 So, now that we have seen linear transformations, we will define a new sort of objects called linear functionals. So, suppose you have a vector space finite dimensional vector space over a field F and we define this object V prime in the following manner which is a mapping from V to the field such that F is linear, this is what we shall now study. Some examples first to at least satisfy yourself that such objects do exist. So, for example let us say V is nothing but say R m cross n. So, it is a vector space of m cross n matrices alright and define F rather F from V to R like. So, how if there is an object A here or rather let us take an object x here it gets mapped to trace you know what is a trace right trace is the sum of the diagonal entries of a square matrix. So, it is the trace of A x for sum fixed A belonging to R n cross m. Can you verify that this is an example? Can you verify that this is linear? See that is all that you need. You need objects in V to be mapped to the field F and you need such mappings in a special manner such that this F is linear. So, what do you have to do? You have to take 2 different x's say x 1 and x 2 and look at the effect of this operation on alpha x 1 plus x 2. So, of course, by the linearity of this trace operation yeah and this A is basically even though x is not square because of the dimensions of A you see it is m cross n. So, this is n cross m. So, this resultant is just n cross n matrix. So, it is a square matrix definitely has a legitimate trace. So, it is going to be a linear map, but remember it is a very special linear map it is not just a map from any V to any W, but the W in this case is fixed it is F. Now, what is the dimension of F over F? Any field over itself? 1 I mean the multiplicative identity is a basis is a standard basis right. So, if I apply the rank nullity theorem right on objects inside this what do you think I can get? What are the possibilities that can come up? Of course, the rank nullity theorem would say the dimension V is equal to dimension kernel F plus dimension image of F. So, what is so special about this particular choice that it makes it you know something apparent immediately. Look at the image of F it is something sitting inside F in this case yeah not R we are allowing any field F right. So, image of F is sitting inside F image of small f that is little f that functional is sitting inside F. So, what could be the possible dimension of this 1 right or any other possibility 0 when can it be 0, but what sort of a functional do we have then in mind if the image consists of only 0. So, it is the identical 0 mapping right. So, if this functional F what to map even a single vector in the vector space V to anything but 0 then what happens? So, let me give this example where is it yeah. So, if there exists V in V such that F of V is not equal to 0 then image of F is equal to F why because of course, if this is not equal to 0. So, let F of V is equal to C which is not equal to the 0 right. Now, if you want any other objects in the field then what you have to do you just have to scale up the F and you will find its pre-image what you are saying is essentially if this F is not identical to the 0 mapping then I can always find a pre-image in the vector space V which maps it to any non-zero element in that in that field. So, if I found 1 non-zero element in the field which has a pre-image then every non-zero element in the field will have a pre-image it is but natural right because of the linearity of F the linearity of F guarantees that. So, suppose you want suppose we want to ascertain the pre-image of alpha in the field then what do we need to do implies let V 1 equal to what alpha C inverse V because C is non-zero right because C is non-zero in the field therefore, C has a multiplicative inverse implies F of V 1 is equal to F of alpha C inverse V because of the linearity we can pull it out. So, it is alpha C inverse F V but F V is just C. So, therefore, this is equal to alpha it is just sort of formal way of writing what we have just argued. If you have at least one vector which maps to anything but the 0 of the field then you have span the entire F with your linear functional. So, the only way that the dimension of this image is 0 is if it is identically the 0 functional which is an a boring case we are not really interested in that. So, in general if you pick out a non-zero functional this number is always going to be 1 it is lower bounded by it is greater than or equal to 0 but it cannot be greater than 1 also because after all mapping into the field and the field cannot be of dimension more than 1 over itself right. So, therefore, we have unless F is equal to the 0 of the 0 of this yeah we have dimension V minus 1 is equal to dimension of the kernel of F straight forward right nothing too great about it it follows from very obvious arguments right ok alright. Now, the next claim is going to be that this object that we have now defined here is not just any arbitrary set of functionals, but it is also a vector space does it require a proof it does not because if you let W equal to F we have already seen L from V to W is a vector space. So, in a special case when W turns out to be just F. So, F is a vector space over itself of dimension 1. So, this being a vector space is not really a surprise it is very obvious is it obvious to everybody that it is a vector space alright. So, I will just go ahead and still write that down vector space over and in fact, it has a very interesting name as well ok with in terms of its relation to V right V prime is the dual vector space 2. So, you wonder what is the relation between those 2 vector spaces that leads us to call them as a dual of one another like why is it a dual why wherein lies this notion of duality in the middle of all this. So, this is clear I am going to erase this sorry such. So, what is it? So, that is so special about this V prime that leads us to give this a special name. So, consider a basis for V given by B is equal to V 1 V 2 V n. Correspondingly consider the set I am not giving it any name yet, but you might be guessing where I am going with this by the nomenclature that I will give it any one of these fellows F i's acting on any one of these fellows V j's from here is equal to delta ij where this is basically what we call the chronicle delta which means that this is exactly equal to 1 when i is equal to j and 0 otherwise ok. So, this is the chronicle delta right. So, this is equal to 1 let me write that using different color just for the sake of completeness is equal to 1 for i is equal to j 0 otherwise that is the chronicle delta definition. So, we have gone ahead and corresponding to a basis here we have gone ahead and defined this set here and we might wonder where we are going with this, but again as I said pay attention to the notation that I am using here and you might be able to guess one step ahead as to where we are pushing all right. The claim is in other words these objects that have defined corresponding to a basis in V these objects remember are coming from V prime right. We see these fellows take objects inside V and map them to of course the special ones and zeros, but they are objects in the field. So, it says that in order to get a hold on any object inside this dual vector space all I need are these fellows that is this is a generating set for every linear functional. Remember what I have done given a basis I have cooked up this set in a very special way because it is related to the original basis in V and now I am saying that this set is a generating set for every linear functional that there is. So, how do we go about proving this here is what we will do suppose. So, this is a quick sketch of the proof suppose f given in V prime suppose f is given in V prime ok that is where we stop all right choose is equal to summation V i times f i. So, this is how I am defining i of course goes from 1 to n what does that mean it essentially means that look this is an object that also has to belong to this. So, this is an object that also belongs to this what I will now show is in other words what I am going to show is that the difference between these 2 fellows maps to the 0 functional that is it takes any object anything that you give it give to this object as an input it is the 0 functional it maps it to 0. This is a 0 of the V prime that is the dual space which means it is a 0 functional and that would then mean that this f can be represented in this manner because if it is a 0 functional identically then this is its additive inverse right. So, therefore, f is equal to f hat right, but if I want to show that what do I need to do if the claim is that it is a linear functional linear mapping after all what do I have to do in order to show that something is a I mean 2 linear functionals or 2 linear transformations agree on every vector in the domain what do I need to do? I just need to show that they agree on any basis set that is exactly what we have seen you can uniquely characterize the action of a linear transformation by seeing what it does to any basis set. So, a linear functional is nothing but a special kind of a linear transformation. So, if we can show that this fellow acting on any basis in this case let it be this basis if every object in this basis set B is taken to 0 then it is indeed identically the 0 functional and therefore, f must be given by f hat and if f is given by f hat then of course, f is spanned by or generated by f 1 through f n in other words B prime is a generating set for V prime any questions on this so far please ask if it is not clear anything is not clear it is just ask feel absolutely free to ask I can start at the beginning it is very important that you understand any linear functional yes I have created an f hat in a very sort of an intuitive choice you might say clever choice depending on how you view it yeah okay exotic choice it is your choice of words not mine but anyway the point is that if you choose this f hat at the very outside what I have done is I have chosen it in a way that is it is spanned by these fellows are generated by this these fellows so that if yeah these are just scalars yeah so now if I can manage to show that this f is indeed equal to this f hat then this f is also representable in terms of fellows in this and nothing but this or as a linear combination of fellows in this so therefore, any object that I pick from V prime is indeed generated by objects inside this set B prime alone that is all I am trying to prove I will prove the linear independence afterwards but as of now let us just try and prove that in order to do that I will have to show that any member in the basis because what is this this is a functional it takes objects in V maps it to the field so if I take a basis for V and show that every object in the basis for V gets mapped to 0 then this is identically the 0 linear functional right so let us try and do that so what is it it is f Vi minus for let us say for V is equal to Vi this is f Vi minus what happens to summation f V ok let us say this is V is equal to Vk so I will just take the kth element so this is f Vi into fi Vk but look because of that chronicle product thing that I have defined this is not really a complete sum this is just one term because unless I is equal to k every other term is automatically 0 is it not so therefore, this just leaves me so f minus f hat acting on Vk is given by this which is nothing but f Vk minus f Vk times delta kk but delta kk is nothing but plus 1 so therefore, this is 0 yeah this is the 0 of the field of course so it is the 0 and this k was not a very special choice right for any k for any k ranging from 1 through n it is taken into 0 so therefore, members in the basis all of them get pulverized and mapped to the 0 of the field so therefore, this mapping this f minus f hat must be the 0 linear functional f minus f hat is equal to the 0 of V prime which means that f is equal to f hat and is contained inside the span of sorry V prime is what I had named it and thus shall remain its name right. So, indeed this choice that I had just pulled out of the hat apparently is now a generating set right. So, if you are trying to push for this being a candidate basis for the dual space the next object that we must get our hands on is to prove that this is a linearly independent set. So, we will prove that in the next module any questions on this what we have here this is clear I hope yes yes we are not calling it a basis yet we are choosing this yeah this chronicle this choice this chronicle delta what do you mean by such thing will not exist like I am just defining this it is the indicator exactly that is exactly the term. So, I would not like to draw a figure here because it is because you know vectors are not after all points right, but if I had this you know very abuse of representation and things like that. So, if you call it V 1 V 2 and sort of things like this then it is this F 1 V 1 is equal to 1. So, it just lights up if you pass F 1 through it this LED glows the others do not glow or so to say it is the indicator yeah right. So, now having seen this our next claim will be that this set is also linearly independent.