 So, the next claim so B prime this is linearly independent once again when we map something to 0 it is very important that you understand what 0 are we talking about the 0 of which vector space this is the 0 of the vector space of linear functionals. In other words to show that summation alpha i f i is equal to 0 v yeah implies alpha i is equal to 0 for all i that is what the linear independence condition boils down to. Yeah of course every field must have the 1 and the 0 right yeah it is 1 and 0 of the field because it is getting mapped yeah if it is all 0s then it is the unimportant like it is the boring case when it is a 0 functional right ok. So, what are we saying here we are saying that ok yeah so this 0 of this v prime yeah so the 0 of the v prime means that every object this means that summation alpha i f i of course i going from 1 through n i going from 1 through n this v is equal to 0 where is the 0 of is the 0 of the field now because I have passed on the argument now. So, this is not in the sense of the functional anymore so this is the 0 of the field for every v belonging to vector space v and that is the most important observation that is it has to be true for every v if it is true for every v I can go ahead and choose some very special v's and conclude certain things about this yeah so choose v is equal to v 1 say for instance and what do I immediately have this means that alpha 1 f 1 v 1 is there anything else really to it because when you take f terms f 2 onwards f 2 acting on v 1 is what 0 so it contains no other term. So, the only thing is this, but what is f 1 v 1 that is 1 so this means that alpha 1 is equal to 0, but is really 1 any special or any more special than any of the other numbers from 1 through n. So, I can just you know choose this i maybe I will take the shortcut while I expect you to repeat this with other values, but I will just take this for any i yeah. So, any i you choose you are led to eventually concluding that alpha i must be 0. So, this means that alpha i is equal to 0 for all i which is exactly what I wanted to show. So, therefore, you have a linearly independent set you have a generating set therefore, by virtue of those 2 claims the big result is that is why I have not erased that part this b prime is a basis for v prime not just that you will see that we call it b prime is a dual basis to b. So, given any basis in the original vector space v I have a corresponding dual basis by virtue of this Kronecker delta thingy here right. I will just work out a very simple example in R 3. So, this is clear right this idea is clear the fact that it says generating set and a linearly independent set and therefore, a basis for the dual space what is it that immediately strikes out the dimensions of v and v prime must be equal therefore, they are isomorphic right, but maybe not in today's lecture the next lecture we shall see there is something very interesting in order to show that these two are isomorphic you had to actually construct basis. So, it is said that there are levels of proof that involve constructions of basis and then there is a higher level of proof which does not go through this basis construction. So, it is like you know it is a good time to read George Orwell by the way. So, there is this Orwellian line that all animals are equal, but some are more equal than others right. So, you have vector spaces that are isomorphic, but some vector spaces are more isomorphic than others and it turns out that there are objects such as double duals. So, you have the vector space v you have the vector space v prime which is a dual to v and then you have a vector space v double prime which is a dual to v prime and it turns out that v double prime is also isomorphic to both of these aforesaid spaces, but in order to show the isomorphism between v and v double prime that is a double dual you do not need to go through any construction of basis. They are what we call as natural isomorphisms. So, isomorphic vector spaces are of course equals in some sense, but in order to prove the isomorphism between v and v prime we had to go through this route of constructing an explicit basis for the second vector space in terms of the first there is this dual v prime as a dual to b when in the next lecture we return and show you that the double prime or the double dual is actually also isomorphic to v we will not require the choice of any basis. It will be a natural isomorphism in some way it might appear a little mystifying at this point, but we shall see that they are actually more equal than this first and not just that if you just expand your sort a little bit you will see that this double. So, the double prime then the fourth prime then the sixth prime and all these happen to be more naturally isomorphic and the first dual the third dual the fifth dual onwards they also happen to be naturally isomorphic I mean it is like some shadow spaces that are happening right their isomorphisms. So, all of them you keep taking duals they keep becoming isomorphic to one another, but that the one hop the skip if you like first to the third or the v to the double dual they happen to be more naturally isomorphic. We will see that proof in the next lecture, but today we shall continue with this and see something interesting as well. So, what is it that we have seen that there is a double there is a dual basis to every basis that we can cook up in the original vector space v. So, let us take the simplest example one can think of. So, let us suppose v is equal to R 3 and let us say b just illustrate the idea b is equal to say 1 0 0 minus 1 1 0 and 1 minus 1 minus 1 I leave it to you to check that this is a linearly independent set the cardinality is 3. So, it is indeed a basis right. So, check to see that b is a basis for R 3. So, by the way what do you think a dual to R 3 would look like? Any guesses what sort of objects will take will ingest objects like these and spit out a member in the field that is a real number. So, if you take this as column vector I put it to you that you take this as R 3 row vectors that is all that is going on all this while we have been talking about this dual, but if you view it in terms of coordinates it is really nothing, but the action of some row vector on these column vectors that will spit out of scalar in the field right. So, now just to explicitly. So, of course I am assuming that you will check and convince yourself that this is indeed legitimate basis for R 3 and once we have the fact that this is the dual space the row vectors is what we are looking for. Any questions here right here? So, now all that we need to construct is a basis for those row vectors those vector space of row vectors 3 tuples right. So, what do we need? We need firstly an object which when we take multiplication with this spits out of 1, but with any of these other fellows it must spit out 0 remember it is a chronicle delta right. So, what must it have over here? See first of all with this and this it must be 0. So, if I am looking for the first object here let us see b prime if I am looking at the first object here my first observation is that the first two entries must be equals should not otherwise this cannot vanish. See when this object this is f 1 acting on v 2 must be 0 right. So, therefore the first two objects must be identical right. What about this? The third object maybe I can just put the third object as 0 it matters not. The first two objects what should I put? Just 1 and 1 check this times this is just 1 this times this is 0 and this times this is also 0 right. What about the second object f 2? I need something that takes this and makes it 0 here, but I need to have 1 here. So, what should I do? And I need to have 0 here again. So, this object I need to make it 0 and these two objects are somehow also badgering me a bit. So, what should I choose? Sorry 0 1 minus 1 exactly right. So, you take for instance this 0 this picks out this and this filters out these two, but their difference is 0. What about the third? So, this is actually the dual basis that I am constructing here. Now you can think of arbitrary abstract vector spaces and you proceed in a similar manner and you should be able to construct duals. If there is a problem just go ahead and go to their coordinate representations that will be simple enough right. Because we have taught you that finite dimensional vector spaces take them to their basic coordinate representations and that is all that you need to understand them. So, what is this? Just to complete the story? 0 0 minus 1 yeah. So, this is indeed the dual basis to this yeah what we have just learned. Any questions on this so far? This is just the example right ok. Now let us look at the following situation v finite dimensional vector space over F b is equal to v 1, v 2, v n is a basis for v. Suppose w yeah I mean I could have started the other way round I could have started with a w taken a basis for w and expended it to a basis for v it does not matter you get the idea. Suppose w which is a subspace of v and subspace given by w it is span of say v 1, v 2 till v k right of course, k is less than or equal to n. So, w is of course, a subspace sitting inside v now we will define a new object w with superscript 0 yeah is defined in the following manner F belonging to v prime such that F of w is equal to 0 for all w belonging to w. So, this is called the annihilator of the subspace w ok why because every time you feed it some object from w yeah every time you feed it some object from w it just takes it to 0 nothing much to read into it except for the definition as of now. Now the first non-trivial claim that shall be made is that this object this annihilator is a vector space of course, over F. So, let me write that in brackets lest it create confusion ok how do we show it is a vector space well you take any two objects F 1 and F 2 I will just a very sketchy proof of this alpha F 1 plus F 2 right if this object where to act on some object picked from w then what should it be because of the linearity we can write this as alpha F 1 acting on that object plus F 2 acting on that object, but each of them individually is a 0. So, therefore, this is indeed the 0 of the field and we are done right. So, the annihilator is indeed a vector space, but the more interesting claim is. So, if your original vector space w which is a subspace of v is spanned by the first k fellows out of this n in the original basis then go ahead look at the dual basis all right. So, maybe I should say that v prime is equal to span of F 1 till F n where F i v j is equal to delta i j like before. So, now if the first k fellows out of this n are a spanning set or a generating set of course, they are linearly independent any subset of a linearly independent set is linearly independent we prove that. So, if this is a basis for w then the last n minus k fellows of this dual basis is a basis for the annihilator if this is true if this is true what can we say about the dimensions of w and it is annihilator yeah the dimension of w. So, if this is true the consequences the consequences that the dimension of w plus the dimension of it is annihilator is equal to the dimension of v and that is precisely I urge you to think about this on for a moment nothing but the rank nullity theorem just think of it in terms of matrices or other things and you will be able to precisely see what I am what I mean by that, but the point is that we have to show that this is a basis is linearly independent under question of course, not because this is a subset of a basis. So, it must be linearly independent the only thing that I need is to show first that this is contained inside span of f k plus 1 till f n and the fact that this span of f k sorry f k plus 1 till f n is also contained inside the annihilator that is the two things we shall prove in order to establish that there is a basis for w naught or the annihilator of w right. So, we will do a detailed write up of this, but I will just give you the idea of how and why this carries forward maybe that will help you grasp the proof better in the next lecture. So, what do we have to do in order to prove that something is spanned by this you take what is the defining property of objects inside this you take any typical fellow sitting inside the annihilator of w it means that that linear functional must pulverize every object inside w. So, when that linear functional acts on fellows which are linear combinations of the first k fellows here it must map it to 0, but look after all every object that is every object that is sitting inside this w naught is coming from where it is coming from v prime remember v prime just like w is coming from v w naught the annihilator is coming from v prime. So, in general an object inside v prime has some special properties and then it belongs to w annihilator. So, any object inside v prime is a linear combination of objects inside this set. Now, if you pass on the special fellows that are residing inside w then they must be pulverized right. So, if you pass on very special fellows like this v 1 v 2 v 3 what do you immediately see if you pass on v 1 to a linear combination of these fellows right or the other way round if you pass on v 1 to a linear combination of where is it the f's yeah these fellows what do you immediately conclude that the coefficient of f 1 must be 0 the others are anyway 0's, but alpha times f 1 v 1 must be 0 otherwise it cannot be an annihilator right. Similarly, for f 2 v 2 therefore, the coefficient of f 2 must be 0. So, when you are saying that any object in v prime for it to also be an object inside w 0 the first k coefficients of these f's must all vanish and therefore, the only non trivial coefficients that will be left with are those of the last n minus k. So, therefore, this is indeed something that is spanned by or generated by these fellows and now it is only trivial to show that if there is any object that is spanned by this and you feed any object from w to these fellows all of them identically vanish because it cannot contain components from v k plus 1 onwards. Any object that is residing inside w is a linear combination of objects from v 1 through v k. So, when any of those objects from v 1 through v k this is the indicator function as your friend rightly said. So, when this indicator tries to check for objects from v 1 through v k they do not light up they are all 0's anyway. So, this is also contained inside this we will write this in more formal terms do not worry about it, but the point is that we will show that this is contained inside this this is also contained inside this and therefore, they are nothing but equal. So, therefore, the dimension of the annihilator given that the dimension of the subspace is n minus k is going to be n minus k where n is the dimension of the vector space. Now, you think about which is the image and which is the kernel and so on and so forth, but I put it to you that this as a consequence of what we shall prove next day is another statement of the rank nullity theorem just put some thought behind this that is the important concept to take home from this. Thank you.