 So, now that we have that cleared out of the way there is another related notion which we shall dwell on for a little longer perhaps in today's lecture and then we will move on to a new topic and that is again along the same direction a dual map. So, we have dealt with dual spaces, dual bases, double jewels, the isomorphisms that underlie these vector spaces and now we talk about a dual map. So, what is a dual map? Suppose you have linear maps from v to u, these are of course vector spaces as we have argued earlier right. Now, let okay, so t belongs to this then t prime this is the dual map. So, let us first see where it comes from and that is going to be interesting to see. So, where does phi come from? Where do you think that does phi come from? See I am defining something new I cannot use too many new things here everything else must be well known. So, this t prime is the map that I am defining now, but I cannot define it in terms of things that I do not understand. So, on the right hand side whatever I have I should know what each of them does. So, where does phi come from according to the definition? Dual of u? Exactly, thank you right. So, this is the way we define the dual map. Now, if I ask you to think about this point a little bit and maybe again related to things that you understand better which are probably matrices what can you think of? What image does this strike in your heads? For example, the moment I said t you should think of a matrix from some n dimensional vector space to an m dimensional vector space and so on. So, an m cross n matrix right. Now, when I talk about this what does it remind you of in terms of matrices? Not really inverse, inverses may not exist as we have seen, but transpose yeah transpose is the answer. The reason why it is important to keep that in mind is because every time you see some results that seem familiar you will know why it seems familiar because it is exactly something you have dealt with transpose of a matrix ok. But nonetheless we will see this in a slightly more abstract fashion without resorting to matrices or without going through that coordinate assignment and stuff alright. So, how does this work? This is of course the composition alright. So, once we have described or defined the dual map it is then befitting that we try and understand this map in terms of its two very important subspaces which are the kernel and the image. But if I just tell you oh the kernel of this is dependent on such and such very complicated objects it might not carry much meaning. On the other hand if I can describe the kernel of this new map that I am calling as the dual map in terms of some known objects in terms of maybe this map then perhaps I will have some better insight. So, that is what our goal is going to be to describe the kernel and the image of this new map this dual map in terms of what we understand about the original map. But please do remember that there is this to be born in mind this is a mapping from v to u this is a mapping from the dual of u to the dual of v ok. So, this is a mapping from functionals on u to functionals on v ok. But if you think of matrices of course it does not look so weird these are basically the row pictures of u these are the row pictures of v right we have seen that when you talk about duals of Euclidean spaces it is basically the columns flip to rows that is what they look like at least right ok. So, with that in mind let us try and investigate the kernel and the image ok that is an interesting observation and you might think oh I suddenly plucked it of course I know it. So, I cheated I just wrote it down like that, but once I do the proof you will see there is nothing much to it really ok. What is it saying? So, let us try and understand first where these subspaces these are subspaces by the way right. So, kernel of t prime is a subspace of what maybe I should just write in small letters here this is v u t prime is coming from l u prime v prime. So, now the first check that they are indeed subspaces of the same vector space after all that has to be otherwise there is no point right. So, let us take a closer closer look. So, t prime is a mapping from u prime to v prime right. So, where does kernel of t prime fit in? u prime. So, this is sitting inside u prime where is image of t coming from? u. So, where is the annihilator of the image of t coming from? u prime. So, at least that preliminary sanity check has been carried out that is very important to do, but now we want to see why this is going to be true ok. So, let us do a quick sketch of this suppose. So, if something belongs to the kernel of this it must be what? It must be a functional right functional in what? Residing where? Inside u prime right yeah. So, suppose this phi belongs to kernel of t prime this means t prime phi is equal to 0. Let me just forget about the argument for now we will bring that argument later is equal to 0, but this means that phi composed with t is identically the 0, but the 0 of what? When you compose phi with this what happens? It means that does not matter what vector v you choose from the vector space v because remember t is picking out fellows from v and mapping to u and then phi is further taking on that fellow from u and taking it to f because phi belongs to u prime. Please try and observe this what is happening here? What is happening through this composition? This fellow is taking a fellow from v and mapping it to u this fellow is then taking that fellow in u and mapping it to f. So, the overall result is that a fellow in v is being taken to f. So, that means, any fellow from the vector space v that I pluck out here is what is happening to it yeah is being taken to 0 is being killed off right because it is being annihilated. So, there you already have the hint of what is to come. So, that means, phi acting on t v is equal to 0 for all v belonging to the vector space v yeah any questions on this so far? This is true, but what are fellows like these also known to belong to? If you can choose any arbitrary fellow v and allow t to act on it what you are essentially saying is that you give me anything from the image of t. So, whenever phi acts on any fellow in the image of t it totally annihilates it right because you are allowed to choose v freely this is identically true means you can choose any v and it is going to go to 0. If you can choose any v like this you can allow t to act on any v it means you can choose any fellow in the image of t. Now, any fellow in the image of t gets pulverized or annihilated by phi right. So, therefore, phi belongs to image t's I have used the whole word image image of t's annihilator. So, I started with something in the kernel of t prime and I ended up showing that it must belong to the annihilator of the image of t. So, one sided inclusion is done, but let me now just flip the direction. So, I will use a different color and again because I am lazy I will just go ahead and tell you to flip the argument and see that this works both ways. You start with something that pulverizes fellows in the image of t. It means that for every v that you can pick out if you let t act on it it belongs to the image of t and such fellows are pulverized by it. That means, phi composed with t acting on any vector is 0 that means phi composed with t is 0 that means t prime by the definition of the dual map t prime acting on phi is 0 that means phi belongs to the kernel of t prime. So, all I am going to do is and just get rid of the suppose here and flip it here right any doubts. So, this is indeed the case that there is both sided inclusion one something belongs to the annihilator of the image of t, it must belong to the kernel of t prime and if it belongs to the kernel of t prime it must belong to the annihilator of the image of t phi no this is phi composed with t it is not phi 0 it is phi composed with t that is why phi. So, first t acts on v and then phi acts on it is the composition right ok. Next perhaps if you are dealing with finite dimensional vector spaces we can also investigate what the dimension of this vector space is going to be right. So, dimension of kernel t prime is going to be equal to dimension of sorry I am going to use the shorthand I am now yeah image of t prime, but just a while back what have we seen. So, dimension of image t plus dimension of image t is annihilator is equal to dimension of what tell me image of t is coming from where image of t is coming from where yeah. So, basically image of t just that is why we do proofs you see we do not memorize results when we did the proof you saw how we went about building this result right we took a basic vector space took its basis extended it until we got the basis for the complete vector space. So, start with image of t as that vector space subspace subspace sitting inside u extended to u right and then the annihilator will complete the number of elements in the annihilator that completes the argument. So, this is dimension u agreed fits in with what we approved right. So, then what happens this is dimension u minus dimension image of t, but now I am going to use the conventional rank nullity theorem dimension u minus what is dimension image of t from rank nullity theorem think of t what is it dimension image t plus dimension kernel is equal to dimension v is it not. So, what am I going to substitute there agreed make sense when I say make sense go back to the matrix picture think about what for a matrix t and its transpose t prime will be the number of rows and columns what will they correspond to the sizes of v or u or which ones and check out if it is indeed true you know that their rank of a matrix and its transpose are the same because the row rank and the column rank are the same. So, just see if this sort of fits in with that picture does it you agree that this checks out see the ranks will be equal the nullities would not be equal of course the nullities cannot be equal right they cannot be because there are different number of columns and rows potentially unless the vector spaces out of same dimension unless it is a square matrix the nullities cannot be the same if the ranks are same the nullities cannot be same unless the vector spaces out of same size or same dimension and that is that difference is exactly what is being captured through this right. So, when I say check out it means basically get your hands dirty and think of matrices right.