 So, when we take two objects in this quotient space, say v quotiented by u, I am not repeating all that I have written like u is a subspace of v and so on and so forth, why now I expect you to be familiar with that already ok. So, now when we take two objects in this, what are we talking about? We are talking about two sets. So, what we are trying to define is an addition operation between two sets and it must make sense because again like you have seen you might choose v, your friend might choose v hat and they might both end up with the same object eventually, but when you are adding the same objects, do they actually lead to the same object again? I mean the way you are defining your addition, if you define 2 and 3 as two numbers and you add them you should get 5 each time, your friend might say let me call x is equal to 2 and y is equal to 3. So, your friend x plus y is also equal to 5 and your 2 plus 3 is also equal to 5 right. Now, it should not matter by what name you are calling that object or your friend is calling those two objects that sums must also agree. So, that is the consistency of that addition that we have to first be aware of, with that in mind we will define the addition in this manner for v1 plus u, v2 plus u belonging to v quotiented by u. We define this addition v1 plus u. So, I will use a different color because this is not this addition, but actually the addition we are defining right that is the definition and the moment we have defined this we have to question ourselves if this even makes sense that is how do we check that? Suppose you take v1 plus u to be this object your friend takes this object to be v1 hat plus u, you take this object to be v2 plus u your friend calls this object v2 hat plus u. However, upon adding your version of the objects that is v1 plus u and v2 plus u if you have ended up with v1 plus v2 plus u like. So, your friend should also end up with the same affine set at the end of the day which is representable by v1 hat plus v2 hat plus u is that clear or have I said a lot of things all at one go. We will go slowly suppose v1 plus u is equal to v1 hat plus u quite possible we have seen that multiple vectors when adding a bias to the subspace can lead to the same affine set there is no unique way of representing it and v2 plus u is equal to v2 hat plus u. Now, this definition is sacrosanct we have defined it in this manner what needs to be checked is that when you are adding fellows on the left hand side of this equality and your friend is adding fellows on the right hand side of the equality this equality must still carry over. So, for you the sum is v1 plus v2 sorry v1 plus u let me carry on using that color for a little while longer plus v2 plus u is equal to v1 plus v2 plus u and on the other hand v1 hat plus u plus v2 hat plus u is equal to v1 hat plus v2 hat plus u what we shall be interested in checking is do these objects check out do these objects equal one another yeah because if these are not identical then it turns out that whether you call something by a name that makes a difference right and Shakespeare was wrong there is a lot to a name then right. So, what do we have to do then yes yeah so everything on this side is defined because this is another new vector yeah you know what this object is it is a set this is another set this is another set. So, this is the way we are imposing the addition operation we are saying that if you go ahead and cook up a set like so an affine set and another affine set if you add them then all that you need to do is just take the individual bias terms the vectors add them up together that gives you a resultant v and it is equivalent to that resultant bias yes yes we are defining that addition yeah okay yeah if you like no that is the reason I use the color coding here because this addition is just the notation, but this addition is basically denoting what you would have wanted in terms of the colon that is the definition yeah alright. So the question is are these identical and it turns out that they are but how do we establish that well we will use that result that we have just proved a while back few minutes back. So look what does this mean if these two are equal can we not say so observe that v1-v2 definitely belongs to you sorry v1-v1 hat I am sorry yeah that is exactly what we have proved if two of those affine sets are equal then the difference between the bias terms must belong to that subspace right and from the second equality we can say v2-v2 hat also belongs to you excuse me. So now if you add these two fellows in you because you is a subspace so it is closed under vector addition therefore v1-v1 hat plus v2-v2 hat must belong to you after all I am taking two objects in you adding them so that must be closed under vector addition. But now I am going to rearrange the terms a bit and write v1-v2-v1 hat plus v2 hat and this obviously rearrangement of term does nothing to change the inclusion of an object in a vector space or not. So this is true but what does this mean if you now call this as some vector p1 and call this as some vector p2 if p1-p2 belongs to you then is it not the case that we have just seen earlier that p1 plus u is equal to p2 plus u yeah. So let us call this say p1 and call this just you know I do not want to confuse you with too many symbols but p1 plus p2 which means p1 plus u is equal to p2 plus u and then plugging back whatever is p1 you have v1 plus v2 plus u is equal to v1 hat plus v2 hat plus u. So therefore indeed the resultant that we and end up with here yeah this object and this object cannot help but be the same right. So therefore this addition is well defined even though objects or elements inside this quotient space might have non-unique representations you do not really need to meddle with that you do not really need to bother with that. Go ahead and represent an object inside the quotient in any way that you like take any two such objects and carry out this binary operation of vector addition it is well defined that is the biggest hurdle towards the construction of a vector space in the form of a quotient. Once you have ensured that it is well defined it is just a trivial matter to check although I would insist that you do so to verify that it is indeed got a structure of a vector space of course we still need to define the scalar multiplication right. So I will do that next so for alpha belonging to the field f yeah just define alpha and maybe just carry on with that color v plus u is just to make you happy I am going to put the definition symbol and it is alpha times v this is remember the usual scalar product in the vector space v. So this I am not using any color because you already know what this means just like this addition you already know is the usual vector addition plus u and once again you can go ahead and check that it really matters not if you have described this by v plus u or v hat plus u because on the right hand side you will either have alpha v plus u or alpha v hat plus u yeah and again by the same token you can argue because of the closure under scalar multiplication alpha v minus alpha v hat must belong to u right that is exactly the crux of the matter. So again you need me to prove this I think you can just go ahead and verify that this yes oh that is what we have just proved you see this is p1 this is p2. So whenever p1 minus p2 belongs to a particular subspace then the affine sets defined by p1 plus u and p2 plus u turn out to be the exact same that is what we have just proved a while back like whenever the difference between two vectors lies in the subspace then by you know adding like describing it as v plus u and w plus u you end up with the same identical affine set that is what we just proved a while back those three equivalences you remember one was v minus w belongs to you is the same as v plus u is equal to w plus u is the same as v plus u intersection w plus u being non empty right. So by virtue of those we are claiming this right ok. So at least we have clarified that addition and multiplication scalar multiplication are well defined you can go ahead and check that subject to these two operations that we have just described performed on objects belonging to this this is it turns out to be a vector space remember this is a non obvious a non trivial set of operations we have defined because these are operations on sets affine sets ok. This is imposed by definition this algebraic structure that we are imposing is not something that comes naturally but it is something we are doing for endowing it with a particular structure right ok. So before we conclude today's lecture we shall define another important object which is the quotient map. So the quotient map is going to be a map that is defined from a vector space that we are very familiar with precisely v to what we have now learnt again needless to say u is a subspace of v I am not going ahead and defining that right. So how is this quotient map defined let me use a color it basically takes objects such as v in this vector space v and maps it to an affine set before we get deeper into this what is the additive identity with respect to vector addition in the quotient space sorry phi what is the additive identity sorry u where do you say that see what do you need for the additive identity just think back on what is the additive identity whenever you add the additive identity you must get back the same object. So if you take v plus u plus 0 plus u do not you get v plus u back right so this qualifies that this is indeed additive identity in v quotiented by u correct it is very important to understand that this is the additive identity in this vector space right ok. Now let us try and understand whatever we can whatever insights we can glean from this quotient map what is the image of this quotient map which set I mean as in does every object inside this have a pre-image it does right. So observations I mean I am not even going to call them proofs because they are just that just observations the observation is that the image of this quotient map is nothing but this right it is true isn't it because you see it takes objects in vector space v and spits out affine sets. So if you are allowed to sweep over all possible v's you are basically covering all possible affine sets that are possible all possible parallel translates of u which is the entirety of nothing but this by definition itself right. So therefore this is indeed the case ok what about the kernel of pi. So when I am talking about the kernel of pi what am I asking for I am asking for that object inside v which maps to the 0 inside this and the 0 inside this I already know is this because this is equal to the 0 inside the quotient space correct. So I am asking for that particular object inside v yeah the entirety of u if I pluck out vectors inside the subspace u they are precisely all the fellows that get mapped to something plus u but then that is this 0 plus u and that is the nothing else nothing else can map to the additive identity that is the 0 of the quotient space which is this. If you take anything outside u it is a translate of u it can never be equal to u because two affine sets are just either that either parallel or they are the same. So the only fellow that is equal is if you choose this v to belong to u so you restrict your choice of v to instead of belonging to anywhere in v to the subspace u of v right. So this is u and do you see where I am going with this all that I need to establish is that this fellow is linear and I will be in a position to apply rank nullity theorem is this linear right you take alpha v maps it to alpha v plus u right. So the scaling holds goes through right if you take v 1 plus v 2 it is just v 1 plus v 2 here I mean the u does not get affected at all. So it is linear just again I am going to leave that to you as an exercise to check linearity but that is a crucial detail because without that linearity you cannot apply rank nullity but having checked linearity if you now apply rank nullity to this what does it turn out to be this is the domain. So dimension v is equal to dimension kernel pi plus dimension image pi but then these we have already observed. So this is by rank nullity theorem this is nothing but dimension u plus dimension v quotiented by u which if I now write down here dimension v quotiented by u is equal to dimension v minus dimension u exactly what we had tried to intuitively ascertain when we motivated discussions about this right remember three dimensional Euclidean space subspace being one dimensional like a line. So therefore yeah what the quotient being isomorphic to yes planes when the subspace was a plane the quotient was isomorphic to lines right. So that is now established by virtue of this quotient map that we have defined and applied rank nullity theorem to get there right. So this is where we will stop today in the next lecture we will carry on a little forward with this and try to study what is known in the literature as the first isomorphism theorem another very interesting result that this quotient spaces and these sort of constructions allow us to delve into and then we shall move into inner product spaces. Thank you because we have everywhere we have invoked rank nullity for that we needed linearity right unless it is a linear map we cannot talk about this rank and nullity and all this sorry if you have found the kernel and image yeah but you cannot say that it is going to be equal to this no this relation is all subject to think of how we proceeded with the proof of rank nullity at every step we were using linearity right we were allowing at the at the heart of it was the existence of a linear map which told us then that if you know what the linear map does to a basis you know what it does to any vector but unless you have that linearity it may not be sufficient to just describe a map over a basis and say that I am done you might need to define it over other arbitrary points. So, that is the niceness of linear maps which allows us to extend and understand them and study them by just seeing their action on a basis.