 So, in the previous class we have tried to look at this new concept called that of dual spaces right. We saw that given a vector space V, if we take a collection of all linear maps from this vector space to the field on which this vector space is defined, then this gets a special name, we call it the vector space of all linear functionals and we denote it by the symbol what is this V prime right and we call it the dual space. Subsequent we also saw that given any basis in this vector space V, you can always find a dual basis here. We also showed you explicitly how the members in this original basis and the dual basis are related. So, that given this basis you can cook up this particular basis alright and thereby we have also shown you that the dimensions of this and this must be same of course you did not need that proof because prior to that if you recall we had shown you that if you had V and W and this is a vector space whose dimension is exactly equal to the product of the dimensions of V and W. In this case the dimension of a field over itself is 1, the dimension of V is n, so n into 1 is n right. So, therefore this is obviously going to have the same dimension as V right that is just another observation that you can notice right ok. Then we introduce this notion of the annihilator ok. So, it is very important to know where all these spaces are residing ok. So, what is the annihilator? We said that we start with this vector space V alright and you can take any subspace or for that matter you can actually take a subset yeah something probably we did not make very explicit in the previous lecture you can take a subset of V that subset need not be a subspace, once you take the annihilator of that particular subset in question that becomes a subspace residing inside this right. So, we have shown that suppose you have or did we call it W probably we called it W. So, let us go with that notation suppose you have W which is contained inside V right and then we say that this W not is defined as belonging to V prime such that F of W is equal to 0 for all W belonging to W right. So, this we showed is a subspace not just that we did not stop there we then went on to show that in fact we have not shown that we will show that today. We said that look if your W has a basis say W has a basis given by V1 V2 till Vk alright extend. So, let me write suppose here suppose W has a basis given by. So, extend it to V1 V2 till Vk Vk plus 1 until Vn extended to this basis for V ok. Now look at the dual basis B prime given by B prime is equal to F1 F2 until Fn and then we claim that Fk plus 1 Fk plus 2 till Fn is a basis for the annihilator of W. So, this is what we are going to try and prove I have already given you hints about why this should be true in the previous lecture, but we will write it up more formally today and explain why this should be true and then we will get to the consequence of that which also I have already mentioned in the previous lecture right at the end all clear so far ok. So, why is this true let us try and see ok. So, we will do this in the following manner if this is a basis for this what we shall try and show is look already one part of being a basis is fulfilled that is it is linearly independent because the entire thing this is a basis. So, this is linearly independent set any subset of a linearly independent set is linearly independent. So, the only thing we need to show is that this is a generating set for this we will do this we will show that anything that belongs to this must belong to the span of this and anything that belongs to the span of this must belong to this. So, it is both way inclusion and that will prove that indeed it is a basis all right. So, suppose V ok. So, you would start with any arbitrary V all right that belongs to W all right then what can you say look that V because it comes from W it must be a linear combination of the first k fellows alone right then V is equal to summation alpha i V i i going from 1 through k consider any f belonging to span of f k plus 1 till f n all right and look at f V which is equal to what what can I say about this if this is coming from a span of this then this is easily written as summation beta i f i i going from where to where k plus 1 through n right let us keep this V inside for the time being. But now in the next step I am going to write this V as this. So, what happens then just look at this i going from k plus 1 through n beta i f i summation let us change the index may be alpha j V j j going from 1 through k right but again because of the linearity I can pull it out. So, I have summation i is equal to k plus 1 through n beta i alpha rather summation g going from 1 through k alpha j f i right V j yeah. But look can i and j ever be equal because of the way the summations are running j runs from 1 through k i runs from k plus 1 through n and f i V j cannot be anything other than 0 unless i and j are equal. But this the way the summation has been taken here rules out such a possibility I know it sounds a little convoluted because I have taken both the sums are a time. But I hope that it is clear why this must be 0. So, what have I proved I have proved that if I take any vector which belongs to the span of these fellows then that vector must be in the annihilator for W right. So, therefore, this means f belongs to annihilator of W which means I have proved one side what the fact that the span of these fellows is contained inside that is one side of the inclusion done. Now, suppose I start with the other side. So, naturally this is contained inside what just a while back I mentioned this right right no doubts about this. So, in general let me just forget about this intermediate step and let me say let f is equal to summation let us say gamma i I have grown tired of alphas and betas. So, gamma i f i i going from 1 through n at most it can contain all of those n fellows because it is coming after all from this dual space of linear functionals. Now, what is it that cuts it out as special that defines the fact that this f must be in the annihilator of W the moment you give me any vector from W this function must vanish must kill it off must annihilate it right. So, what can I say about this consider. So, let us say say V 1 V 1 is definitely a legitimate member in W because of course, I have erased it now V 1 through V k were indeed members of a basis for W. So, at least V 1 through V k themselves are members of W right. So, if I consider V 1 then this V 1 must be annihilated by this, but if V 1 needs to annihilate this then what must happen. If you pass on V 1 as an argument here look at f V 1 what is that equal to will any term other than f 1 survive no right because f 2 of V 1 is 0 f 3 of V 1 is 0. So, on only f 1 V 1 is going to survive and V equal to 1 which will just filter out gamma 1. So, that is equal to gamma 1, but because it is from the annihilator of W this implies that gamma 1 must be 0 right. Similarly, gamma i is equal to 0 for i is equal to 1 2 until k which means that this summation essentially no longer remains a summation of all those n fellows, but rather it must be restricted to be a sum of only the last n minus k fellows. Because each of the coefficients of the first k fellows must vanish by identical arguments if you now pass on V 2 gamma 2 must be 0 if you pass on V 3 gamma 3 must be 0. Similarly, if you pass on V k gamma k must be 0 beyond that I cannot argue because beyond that V k plus 1 onwards they do not belong to W any more right. So, this means f is equal to summation i going from 1 through sorry not 1 k plus 1 through n gamma i f i which means that f belongs to the span of f k plus 1 until f n right which you may now complete by drawing the inference that this means the other side of the containment is also established right. So, in view of these two containment we have the fact that this turns out to be a basis for the annihilator of W and therefore, so this part is clear any questions up until this point ok. So, therefore, we now have this very interesting result which says that you start with any vector space V look at any of its subspaces say W of course, we are talking about finite dimensional vector spaces in which case we can go ahead and enumerate the basis elements and order them in such a fashion then we can always say that the dimension of W plus the dimension of the annihilator of W is equal to the dimension of V. If you like that is also equal to the dimension of V prime basically summarizing all the results we have derived since the previous lecture in terms of dimensions of course. Any doubts on this point so far yes questions ok. So, I promise that I will talk about extending the idea of this dual to a double dual and so on and so forth and I will just briefly touch up on it because there is a very interesting thing that happens when you go from first we went from V to V prime and now when you go from V prime to V double prime which is a double dual basically the double dual is defined in an analogous fashion to the dual itself. Now, you can sort of forget about this and just say that look at objects in this vector space this is a vector space we have seen if this is a vector space just look at it as an independent vector space in and of itself and define functionals on this vector space. So, the vector space of functionals on this vector space is the double dual of the original vector space. In other words this is nothing, but all the functionals on V prime and now you can like you know really push your luck and say oh analogously I can go on extending of course, you can there is no limit to extending this to double dual triple dual so on and so forth see because every time you come up with a new dual a case a dual or something it is of vector space in its own right and therefore, you can always define linear functionals on that vector space and go ahead and cook up another dual. But what makes it interesting is to look back every once in a while and see the connection between these two which may not be very obvious or apparent, but it turns out to be a very interesting relationship in this case because as I said you know that analogy all equal, but not all are equals right some are more equal than others. So, as I said these two are more naturally isomorphic these are all isomorphic right because we have argued that dimension of this and the dimension of this must be same by the same argument you can extend dimension of this and dimension of this will be the same. So, they are all isomorphic undoubtedly and because it is a transitive relationship. So, this is also isomorphic to this you can already guess that, but in order to cook up an actual isomorphism that is a linear bijection between this and this unlike for the case of the bijection between these remember we had to cook up the dual and the double dual and relate the v with the f you do not need any construction of basis it is a very natural way of relating elements in this and elements in this. So, we will just take a close look at that ok. So, what does this fellow contain I will draw a figure again contains vectors like v 1 v 2 some v n and you know tons of other fellows right this fellow it contains those functionals f n and lots of other functionals that are linear combinations of those fellows. So, what should this fellow contain this fellow should contain possibly some gamma 1 gamma 2 gamma n I do not know, but what are these gamma supposed to do each of these gammas are supposed to act on every member here right and spit out what exactly some scalar right. So, you might wonder what is it that is so special about these two sets and here is what makes it so interesting. So, what I am going to do is I am going to replace this with gamma v 1 gamma v 2 gamma v n like. So, in general if you have a v here I will have a gamma v here and I am going to talk about the mapping that takes me from v to gamma v, but I have not defined this entity called gamma v yet. So, straightforward way of getting this gamma to spit out exactly a scalar by acting on fellows here is this we just define gamma v acting on f results in f v and what is f v f v is a scalar right any functional here acting on a fellow here spits out a scalar and I am just going to define this gamma v in this manner such that gamma v whenever it acts. So, gamma each for each gamma v you can sweep it over every member here because now for this gamma this is now my domain and my range or my co domain is the field f. So, this gamma must take in fellows here and spit out fellows in the field just check whether it does it every functional that you give me from the dual yeah this fellow this gamma v just passes on the argument v to that particular member in the dual and that will result in a scalar. So, it is a very natural way the association between v and gamma v happens in a very natural manner, but I have not yet proved or I am not done showing that this is indeed this relation that I am claiming here the so called natural isomorphism I have not yet shown that this is indeed going to be linear or 1 to 1 or on 2 and all those things that are required right, but when I do that in a few minutes from now maybe a couple of minutes you will see that I have not used any basis argument here not explicitly at least and that is what is so beautiful about this. So, in order to check that this is indeed a linear mapping what do I have to do I will have to check if v maps to gamma v then does let us say alpha v plus alpha v 1 plus v 2 let us say. So, does this map to gamma alpha v 1 plus v 2 that is the question of linearity that is the question of linearity. So, what happens is this true just let us pass this on. So, what is gamma alpha v 1 plus v 2 of f is just f of alpha v 1 plus v 2, but because f itself was linear that sees me through because now I can say this is f I can pull the alpha outside v 1 plus f v 2. So, what does this mean this is alpha gamma v 1 plus gamma v 2 acting on f there is no doubt about this of course, there is another kind of. So, because these have multiple identities you see. So, there is another way of this another level of linearity that you have to check this is one level of linearity here, but the other level of linearity is it is acting on fellows f here. So, if I have chosen f and g and then what would have happened. So, on the other hand if I had chosen for example, gamma f 1 plus f 2 gamma v what is that even there that linearity would hold or let us say just some alpha bar yeah I just I am just going tired of alphas and betas in the same symbols. So, let us just call it alpha bar. So, then what is this can you not relate it in the same manner right this is basically gamma v and you can pull out the alpha bar also why is this possible to do what is that the art of this kind of a linearity here in this argument here can you argue why this must be true. So, what I am saying here is basically this f 1 plus gamma v f 2 how do I argue about this. So, basically this means that you have to take this fellow and pass it on an argument of v 1, but because these are both individually linear functionals again it really matters not right you see that there are two different flavors of linearity here coming in, but both of these are they check through right. So, the only thing that we now need to worry about in showing isomorphism that is relating every v here to a gamma v here is to ensure that it is 1 to 1 I would not check the on toness because I know the dimensions are equal checking for injectivity when the dimensions are equal suffices for surjectivity as well. So, now if I want to check the 1 to 1 nature I need to check the kernel right. So, what is the kernel of this relationship yeah. So, when I am saying that I want this to be identically 0 what v irrespective of whatever function whatever the argument f is right. So, the question I am asking so it is very important to frame your questions correctly once you have framed the question correctly everything becomes very obvious. So, question is which vector v in the vector space v leads to gamma v f is equal to 0 for all f now it will make sense because it has to identically vanish. So, therefore, it must be true for all f of course, for all f from v prime is that clear. If it turns out that only the 0 vector is the answer then I would have shown that this relating v to gamma v is not just a linear relation, but it is also 1 to 1 and because of the equality of the ambient vector spaces also on to therefore, and thus a bijection a linear bijection right. Any questions on this? So, is it true that only v is equal to 0 satisfies this? Well of course, if v is not equal to 0 one power always be able to find some f. So, that this is not 0 I have to make this vanish for every f unless v is equal to 0 v is equal to 0 is the only vector such that f v is equal to 0 for all f because otherwise remember what this fellow is? This fellow is nothing but f of v and I am saying that you have to find me a v ok. So, in other words in other words find v such that f v is equal to 0 for all f that is the equivalent condition. I am just trying to break it down piece by piece into simplified terms that are obvious. So, this is clear when I am asking for this I am basically asking for this. If I am asking for this I am asking for nothing, but v is equal to 0 because if v is not equal to 0. So, unless v is equal to 0 there exists f such that f v is not equal to 0. Why? If v is not equal to 0 then go ahead and choose this v as the first element in a basis. If nothing else works just the brute force argument if v is not equal to 0 any non-zero vector can be the first element in the basis and then go ahead and build an entire basis from it construct its dual basis. In the first element in the dual basis itself will not vanish right by the way we have cooked up the basis and the dual basis f i v i is 1. So, f 1 of v 1 is going to be 1. So, if you have found a non-zero vector v choose that as v 1 and build your entire basis from v 1 through v n choose the corresponding dual basis accordingly f 1 through f n then f 1 is exactly one such functional for which f 1 v 1 is not equal to 0, but that would have violated this requirement. So, you cannot have indeed unless v is equal to 0 you cannot have any other vector which identically vanishes for every other functional right. So, therefore, the kernel is trivial yeah therefore, this is a linear 1 to 1 mapping from v to the double prime the double dual of v and we have not used any basis construction, but we have just given you a recipe for getting a bijection why is it a bijection again to reiterate because v and v double prime have the same dimension. So, injection implies bijection linear injection implies bijection right. So, therefore, they are natural isomorphs or isomorphisms between these two ok no need for going into descriptions about what their individual basis are right ok we will move to the next module.