dual space stuff (CH 7)

The matter of dual space and dual basis. PhL 8.15.15 I think this is something I already know about from tensor doc. Recall there from page 75: [ pages 82 and 83 ] ______________________________________________________________________________ Notes on Reciprocity (Duality) 1. Suppose some set of vectors bn forms a complete basis for x-space. Can one find a set of vectors Bn that have the property Bm bn = δm,n ? // duality relation; Ba and ba are reciprocal (6.2.8) As shown below, the answer is normally "yes", and the vectors Bn are uniquely determined by the bn. One says that the set {Bn} is "dual to" the set {bn} and vice versa. If we regard bn as a basis, then Bn is the "dual basis", and Bm bn = δm,n is the "duality relation". Another terminology is that the vectors Bn are "reciprocal to" the vectors bn and vice versa. 3. One can solve for the Bn in terms of the bn. Each Bn has N components, so there are N2 unknowns. The duality relation Bm bn = δm,n is a set of N2 equations. This is basically a Cramer's Rule problem in N2 variables. Since the bn form a complete basis, one can expand Bm on the bn with some coefficients we will call w'mn ( at this point w'mn is unknown), Bm = w'mnbn . (6.2.9) Then from (6.2.8), δm,k = Bm bk = w'mn bn bk . // bn bk = ij(bn)i(bk)j . (6.2.10) Define matrix W' by, W'nk ≡ bn bk (6.2.11) and note that W'nk is symmetric. Then (6.2.10) says δm,k = w'mnW'nk or w'W' = 1 or w' = W'-1 . (6.2.12) Assuming for the moment that detW' ≠ 0, the solution is given by w' = W'-1 . The (5.6.4) "Digression" showed that (A-1)T = (AT)-1 for invertible A, so w'T = (W'-1)T = (W'T)-1 = W'-1 = w' and therefore w' is symmetric as well. Since W' is known from (6.2.11), w' = W'-1 and the Bm = w'mnbn of (6.2.9) have been found. Finally, Bm Bn = w'mibi w'njbj = w'miw'nj bi bj = w'miw'njW'ji = w'miδni = w'nm . (6.2.13) __________________________________________________________________________ So in this tensor doc context, I can regard the bi = Sja vi as some basis vectors in the vector space spanned by the bi, and then I can regard the Bi = Sja λi as a set of row vectors that span the "row space". You can think of each row vector application as a linear functional (which is a special case of a linear map) Bm x = number Bm : Rn → R x in Rn λm x = number λm : Rn → R In fact any row vector acts as a functional in this sense, not just the basis vectors. Fine. y x = number y : Rn → R y in Rn* x in Rn Sja wants to refer to the vector space of column vectors x as V = subset of Rn spanned by the bj = ej. Sja wants to refer to the vector space of row vectors y as V* = subset of Rn* spanned by the bj = ej. Then the space V* is "dual to" V, and V* is "the dual space". This all seems quite innocent. [page 84] Example 7.2 . Here V is a space of real functions (not just a piece of Rn) in the Stakgold sense defined on [a,b]. The functional of interest is μ[f] = ∫f dx which gives you a real number. Somehow I*f = integral value. Sja is only here giving an example of a linear functional, he says nothing any "dual space" to V. [ below I show that this particular functional is just one of a infinite number of such in V* ] Example 7.3. In this example, the functional is μ[x] = v x for some fixed v where now we have the "standard inner product" used in Rn. I have been using this notation all along, while Sja has been avoiding it. Now, suppose vi are the basis vectors of V of dimension n. (bi in tensor doc). Sja expands a general vector v as v = Σj cj vj = Σj λj(v) vj cj = λj(v) So this object λj(v) is a row vector of constants cj which when linearly combined with the vj gives some vector v. I would say then that λj(v) = v vj and then v = Σj (v vj) vj But I think his definition does not require the existence of an inner product, it just requires a vector space. Working again with his definition, it seems clear that λj(vj) = 1 λj(vi) = 0 i≠ j and you arrive at this just looking at the linear combination. So we then have this set of coordinate functions λi and each of these is a functional mapping Rn → R such as λj(v) = cj. OK, so I think the idea is to try to avoid talking about dot products to be general to any vector space, not just a Hilbert space. Wiki and Stakgold I looked at wiki at this point. It says that the space V* is the space of all linear functionals on V. So let's look again at the above examples. When V = Rn which is a finite dimensional space, I think this is the most general linear functional you can write: v = Σj=1n cj vj = Σj λj(v) vj cj = λj(v) Now exactly what is the linear functional here? I would say that each of the λj functions is a linear functional. I show linearity below. So for each j we have λj: Rn → R. Thus, we are aware right away of a set of n different functionals on V. Below we show that these λj functionals form a basis for V* and therefore V* has dimension n. Wiki would say that the elements of V* are called covectors, so therefore each of the λj functions would be a covector. Wiki allows that λj(v) = [λj,v] = <λj,v> are other common notations. Wiki says for finite n V, you can take ei as the V basis and then ei is the V* basis where eiej = δij, so this is the same as my tensor doc idea. The dual space is the space of row vectors. Wiki goes on to talk about V being ∞ dimensional, but does not really address any space of functions. But: Suppose V instead of being Rn is some infinite dimensional function space. Consider μ[f] = ∫f dx where f lies in V. Here μ is a linear functional and we have μ: V → R. So what is the dual space V* in this case? I have displayed only one linear functional, but V* is the space of all such. So I don't have an answer here. Notice that μ[f] = ∫fα dx for α ≠ 1 would not be a linear functional. μ[f] = ∫f dx s(x) might also be a linear functional where s(x) is a weight function. Stak talks about this subject. In Stakgold, a functional is called T[x] where x lies in V. When V is a Hilbert Space, you can state the famous Riesz Representation Theorem Stak chap 2. " Any continuous and linear functional which acts on any S in any Hilbert space V can be "represented" as T[x] = <x,f> (for all x in V) where f is some unique element of V " Here we have x being some vector in V, and the theorem claims that any linear functional can be expressed as the inner product of x with some other unique element of V which above is called f. Thus, the number of linear functionals would be the same as the number vectors x in V. For V = Rn both these numbers are infinite, and you can think of f as being the row vectors.  For a function space where x is a function, I might restate this theorem as " Any continuous and linear functional which acts on any S in any function Hilbert space V can be "represented" as T[f] = <f,F> (for all f in V) where F is some unique function of V " . Then maybe T[f] = ∫f(x) F(x) dx Then the example above happens to use the function F(x) = 1. So then my conjecture of weight functions would be correct. Note that Riesz is in the context of a Hilbert Space, not a simple vector space. Now back to Sjamaar: Lemma 7.4. The λi form a basis for V*. Now what does λi mean without an argument? It is a function so can only exist in a space of functions. He would have then to show that if λ were an arbitrary function, you could write λ = Σ di λi where di are some constant coefficients. If we could construct di, then maybe we have our proof. So evaluate this arbitrary function λ for one of the V-space basis vectors λ(vj) = Σ di λi(vj) = Σ di δi,j = dj So there you have dj constructed. We then have λ = Σ λ(vj) λi and we have then shown that λi is a basis since it can generate any function λ by linear combination. So I guess this looks good, and again we have avoided mention of any inner product. If you had an inner product, you could write λj(v) = v vj but we don't want to assume an inner product, so we just have these functions λj. Fine by me. Because the set of n functions λi spans V*, we must have dim(V*) = n. Fact : The V* basis of functions {λi} is dual to the basis of vectors {vj} of V. Note that λi is a scalar function of a vector just as λj(v) = v vj would suggest were there an IP. Note that the functions λi are in fact functionals since they map to R. Are the λi linear functionals? Let's investigate: v = Σj λj(v) vj (αv) = Σj λj(αv) vj = α [ Σj λj(v) vj ] = Σj (α λj(v)) vj = Σj (λj(αv)) vj Since the coefficients are unique, we have λj(αv) = α λj(v) first part of showing linear. Next, v + v' = Σj [λj(v) + λj(v')] vj = Σjλj(v+v') vj λj(v+v') = λj(v) + λj(v') So YES, the functions λi are linear functions and are linear functionals. No dot product needed. Example 7.5. Let ei be the basis vectors of V, as usual. We know that dxi = eiT which lies in the dual space Rn* . It seems pretty reasonable that the {eiT} form a basis of Rn* and in fact eiTej = ei ej = δi,j so then we can say: " the set {dxi} form a basis in Rn*". I guess one could write v = Σ viei general vector in Rn vT = Σ vieiT general vector in Rn* So in this example you can really associate the dual space with simple row vectors, not so much abstract functions. Suppose we have L: V → W where V has basis vi and W has basis wi. Consider Lvj = some vector in W = Σi li,j wi defines some numbers li,j Again, no inner product yet. Lemma 7.6. Imagine this situation for L: V → W vi = basis in V λi = basis for V* wi = basis for W μi = basis for W* Then compute μi(Lvj) = μi( Σk lk,j wk) = Σk lk,j μi(wk) = Σk lk,j δi,k = li,j Not sure why this is interesting, but it is correct. [page 85] Definition: Sja defines the simple property of a k-multilinear function λ(v1,v2.....vk) This is λ : Vk → R, so there are k distinct vector arguments. If there are two arguments, it is bilinear = 2-multilinear. Some simple examples are given. [ these are in fact multilinear functionals since maps to R ] Clarification: The vectors vi could lie in some space V = Rn where n is unrelated to k. In general vi in V. Do not assume that k = n, that is not implied! Comment: I think this is the obvious generalization of a linear functional of one vector variable to many vector variables in the same space V. Of course Vk = V V .. V which is a direct product space as in my tensor doc discussion. Definition: If you add the property that swapping any two arguments negates λ, then you have an alternating or asymmetric k-multilinear function. The big example here is det(c1,c2....ck) is a well-known alternating k-multilinear function. Sja has discussed this animal earlier, and there I broke it down into two properties, each of which determinants have. He gives a second simpler alternating example that is bilinear. Now Sja does a very weird thing. He defines a single function which has the following name: "λ1λ2....λk" The arguments of this function are going to be v1,v2...vk which are the basis vectors of V. And here is how the function is defined: λ1λ2....λk(v1,v2...vk) ≡ det[λi(vj)] The matrix here is this: λ1(v1) λ1(v2) λ1(v3) ..... λ1(vk) λ2(v1) λ2(v2) λ2(v3) ..... λ2(vk) ... λk(v1) λk(v2) λk(v3) ..... λk(vk) This fancy function is a k-multilinear function because that is the way determinants work. It is also alternating since THAT is also how they work. This function λ1λ2....λk is called "the wedge product of the λ's" function and one writes then [λ1 ˄ λ2 ˄....˄ λk](v1,v2...vk) ≡ det[λi(vj)] I see the mechanical definitions, but I have no idea how this relates to anything I know about forms. Question: Can we identify λi(vj) = δi,j ? Answer: Well, in the current context, the vi are arbitrary arguments of the multilinear function and are general vectors in V. They could be, but are in general not, the set of basis vectors. Only if we happened to evaluate the wedge thing at the basis vectors would be make such a claim. AkV = set of alternating k-multilinear functions. This AkV is in fact a vector space, and as such, it should have some basis. Comment: Recall that V* is the collection of linear functionals on V -- the dual space. For finite V with basis vectors vi we found that this dual space is spanned by the set of functionals λi. Each of these functionals is a function of v where v lies in V. Now we generalize so that V is replaced by Vk = V V .. V. What is the corresponding dual space? Perhaps it is the set of multilinear functionals on Vk. I don't know how many there are right now, but write one of these as μ(v1, v2.....vk). I presume then that the set of all possible such multilinear functionals would be the dual space (Vk)*. That would seem to be the logical generalization. For reasons not yet clear, Sja wants us to consider only multilinear functionals μ which have the alternating property, and instead of (Vk)* he refers to this restricted space as AVk. So before the reasons for being interested in sort of dual space are given, Sja wants to develop some properties of this space, and only then can he explain the motivation. Sja is not referring to either (Vk)* or AVk as a "dual space", but either seems a reasonable generalization of the dual space concept. Naively you might think that (Vk)* = V* V* .. V* and then maybe the basis vectors of (Vk)* would be λi λj ..... λk i = 1..k, j = 1..k etc I will come back to this idea later. This approach however does not work in that alternating property. But maybe this would be the normal dual space of (Vk)* . I think it is. Properties of the space AVk. 1. It is a vector space. Well, (Vk)* is a vector space since it is a dual space, and if restrict to multilinear functionals which have the alternating property, you still get a vector space. Key fact is that if you add two alternators, the sum will be alternating. 2. As a vector space, AVk must have some basis vectors and must have some dimension. [page 86] Define I as our usual increasing multi index. Example 7.11. Here V = R3 with the usual ei basis vectors and the usual eiT dual space basis vectors.