 Next, we will study the image of this dual map, ok. Again, now that you have already seen that earlier result, this should not come as too much of a surprise. Should I leave this to you as an exercise to try? Perhaps I should, I mean, these you will find in standard textbooks as well, right. But the important thing is that this tells us a deeper story, these things they fit in. Of course, in terms of matrices again they are very easy to see. But you see sometimes it may be easier to check injectivity and sometimes it may be easier to check surjectivity. Now, what if you are faced with a situation where a given map, for a given map it is difficult to prove injectivity. It turns out there is a way out, you can just go ahead and cook up the dual map and prove the surjectivity of the dual map that is as good as proving the injectivity of the original map. Come to think of matrices again, it is very obvious, it is a special case. But in general, why is this going to be true? So, we have this result. So, maybe I will leave this to you as an exercise, ok and also the dimension. So, you will also be able to show that the image of T prime's dimension is going to be the same as the image of T's dimension. Please do these exercises, ok. It is very important that you practice, get some practice, ok. So, these two exercises I leave to you. But I am going to just talk about this equivalence of injectivity and surjectivity between the map and its dual, ok. So, T is surjective. This implies T prime is injective. Why should this be the case? So, again let me just refresh our memory. It is a map from V to U and T prime is a map from U prime to V prime, right. So, if T is surjective, what can you say? It implies, implies what? What is the definition of surjectivity? The image of T should be equal to U, right, which means that the image of T's annihilator should be trivial, but where, where should it come from? U prime. Thank you. Maybe I should just, you know, put it in bracket. So, that is, but what is that equal to? What is this equal to? Just what have we proved this object to be equal to, right? This implies kernel of T prime. Does it fit? I mean kernel of T prime should be from where exactly? U prime and that is indeed the 0 of U prime, but what does it mean to have a trivial kernel? The fact that it is 1 to 1, right. Again, is this a both ways assertion already? I mean, can I just use that same old trick? So, no matrices, no columns, no rows, nothing on all, all that we have done earlier, right. Once you identify that T and T prime are just related to each other like matrices to their transposes. See, these proves become much more elegant and sophisticated and they require much lesser work. That is why the idea of duals is very important to understand, ok. So, we also have the other result. Maybe that also I will leave you as an exercise to you. The other part which is that T is injective and T prime is surjective. The fact that those are equivalent. You can also try and prove that. So, please complete these exercises. When I leave them as exercises for you, it is absolutely important for your complete understanding that you go ahead and practice them and do them, ok. So, I will just leave that other part as an exercise as well, ok. So, these are the three exercise problems. We will be sharing another problem sheet with you, maybe by tonight or tomorrow, but you will not be expected to submit them before your mid-sem. However, they should help you in preparing for your mid-sem. So, these three, I am not sure if those are, these three are part of that assignment sheet, but even if they are not, please complete these, all right. So, with that we will bring this brief foray into duals and double duals and dual maps and stuff to a close and we will start a new topic. So, so far we have seen sums of subspaces, intersections of subspaces, those individually also turned out to be subspaces. Now, we will look at something else which is products of subspaces and they will also turn out to be subspaces in their own right, ok. So, already please be aware that when we talk about products of subspaces, the constituents that go into making this product may seem like very familiar objects, but the eventual objects that come out as a result of taking this product are vastly different from those individual constituent blocks, right. The same will hold when we next go for quotient spaces, right, just because they are built out of familiar objects, they are completely new objects in their own right. See for example, we often get into this idea of thinking or I am kind of misled into this way of thinking where we say, ok, this is the three dimensional vector space, yeah, you have this x, y and z. So, the x, y plane is a subspace of R3, fair enough, till there it is ok, but then if I say that R2 and R3 is R2 a subspace of R3, what would your answer be? Absolutely, you cannot do all that algebra, see. Geometry is one thing, when you say a x, y plane is a subspace of the Euclidean space R3, that is fair enough, but if you want to represent that x, y plane in R3, you will still require third three tuples with the third tuple being zero every time, you cannot just chop it off, that is why you have this powerful idea of isomorphism. See, when you say that x, y, 0, this is definitely isomorphic with x, y, I have no issues with that isomorphism at all, but I certainly have issues when you say that x, y, yeah, this belongs to R3 because it does not, see the point. Therefore, R2 is not a subspace of R3, alright. The reason why I launched into this is because of the following reason, when you think of products, you have already seen that, you have seen things like these products, the Cartesian products for example. So, for instance, when you define a function, when you say that I have a function f of x, y, alright, such that we normally say it is two variable function, we say it is defined from R2 to R, like so. Now, what we essentially mean is, it is a mapping like so, you could have x and t, of course, t then we would say it R plus, anyway, we will let us not get into that, but the point is what we are saying is this, there is an ordering, you cannot flip the order of x, y. For example, if you take a function x, y, x cubed y plus 3 x, y cubed, you cannot interchange, I mean f x, y is this, f y, x is not equal to f x, y, right. The order matters, right. So, it is this ordering and this is what we essentially mean, but often we write like this. When we write like this, loosely speaking, we are using that isomorphism argument and this is exactly the product, but it would be unfair to say that this, this is equal to this. In the strictest sense, we should not say that, we should resist the temptation of saying so, we should say it is isomorphic. So, it is these products that we are now going to look at. So, what do we mean by these products? They could be very exotic objects, might not be even founded in reality. For instance, I can just say R 3 product with R x 2, okay. Let V, I do not yet know whether it is a vector space, remains to be shown that it is, but let us say this is a product that I am cooking up. What do objects inside this look like? So, they look like basically two objects, one of each kind, one object which is, so there is an ordering here, one object which is V and the other object which is P of x such that this V belongs to R 3 and this P of x is a polynomial of degree no more than 2 with real coefficients. So, that is just an example. If I now generalize this, what I shall have is, it is a collection of objects, k objects, one of each kind with the first object coming from V 1, the second from V 2, until the kth coming from V k, okay. So, it is a list. So, this is V 1, V 2, V k such that V I belongs to R 3, V 2, V I and very importantly, so they have to be defined over the same field because otherwise you cannot go ahead and carry out your algebra, okay. So, this is the object we are going to study now and the first claim as with objects such as these that we are going to make is that this is also a vector space in its own right over that same common field. So, how do we show that this is indeed a vector space? How many of those things we have to check? First, we have to know what it at all means to kind of define the addition and scalar multiplication after all unless we have defined them the question of checking whether it is a vector space does not even arise just a set alone does not make a vector space. So, what are the ways in which we define this? It is the most natural way see that is why we often take it for granted but nonetheless we should still define it. So, what is the vector addition defined on this? The vector addition is as follows we will take. So, this is the plus that is defined in this manner v1 v2 until vk plus u1 u2 uk. So, this is the plus we are defining remember yeah this the plus that is being defined and the plus that is being used in the definition ok I will do that later is going to be v1 plus u1 this plus is coming from vector space v1 as defined in the vector space v1 then you have v2 plus u2 this is as per the definition in the vector space v2 and so on until you have vk plus uk where the last one is defined as per. So, this is the addition the way you define the addition operation accordingly by the same token if you take the scalar multiplication. So, then suppose you take lambda the scalar multiplication that you are defining now with v1 v2 until vk and on the right hand side you will be using those definitions which have already been made for you in the original vector spaces. So, lambda times v1 this is as per the scalar multiplication in v1 then lambda times v2 so on till lambda times vk. So, these are the rules of addition and scalar multiplication now according to these rules our claim is going to be that this is a vector space and you will see that it follows very naturally what do we have to essentially check it just boils down to that check for sub spaces again convince yourself that you take any two objects that have this longish looking list like things and take alpha times the first one plus the second one and check that it is closed under that operation and that closure will suffice to prove that or suffice to guarantee that this is indeed a vector space nothing else because all those associativity and other properties are already inbuilt into this structure. So, we have been doing this on and off so often that we take this for granted, but the interesting thing the first sort of interesting thing that pops up is after this step. So, again this I am not going to really prove this that this is a vector space by now you should be able to follow the pattern and also convince yourself that this is a vector space. But what this will lead us on towards next is what about the dimension of this product? Are there things that are lurking inside? Is there something more interesting about the dimension of this products of subspaces? Knowing the dimensions of the individual subspaces can we say something about the dimension of the overall product? So, what is your guess? Multiply you mean multiply the dimensions? Yes, it is actually more like the logarithm it is where you add. So, the thing gets multiplied, but the dimensions get added. The proof is it will take a bit of writing up, but I will give you an idea we will do the formal writing up in the next lecture. I will tell you how the proof follows and it is very straightforward it is incredibly simple once you see it. So, first check out for all the elements in the basis for this one. So, let me use a different color exactly. So, what you do is you cook up a basis for this one B1, cook up a basis for this B2. Of course, we are assuming all finite dimensional vector spaces right and what do you do? What do you think you have to do? Exactly. So, pad the rest with zeros you know. So, you take the first element in the basis of V1 put it here pad the rest with zeros that is your first element in the overall basis. Take the second element in this basis still keep padding the rest with zeros until you have run out of all the fellows in the basis for V1 right. Then once you are done with that move on to the basis of this now keep this one as zero padded keep all the others to the right of V2 as zero just choose one element at a time of the basis. So, you have first let us say this one has m1 this one has m2. So, this one has mk. So, you see you will end up with m1 plus m2 plus m3 plus dot dot dot until mk. So, you have summation mi i going from 1 through k yeah is equal to dimension V. Of course, I have not proved that this is linearly independent we will do that and that this is a generating set, but this is the way yes no it would not because they are not appearing in the same place. The position the order matters see I told you it is an ordered when you take Cartesian products is the order that matters. So, having a I mean see if you if you think of it that way you look at the standard basis in Euclidean space 1 0 0 0 0 0 0 1 0 0 0 it is after all the one that is changing position, but the changing position of the one itself makes a difference in linear independence right. So, it would not matter if I take the same vector space over and over and keep taking its product with itself it will still bloat up the dimension of the original vector space because by playing around with this position it keeps bloating up the dimension. So, it would not matter we will write up this proof in a more organized fashion I have given you a recipe for constructing the basis for the product subspace, but we still have to show that that said that I am just I just proposed to you is indeed also linearly independent and is a generating set for this product. Once I have done that then that proof is complete that this is indeed of a dimension that is equal to the sum of the individual dimensions and then we will see subsequently in the next lecture relations between these products and the direct sums that we have studied and we will see something very interesting that there is a isomorphism between them because you see in the direct sum at least with 2 you have seen that the intersection is trivial. So, it is just the sum of the individual subspaces and if you have solved the exercise in the second assignment you have seen it is generalized for m direct sum of m subspaces as well. So, already the dimensions match. So, direct sums and this products they are isomorphic we will see all of that in the next lecture. Thank you.