 Suppose, s is just given by 1, 2, 3, 4, 5, 6 right. So, it is just the set of first six positive integers first six natural numbers right. Now, just think about it the moment I am giving you a particular function what am I essentially telling you? What I am telling you is right. If I give you these six numbers I have told you all there is to know about this function because this function is defined only over s. I do not care about what it does over other points because that is not part of its domain right. So, I could have just carried on for this with this till any n. You immediately see that this is analogous to r 6 if my f is r then this is nothing, but very similar or a term that we will come to know later as some isomorphism similarity in structure. This is very similar it is just a column of six numbers it is what you do in r 6 is it not right. So, although this looks or it comes from a very different definition you can immediately associate this with the typical n tuple of numbers or in this case six tuple of numbers that you call vectors in Euclidean spaces. What if I kept this f as the same, but I chose this s to be 1, 2 till you know all possible natural numbers. What do you think it becomes then? Any idea? So, this is the first example is the second example of course, the field is still that of real numbers isn't this the collection of all real sequences right. It is a collection of all real sequences that is what a sequence is right. I give you the nth I give you the number n and you tell me what is the nth term of that sequence that is what a sequence is that is how you define a sequence right. So, in other words sequences that look nothing like what you are used to seeing as vectors are also part of a vector space if you know how to define them in a proper manner right. The third example again let us say f is equal to r and s is equal to say i comma j and ordered pair such that i varies from say 1 through m and j varies from 1 through n. Can you tell me what this is a collection of m by n matrices right. So, these are all examples of vector spaces. In other words vector spaces do not have to look like those columns of numbers anything that satisfies it is I leave it to you to check. I mean it is readily verifiable you do not have to put in much effort, but you can readily check that each of these things where the first one obviously you can make it look like a column of numbers, but the other two they do not look anything like it and yet they come under the same framework by dint of our definition of what a vector space is and by showing that these indeed meet those criteria right. The second important notion after learning a vector space in order to be able to answer that question of x is equal to b is to formally understand what is a linear combination. We have used this on and off and we kind of implicitly assume that we understand what linear combination is, but let us nonetheless write down what we mean by a linear combination. You can think of several others in fact there will be some exercise problems where you will be required to deal with a lot of different types of vector spaces. So, you at least see that the spaces of functions the collections of functions can very well be vector spaces. Yes, no it does not have to be finite the sequences can be infinite countably infinite. So, these are infinite sequences, but if you add to infinite sequences you will get another infinite sequence. So, it is closed under the rule of vector addition. I leave that to you to check that is the reason I have given you these three examples. You should be able to convince yourself that these indeed fit in with the definition of vector space that we have given as per our understanding. You know how to add sequences. So, that is why I have not explicitly told you what the rules of vector addition and scalar multiplication are. Multiplication of the sequence means every term has to be multiplied by that particular scalar. No you can then you do not call it a sequence. So, I was interested in showing you that sequences are also part of a I mean they form a vector space. So, if it is uncountable then you cannot match them with indices. When we talk about a sequence we have some mapping with a set of integers or natural numbers that is that is the only reason we call them sequence. You do not say a 1.73 th term of a sequence. You only say first term, second term, third term, fourth term and so on right. You can think of it in any manner. It is like it is mapping every ij to a real number. This is its domain you see. We are mapping we are interested in functions that take elements from the domain that is from s and map it to this. So, corresponding to every ij pair that you give me this is a bracket. So, corresponding to every ij pair that you give me I am going to be able to tell you a real number. You can write them as a matrix. You can represent them as a matrix. I mean that is what we do when we write them. But yeah for every ij you can think of it like a things over a grid. So, you give me every ij for i ranging from 1 to m, j ranging from 1 to n. So, at every grid junction you have some real number sitting. Now of course, that array of numbers the best way we know how to represent that is through matrices. So, we will just talk about what is a linear combination. Although again I am pretty sure you know what it is. Consider a set V sum V equal to summation i going from 1 through n C i V i with C i belonging to the field and V i belongs to V with of course, n being finite not infinite in this case. You cannot take infinite number of vectors and combine them. Then V is said to be a linear combination of a set V i i going from 1 to n. That is it simple. You have the field. So, from the field you are picking out the scalars. You have the set of vectors from which you are picking out the V i's. Of course, you have to pick out only a finite collection. You cannot take an infinite number of objects and talk about their linear combination. It has to be always finite. So, that is why I mentioned this total number in this sum must be finite. That is what a linear combination is. So, now you see that question we asked about whether those two functions combined anyhow in a linear fashion to yield the third function makes sense. It is the same type of question. Of course, how to answer that question in general we will not deal with right away. But at least you have the idea of what a vector space is now clear. You have the idea of what a linear combination is and we will hope that we will now be able to cast on unknown problems or seemingly unrelated problems in the same fashion as we solved Ax is equal to b. That is the goal we will be driving at. So, now as an immediate consequence of the way in which we have defined these operations I would like to prove a few seemingly trivial results. But we will see later very soon in fact that those are going to be very useful and those results are also not very trivially obvious even though the operations we do here might seem a funnier at times. So, the first result is the following. If you have x y z belonging to a vector space you have x plus y is equal to z plus y implies x is equal to z. Of course, it goes both ways the implication goes both ways it is an if and only if. But which part do you think is maybe requires a bit of thought. Let us say I start with x is equal to x plus the 0 of the vector space. This is guaranteed to exist because of the vector space definition that there must be the additive identity alright. So, now this additive identity can also be written as what? What can it be written as? Can I not write this as so far so good because any element comes from the vector space it is additive inverse must be there because it is an abelian group. And therefore, its sum with its additive inverse would lead to this. Now I can combine this as x plus y plus minus y. Now if I assume that x plus y is equal to z plus y. So, I am going to go in this forward direction the reverse direction I put it to you is straight forward. So, if I assume that x plus y is equal to z plus y I can replace this x plus y with z plus y can I not. So, this is z plus y plus minus y then I open up the bracket and use associativity. So, I have z plus y plus minus y which is equal to z. So, if I start with the assumption that x plus y is equal to z plus y then I am led to conclude that x is equal to z right. But see all these steps that we have done here which might seem like a bit of tomfoolery to begin with it really is not. It allows us to extend our notion of what we did since we were kids we were counting marbles to these abstract spaces because we have imposed this. Had we not imposed these conditions or this on this structure we would not have been able to do this. Things that we take for granted things that we do as trivially obvious are not that obvious right. At least I hope that much is evident from this even if much of it seems like you know much I do about nothing ok. Let us take another instance we will do a bit of this seemingly trivial results we will try to see why these are true. We will also try to see suppose alpha belongs to the field f then and let us say x and y belong to the vector space v then we will see that alpha times x is equal to alpha times y is implies and I have only proved one side by the way here also I am going to do the same thing I am going to urge you to check the other side is true. Any way if alpha is equal to 0 we are going to shortly see that it is going to be 0 on both sides right. So, for now we will assume that alpha is equal to 0 later on we will see that even if alpha is 0. So, I really do not have to put this caveat here ok. So, we are going to again try and see this. So, how do we start with this? We will see that x is nothing, but because I am assured about the existence of the multiplicative identity over the field I can write it in this fashion. See this is just nothing, but I have replicated the condition on the vector space the scalar multiplication where I have said that if you have the multiplicative identity from the field and if you do a scalar multiplication of that element that multiplicative identity in the field with any vector it gives you back the same vector that is the only property that I have invoked already in the definition now I am just rewriting it here, but that also means does it not that if alpha is not equal to 0 I can very well write this as alpha inverse times alpha times x no problems there this one can be written as alpha inverse times alpha. And now again I am going to use another property which I have imposed. So, alpha inverse this is alpha x it matters not right, but I know suppose that alpha x is given to be equal to alpha y. So, in the next step I am just going to write alpha inverse I am going to write this as alpha y so that again I am going to go back it is just a bit of how to change these operations around a bit and this again gives me 1 times y which again by the definition is y. So, given that alpha x is equal to alpha y I must have unless alpha is equal to 0 by this line of reasoning x is equal to y right. So, second apparently seemingly simple result proved I mean I should not even call them or dignify them by calling them a proof these are just bits of exercise I would say, but it is sort of sharpens our ability to see things as to what we can and cannot do. Let us look at let us say we have alpha times 0 vector is equal to the 0 vector for all alpha coming from the field. We have done something similar. So, how do we go about trying to show this please complete the other sides of the proof in those 2 cases. How do we prove this? So, this 0 is just another element in the vector space v. Therefore, when the 0 is added to itself which is itself the identity must give me the same thing. So, can I not write this as alpha times let us just start with this alpha times 0 it is alright. I have just used the property that this is the additive identity. So, when I add this with it itself I get back this. So, this is my first step to begin with. Now, I can write alpha times 0 v plus alpha times 0 v is equal to alpha times 0 v, but where does this element come from because of the closure under multiplication and scalar multiplication this must be another object sitting inside the vector space v right. So, therefore, it is additive inverse must also exist. So, let me add the additive inverse on both sides. See I can take or take out or add identical elements on both sides of the equal to sign like I have just proved a while back right. So, I am not using something that is undefined. I have already proved that x plus y is equal to z plus y therefore x is equal to z and it goes both ways. So, that is the only property I am using I am going to use likes to both sides of the equality that is all I am doing. So, I am going to just add minus of alpha times 0 v plus alpha times 0 v plus alpha times 0 v is equal to alpha times 0 v minus alpha times because of the commutativity of the addition operation. It matters not if I do it on the right or on the left. So, the right hand side of course gets pulverized to just 0 v what happens to the left hand side. I can combine these two because of the associativity and I am just left with 0 v plus alpha times 0 v is equal to 0 v but this is just the additive identity. In other words I have alpha times 0 v is equal to 0 v in exactly the same fashion. I leave it to you as an exercise to verify that when you pick out the additive identity of from the field and do a scalar multiplication with any v you are going to get the additive identity in the vector space for all v belonging to v. Just quick sketch of how you will do this you will use the same thing over the field 0 plus 0 0 is the additive identity on the field. So, 0 plus 0 then use the distributivity the same things exactly the same things which means what based on these two observations let me use a different color. So, based on this observation and this observation what can we say that if when you are doing a scalar multiplication if either of those two happens to be a 0 if either the vector is a 0 or the scalar is a 0 in any case you cannot have anything but 0 right. What if you are told that the scalar is not 0 what can you conclude when you are told that there is a scalar. So, here is the situation if I tell you so by the way this I leave as an exercise just replicate these steps it will be a good exercise to mimic whatever we have done there. So, if I tell you that alpha times v is equal to 0 v and alpha is not equal to 0. Can you conclude that v must be equal to 0 how alpha has an inverse so you multiply by the alpha inverse on both sides because multiplication of 0 with alpha inverse will only lead to 0 as we have just seen therefore, this right hand side continues to remain 0 the left hand side becomes 1 times v but 1 times v by our definition of the vector space is just nothing but v therefore, v must be equal to 0 right. On the other hand if alpha is 0 then obviously you have already shown through this where it is 0. So, whenever I tell you this situation it is either at the situation that alpha is 0 or the situation that v is 0 right that is the take home story from all of this yeah. So, the next important object that we are going to try to grapple with and try to understand a bit are subspaces just like we had a field and then we went into a sub field what was the property of the sub field though the set of those numbers or objects had to be a subset but not just that the operations have to be inherited from the original set. You cannot define a new set of operations multiplication and addition and still call it a subspace or a sub field in that case same thing here you have in the parent vector space v you have f you have addition and you have this from there you are trying to get to this alright the only thing is that this w is now a subset of v and this must also satisfy all those properties of the vector space right. So, it cannot be just any arbitrary you can immediately imagine that it cannot be any arbitrary subset right for example, the subset of let us say the real numbers. So, real numbers are a field but also we have also hopefully try to convince you that any field is itself also a vector space. So, the real numbers is also a vector space the set of real numbers is also a vector space. Now, suppose you take the set of real numbers and you take any subset of the set of real numbers that does not contain 0 can it be a subspace because you see any vector space must have the additive identity right also it must needs have the multiplicative identity in itself right because the scalar multiplication. So, what is the 1 here that is the number 1. So, if you are taking the conventional addition and multiplication of real numbers you must have those objects right. So, you have to reason out accordingly and think about what all objects must be there in this specially chosen subset. So, that this subset can be deemed with this combination of a subset the original field the addition inherited from there and the multiplication inherited from there to meet all those properties immediately you will see that it seems like maybe we are putting in more effort I mean already we know something about this structure. So, can we not reduce our efforts in other words if we already know that there are certain things that this fellow meets can we not reduce our check see what we have to do is check for those properties that we have enlisted. So, 5 properties for the Abelian group and then a bunch of properties that distributivity the existence of the multiplicative identity in the field when acting on any element in the vector space leading to the vector space element itself and so on and all those properties you have to check every one of them of course, you can do that in its own right a subspace must also be a vector space obviously that is what it is, but we wonder if there might be some easier way to get a handle on this once we know that this already exists as a super structure this is like a substructure maybe we can do with a fewer number of checks that is our hope all right. So, in order to do that I will just enlist all the properties in sort of shorthand kind of notation ok. So, you have with respect to addition you have closure you have identity you have inversion you have commutativity and you have associativity. Now, when you bring in the multiplication into the mix you also require closure you also require there exists one in the field such that one times V is equal to V for all V in ok. Now, I am testing it out for this W let me just call it small w this right what else I need the property that alpha beta times W is equal to alpha times beta times W where alpha beta come from the field and W comes from W and finally I am just going to write in words, but you know already that there is two kinds of distributivity yeah this distributivity of kind a and kind b you know what it means. So, I am not just mentioning it right. So, now it stands to reason that we should try to see that by dint of our knowledge about this original super structure how many of these things can we take for granted and do away with investigating right. Then the rest of the things we should try to check and try to come up with a an efficient way of checking whether this subset having inherited the following operations here into itself and also inheriting the same working on the same field is also going to be a subspace or a vector space in its own right that is the question ok.