the meaning of lambda-i

More on the meaning of λi in the dual space. PhL 8.24.15 Every time I think I understand it, it just slithers away like water in my hand. It slithers away when I try to write down a coherent explanation. Here I think Plan A3 is a coherent explanation, after failing with some earlier attempts A0, A1 and A2. We shall see if it stays put. Plan A0. 2 Plan A1. 3 Plan A2 4 What Sjamaar says 5 Plan A3 6 Plan A3 Restated 7 Supporting Work 8 Overview In the following approach (called "A") I try to find a specific set of vectors v(n) which have coefficients c(n) and then I define C(n)(v) ≡ c(n)T v and try to show that C(n)(vj) = δn,j so λn = C(n) . In Plan A0 I use v(a) where (a) is a continuum index and things done don't work out because of the fact that "a" is continuous, although I am able to come with at least some functional C(a). In Plan A1 I instead select v(a) = v(n) = en so now the selected vectors have a discrete integer index. At the same time, I choose the specific basis vi = ei . This approach does lead to C(i)(ej) = c(i)j = δi,j and this explains Sjamaar as λi = C(i). The problem is that Sjamaar comes up with his λi without having to say vi = ei, so he is more general. In Plan A2 I try setting v(n) = vn instead of en, in order to be more general. But then C(n)(vj) = (vj)n and this things don't work, since I am looking for C(n)(vj) = δj,n. This forces me to take vn = en which I don't want to do. In Plan A3 I make use of the reciprocal base vectors qi where qi vj = δi,j . Specifically, I suggest that maybe C(i)(v) ≡ qiTv is the correct C(i). This is the correct solution and then C(n)(vj) = δj,n so I then identify λn(v) = C(n)(v). In order to get this solution from the methods of Plan A1 and Plan A2 for general basis vi , I have to obtain the v(n) as the solution of an obscure eigenvector problem. This solution then allows the basis vi to be arbitrary, as Sjamaar says. I restate this plan in Plan A3 Restated. In Supporting Work I show that the qi are a basis if the vi are a basis, and that the λi are a complete basis for the functional space V*. Plan A0. This is a generic idea where you label selected (non-basis) vectors in V this way: v(a) 1. Consider v(a) = Σj=1n c(a)j [vj] Here the vectors {[vj]} form a basis for vector space V, possibly non-orthogonal. The notation [vj] has j being a label, not a vector index. The quantity vi as the ith component of vector v is not the same as the quantity [vi] being one of the basis vectors of V. Don't get confused about this! The object v(a) is some specific vector in V and we have written it as a linear combination of the {[vj]} with real coefficients c(a)j. Some other vector v(b) in V would have some other coefficients c(b)j . 2. We can think of a column vector c(a) and a corresponding row vector c(a)T. 3. Consider this scalar product c(a)T v where v is some arbitrary vector in v. This dot product is a real number in R1. 4. We might then define the following linear functional C(1) which acts on an arbitrary vector v in V: C(a)(v) = c(a)T v = Σi c(a)i vi vi = component of v We do not bold the functional C(a) because that is the name of a single scalar function. We can write C(a) : V→ R 5. One might say that the scalar linear functional C(a) "is represented by" a vector object c(a)T . If we want to know about the action of C(a), we need to have c(a)T and then the answer is C(a)(v) = c(a)T v. 6. The "dual space" V* of V is defined as "the space of all linear functionals on V". Therefore, we have found one element of V* : we know that the functional C(a) V*. The row vector c(a)T is NOT an element of V*. Had we started with specific vector v(b) we would have generated another functional named C(b) and we would then have C(b) V*. It certainly seems then that for each vector v(a) in V, we obtain a scalar linear functional C(a). This makes you think that maybe the vector space V and the dual space V* have the same number of elements! 7. There seems to be a 1-to-1 relationship between the set of vectors v(a) in V and the linear functionals C(a) in V* (both sets are uncountable). There also seems to be a 1-to-1 relationship between scalar linear functionals C(a) and row vectors c(a)T . So we have "three sets of things" to think about. I suppose you could say that c(a)T is an element of a space called VT. So then we have V, V* and VT. 8. The following facts are true: C(a)( v(a)) = Σi c(a)i (v(a))i C(a)( v(b)) = Σi c(a)i (v(b))i C(b)( v(a)) = Σi c(b)i (v(a))i C(b)( v(b)) = Σi c(b)i (v(b))i 9. Now how do we make any connection to the object λi on page 84 Sjamaar? It is referred to as a "covector of V* " so that means it must be a linear functional like C(a). But it has an integer index i, which differs from the continuous index a. We are given this fact: λi(v) = ci where v is an arbitrary element of V. But all this really says is that λi is a linear functional on V. Let's compare this with our result above, C(a)(v) = c(a)T v = Σi c(a)i vi Then look pretty far apart, don't they! Plan A1. Suppose we go back and instead of selecting all our v(a) vectors as arbitrary vectors in V, suppose we choose them as the usual unit vectors in V. Then for example, v(1) = (e1) = (1,0,0...0)T Then we could at least have numeric indices on things instead of a continuous index. Tracking from the above, v(1) = Σj=1n c(1)j [vj] Suppose now, at the same time, we select the basis vectors of V to also be the ej so [vj] = (ej). Then the above expansion becomes (e1) = Σj=1n c(1)j (ej) More generally, we would have v(n) = (en) = Σj=1n c(n)j (ej) It is obvious looking at the above, that in this case we have c(n)j = δn,j This is a statement about the vector c(n) in the space VT. Now consider C(a)(v) = c(a)T v = Σi c(a)i vi so C(n)(ej) = c(n)T v = Σi c(n)i (ej)i = c(n)j Now we combine the last two results above to get (changing integer n to integer i) C(i)(ej) = c(i)j = δi,j This looks very much like Sjamaar's page 84 statement that λi(vj) = δi,j But in order to get the above result, we needed to have [vj] be the specific basis ej. Maybe we can redo things to avoid that necessity. Plan A2 This time, assume a generic set of basis vectors {[vi]} for V. And this time, assume that v(i) = [vi] = one of the basis vectors as our selected vectors with which to define functionals. Then write v(n) = Σj=1n c(n)j [vj] v(n) = vn Then in this case it seems quite clear that we must have c(n)j = δn,j just as above. What do our functionals look like now? C(n)(v) = c(n)T v = Σi c(n)i vi which we evaluate at [vj] C(n)(vj) = Σi c(n)i (vj)i = Σi δn,i (vj)i = (vj)n Now we nothing at all about (vj)i for a general basis vector [vj] so we cannot continue our development here to where we think we want to go. However, suppose instead we evaluate at the ej vectors: C(n)(ej) = Σi c(n)i (ej)i = c(n)j = δn,j Conclusion: In this Plan A2 we have done the following: defined the functionals C(n) based on selected vectors v(n) = [vn], where {[vi]} is an arbitrary basis of the space V. These vectors need not be en. Specifically, we expand these v(n) to find their coefficient vectors called c(n), then we use c(n) to define C(n) as C(n)(v) = c(n)v. We then find that C(n)(vj) = (vj)n which is not too helpful C(n)(ej) = (ej)n = δn,j So in the first step we can choose either to have the {[vi]} be arbitrary, or we could select them to be the basis {[ei]}. Regardless of what we do on that matter, the only way to get δn,j is as shown above. Rewrite this way C(i)(ej) = δi,j Now the link to Sjamaar must be this: λi ≡ C(i) = a linear functional in V* Then we have λi(ej) = δi,j and this seems more specific that the claim made on page 84 A. What Sjamaar says So how does Sjamaar obtain what seems to be a more general result? Let us examine each of his densepack steps: (1) v = Σjcj(v)[vj]. I have to show that the cj are associated specifically with v. (2) define a covector λi(v) ≡ ci(v). But a covector must be a linear functional, so he must be saying here that λi = ci = a linear functional. Alternatively, he is saying that λ(v) = cT(v) = a vector in VT. We have to put a label on ci because it is specific to v. I feel Sja notation is simply no good. Does he do something different in his new update? Yes, he has completely changed the discussion there!!! Here is his new discussion: But this is the exact same discussion where he has taken vi → bi as the basis (letter b) and λi = βi. He now is writing b = Σjcj(b)[bj] so again we need a b label on the cj(b). Plan A3 Another interpretation of Sjamaar. I know that, given a basis {[vi]} of V, I can find a unique set of dual basis vectors qi which satisfy this rule qi vj = δi,j for all i,j The vectors {[qi]} are really an alternate basis for V , not for V*. Example: Think of the 3-piped in R3 spanned by the en which form a basis. The three area vectors En form an alternate basis in R3, the reciprocal basis. Now how can I define a functional in V* from this? C(i)(v) ≡ qiTv Very good. Once the vi are known, the qj are completely determined. We then have at once C(i)(vi) ≡ qiTvj = qi vj = δi,j So then we can define λi(v) ≡ C(i)(v) ≡ qiTv and then we have our functional λi and it does the general result λi(vj) = qiTvj = δi,j just as he says! Note that the C(i) defined in this Plan A3 are not the same as those defined by me in Plan A2. For Plan A2 I had C(n)(v) = c(n)T v where c(n) = en. In Plan A3 I have c(n) = qn . I will now show that Plan A3 is NOT the same as Plan A2 where you just select v(i) = [qi] : Redo those Plan A steps again: We now try this third choice for the reference vectors v(i) = [qi] v(n) = Σj=1n c(n)j [vj] v(n) = qn Then comparing we have qn = Σj=1n c(n)j [vj] // this does not say qn = cn Then for one thing we can do this: qn vi = qn ( Σj=1n c(n)j [vj]) = Σj=1n c(n)j qn vj = c(n)i = δn,i What then is our functional? C(n)(v) ≡ c(n)T v = Σi(c(n))i vi = vn But this is the wrong functional to choose. If you choose this, you get C(n)(vi) = (vi)n ≠ δi,n So again my method does not produce the λi of Sjamaar. So what vectors might I select for the v(n) to make Plan A2 agree with Plan A3? Consider: v(n) = Σj=1N c(n)j [vj] qi v(n) = c(n)i v(n) = Σj=1N (qj v(n)) [vj] This looks like a messy system of N equations in N unknowns that I would have to solve for v(n). I just showed on scratch that you end up with an equation of the form v(n) = Av(n) where A is a matrix which is a function of the qi and vi vectors. This then is an eigenvalue problem. Leave it there! Sjamaar's discussion is still no good in my book! Here is my replacement: Plan A3 Restated Let {[vi]} be some arbitrary basis for V. Find the vector set {[qi]} called "the dual basis" for which the following is true: qiTvj = δi,j I know from tensor doc that the solution set {[qi]} exists and is unique. In general the {[qi]} form a basis for V. Define a linear functional λi as follows: λi(v) ≡ qiTv Then it follows that λi(vj) = qiTvj = δi,j just as Sjamaar states in page 84 A. If we happen to choose vi = ei, then we will find that qi = ei as well. Comment: My presentation works specifically for Rn where I know how to compute the qi if I am given the vi starting basis. But suppose things are more abstract for space V ≠ Rn. Then maybe you can still define the λi completely by stating the action of λi on all the basis vectors vi. So the definition you make is that λi(vj) = δi,j . This I think is the way Sjamaar states things. However, it might not be obvious that you really can impose that definition and obtain a set of λi that work. For Rn I know it works, and Rn is generally where Sjamaar wants to work. Supporting Work (1) show that the qi are a basis in some space. That space must be V, so qiT basis for VT. (2) in tensor doc I seem to have omitted a proof that if the bi are complete, so are the Bi. Do that proof and think about maybe adding it to tensor doc. (3) Show λi is complete basis of the dual functional space. (1) Assume that the bi are a basis. Then we can write Bm = w'mnbn Want to show that one can write any vector v in V this way v = Σn αn Bn so the αn must exist for any v. Try this constructive derivation: try, αn = (bn v). Then we would have v = Σn (bn v) Bn For this equation to be valid, the components of both sides on the complete bm basis must be equal. If they are equal, then we have proven what we want. So bm v =?= [Σn (bn v) Bn] bm = Σn (bn v) δnm = bm v But this is obviously true, so the dual basis is a complete basis for V. Conclusion: (1) qi are a complete basis for V just as vi are a complete basis for V. (2) qiT are a complete basis for VT (3) If you think of VT as the "dual space" whose elements are row vectors, then you can say that qiT form a basis in the dual space. However, above I am thinking of the dual space as a space of functionals, not vectors, and I have λi(v) = qiTv as the functional λi. (3) Proof that the functionals λi are complete in V* Must show that for arbitrary λ in V* you can write λ = Σiciλi. If you can find the ci that work, then you are done with this proof. Now consider λ(v) = Σiciλi(v) = Σici (qiTv) λ(vj) = Σici (qiTvj) = cj So if you can write λ = Σiciλi, then the coefficients must be ci= λ(vj). We have no issue with whether λ exists or not, it is just a question of whether we can expand λ as shown. Thus, we can for sure compute the coefficients ci ≡ λ(vj). But then these coefficients do the job! Sja shows also unique solution. So I think with this use of the dual basis qiT of VT, I understand that the λi form a complete basis for the dual space of functionals V* !