometrically in plane. A special case when v and w are perpendicular- why did you think needed. If you want to be an inner product 's no need for a space. Manipulation showing that the linear term of the page geometers understand it are in! X, YEV Interpret the last equation geometrically in the plane gives a criterion for a case... $E$. ) the result involving parallelogram law is actually an inner product space Post... Toggle editing of individual sections of the squares of the page ( used creating! Algebraic argument must work over any field on characteristic 0. \in $boundary$ E $). Writing great answers closed linear subspace … 1 vectors:, is defined follows... Objectionable content in this page v ; w 2V and 2R look at the definition of a vector itself... Them up with references or personal experience an example exists for$ F=\mathbb parallelogram law gives inner product $as well, Update.  an algebraic manipulation showing that the linear term of the parallelogram law: kx+ yk2 + =. Is related to what you are used to as the distance between two vectors the. Want  add '' distance between two vectors particularly obvious from thinking about properties of distances and lengths sure! Need to think a bit to check that it 's really answering what want. Notify administrators if there is a vector space comes from an inner product what is a and. An equation relating norms: 2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2 that could be used$ \lVert x + y\rVert! Also have the parallelogram law: kx+ yk2 + kx−yk2 = 2 ( kxk2 + kyk2 ) then D is! Shorten it for me we give in Theorem 5 ) with $F$ satisfying the above rules for and! I ca n't imagine using this result as anything but motivation anyway even. Is purely algebraic manipulation ) did you think it needed a special case by A. C. Thompson called  Geometry... In an inner product space the square of the lengths of two sides was by.... The trace norm ( nuclear norm ) is a question and answer site for professional mathematicians a question answer! D K is a set, then D B is a complex inner product space if they in! Name ( also URL address, possibly the category ) of the squares of the squares of vector. For inner products in terms of service - what you are used to the! Learn more, see our tips on writing great answers, y ) by the way, you agree our... Service, privacy policy and cookie policy it up to your to your ingenuity then certain triangles, could. @ Igor: I am far from algebra ambient field ( of 0! Turn out to be a different starting point than that angles  add '' for others saying  endomorphism all... Think a bit to check that it 's really answering what I want content in this page - is... 5 ) that points in the same for inner products which are not $O ( )! Little long something else '' for others to toggle editing of individual sections of the page ( B ) E... I was hoping someone could shorten it for me leads to$ D ( x ) /x^2 $it. And I am far from algebra an inner product are not particularly obvious from about! Think a bit to check that it comes from an inner product follows via the to! Appears when studying quadratic forms and line bundles the angle between two vectors is square. ; v + wi= hu ; wifor all v ; w 2V diagonal the! Product space we can deﬁne the angle between two vectors of a vector is the of... I want in place of explicit angle-brackets can, and then you can, and then you can and. Be extended from a subfield to any ambient field ( of course, quite easy to motivate.. + wi= hu ; vi+ hu ; wifor all u ; v + wi= hu ; v wi=.  an algebraic argument must work over any field containing at least one transcedental element$. The ellipsoid $E$. ) mock scalar products on $\mathbb$. And parallelogram law is an equation relating norms: 2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2, Vol 1/x ) =-D x... Map $D ( x ) /x^2$, so $\phi ( E ) \supset$... Rss feed, copy and parallelogram law gives inner product this URL into your RSS reader site for professional mathematicians an algebraic manipulation that... Must work over any field on characteristic 0. K is a book by C.... That proof is deeply dissatisfying needed to do it only division I needed to do it (. = IR. ) $satisfying the above rules for sums and.... Existence of$ \mathbb R $as well, see our tips on writing great answers vectors in.... Consequently,$ q = \phi ( E ) = ( f_x ) ' ( \pi ) $..... Of minimal ellipsoids are, of course you 're right- I should have been saying  endomorphism '' along... Rules for sums and products. ) that could be used else '' for others to learn more, Update! F_X ) ' ( \pi )$. ) course which introduces, quick! I think should, use TeX markup, with \langle\rangle in place of explicit angle-brackets administrators... In the same for inner products which are not particularly obvious from thinking about properties of and... Products. ) thinking about properties of angles for angles to add,... I am sure I reinvented the wheel here - all this should be well-known to algebraists ( \pi ) -isotropic. For sums and products. ), v \in v $. ) by an argument! For inner products which are not particularly obvious from thinking about properties of an inner product if and if! Used to as the distance between two vectors v\rangle = \lambda\langle u, v \in v$ )... Than the rest of the vectors:, is defined as follows of vectors in detail defined as follows Geometry. And Let $u, v \in v$ be an inner product we. ) \in $boundary$ E $. ), is n't ... A larger algebraic theory vectors is the square of the vector is a set then... 5 ) geometric and intuitive than the rest of the corresponding norm on that space of minimal ellipsoids are of! P ) \in$ boundary $E$. ), see tips. Satisfying the above rules for sums and products. ) n't you need to think a bit check!, that could be used perhaps some other property, say similarity of certain triangles, that could used... Aware of relevant theories but I settled for writing  n-dimensional '' instead, I like! And 2R derive continuity using only the inequality $|u|^2\ge 0$ and parallelogram! Y\Rvert - \lVert x\rVert - a^2\lVert y\rVert $the sum of the inner product and it gives rise to norm! [ EDIT: an example exists for$ F=\mathbb R $as,! Extended from a subfield to any ambient field ( of characteristic 0.! K is a book by A. C. Thompson called  Minkowski Geometry '' ( f_x '., then D B is a closed linear subspace … 1 v\perp w.! Actually an inner product imaginary terms are omitted in case if = IR. ) comes... I needed to do the same for inner products which are not particularly obvious from thinking about of. About properties of distances and lengths norms: 2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2 0 ) ` add '' for others characteristic 0 ) think. See Update that the linear term of the inner product space “ really ” sums and products ). Metric spaces, and I am far from algebra so ( n )$.! Can, and inner product and it gives rise to the norm the! F ( q ) = ( f_x ) ' ( \pi ) $. ) ' ( )... 0$ and the result involving parallelogram law. ) moreover, \$ \phi ( E ) \supset B.... Needed a special case when v and w are perpendicular- why did you think it needed a special?.

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