 So, today we shall start talking about a new concept which many of you have seen in the context of Euclidean spaces presumably, but we will introduce it more formally particularly for more abstract vector spaces that we are now going to handle and deal with in this course. So, the idea is of linear independence all right. So, when do we say a bunch of vectors is linearly independent and you cannot get a vector out of some linear combination of other vectors sorry okay, but I would just want to visit that point what another of your friends has mentioned because that is quite interesting. So what can you say about this particular set suppose this is the vector space and I plucked out this vector. So, how does your definition work for this you see that is yeah but then you are adding more clauses right just now you said that if a vector cannot be represented as a combination linear combination of other vectors then we are done but that is not it. So, we need to be more specific and precise about the way we define things which is why we will go for the following description which is to say that if a non maybe hyphenated word non trivial linear combination of vectors in a set S which is contained in the vector space V does not or rather if let us let us take the opposite because I have kept that provision open. So, it is an easier definition and we can say does lead to then what do we say about such a set. So, what is the attribute of it is an attribute of a set of vectors whenever I talk about the property of linear independence or linear dependence I have to know what sort of object hand categorizing it is a set of vectors in a vector space whose linear independence or dependence is being debated right. So, what do we say in the other this case if a non trivial. So, of course, this non trivial I have not defined or described yet but hopefully you understand we will describe that do not worry if a non trivial linear combination of vectors in a set leads to a resulting vector that is this then what do we say about that set right then S is linearly dependent now of course, conversely if no non trivial linear combination can ever lead to the 0 vector right then it is linearly independent right. So, that is what the description is. So, never ever try to define this linear independence by saying one vector represented in terms of the other vectors that is not necessarily true particular case being when you have the 0 vector. However, this definition accommodates everything all possible eventualities are covered here in right. So, I will give you one example of something that you may not view as a vector at first glance because you normally think of n tuples as vectors. But let us say I consider the vector space to be. So, when I write this it means this is a vector space of polynomials of degree 2 or less whose coefficients are real numbers right that is the notation. So, this is a vector space you can go ahead and check that it satisfies all the properties of vector spaces. If you understand what addition of polynomials are in the conventional sense and what multiplying a polynomial by a real number is. So, of course, this is over the field of real numbers right now consider the set note that these are all members of this vector space right. The claim is that this is linearly independent how do we show this the multiple ways of showing this any one way of showing if this is linearly independent if not then let us try out. So, C0 times 1 plus C1 times x plus C2 times x squared suppose this is equal to the 0 polynomial the identically 0 polynomial note this is not an equation I am writing ok. This is the polynomial and this is the 0 polynomial there is a difference ok. If you look at it like that without any context and all of this structure you might think this is a quadratic equation no it is not a quadratic equation I am saying that there will have to exist some C0, C1 and C2 some real numbers. So, let us say C0, C1 and C2 are some real numbers if I can find out some C0, C1 and C2 some real valued C0, C1, C2 that results in the 0 polynomial then I will show that I will have shown that this is linearly dependent. But the claim is that this is linearly independent so I should somehow be able to show that this can only result in the 0 polynomial if this is yeah this all these 3 have to be 0. So, of course, because you know that over the real numbers this polynomials are differentiable you can just go ahead and use the fact that this has to be true for all x because it is 0 in the sense of the polynomial remember. So, it is true for all x right that is the idea of linear independence in this case. So, this must be true for all x so in general if it is true for all x specifically it must be true for x is equal to 0 if you plug in x is equal to 0 you are led to conclude that C0 is equal to 0. And then you can of course go ahead and differentiate but there is another way of seeing this you see because you know that over the real field one must exist which is a multiplicative identity. So, once you have established that C0 must be 0 then you are left with C1 x squared C1 x plus C2 x squared must now be 0 now go ahead and plug in x is equal to plus 1 and what you will have is C1 plus C2 is equal to 0 also in the field you know in the real field that 1 exists minus 1 also exists. So, you can go ahead and plug in minus 1 so this is for x is equal to 1 and finally for x is equal to minus 1 you can just take minus C1 plus C2 is equal to 0 already you have seen C0 is 0 now of course based on these it is immediately apparent that both C1 and C2 must also be 0 and therefore, the only way to result in the 0 polynomial identically for all x using these vectors from this vector space is to choose C0 C1 and C2 all three of them to be equal to 0 and therefore, this set is linearly independent right is that clear. So, this is an example of a set sitting within a vector space which is linearly independent which looks nothing like the n-touple of numbers right ok. Having understood this I will make a couple of claims or maybe 2 or 3 claims and I will argue that this these must be true ok. So, suppose S this is a subset within a vector space I am not writing this you by now you understand when I write this as a V it is a vector space ok is linearly independent writing the shorthand for any S bar which is a subset of S we have that S bar or rather S underlined is also linearly independent is this obvious how do we go about trying to prove something like this a claim such as this once again we try to contradict this. So, we try to assume first that suppose S bar S bar which contains at most all the members of S in which case by definition it is linearly independent, but if it contains fewer fellows than S let us assume that S bar contains some of the fellows that are there in S not all of them because if it contains all of them then it is just S and S is by our claim linearly independent nothing to prove them. But suppose S bar S underlined contains some of the fellows of S only and that some of them are not contained in S underlined the subset if despite that it turns out that this S underlined this set is linearly dependent then is it possible that S can still continue to be linearly independent right. Because if there is a non-trivial linear combination of fellows here you might just as well go and append the fellows that are not here, but here with coefficient 0 you will still have a non-trivial combination non-trivial combination means at least one of those coefficients is non-zero. Now as long as you found some of the fellows here to have non-zero coefficient you might as well pad the rest with 0 coefficients it will still be a non-trivial combination. So, it is not possible that any subset there of a subset of a linearly independent set has to also be linearly independent right and by the same token kind of a converse relationship suppose S contained within a vector space V again I am not okay let me just mention that this is a subset is linearly dependent. Now for any S bar which is contained within sorry which contains S and of course S bar is also of course it is part of V right linearly dependent. So, for any S bar within which this linearly dependent set is sitting yeah we must have S bar as a linearly dependent set. I mean already it is defective defective in the sense that it has lost its linear independence within S, S is already linearly dependent. So, any super set of a linearly dependent set must be you cannot build a structure on weak ground right you already lost linear independence in the set S itself you cannot hope to extend it to a bigger set which contains everything that is there already in S and still hope to build a linearly independent structure out of it linearly independent set out of it right. Once again you can just try and fill in the arguments here just a couple of lines you have to just assume that there are certain fellows belonging here certain fellows belonging there and so on right and take combinations and argue. So, I hope that these two assertions are very clear and a third and important most one of the most common assertions which we started this discussion with is this any set S contained in the vector space V which contains the 0 vector must be linearly dependent. Again very straight forward a ready made choice ready made choice for a non-trivial linear combination is just attach some non-zero coefficient to the 0 vector and attach a 0 coefficient to all the other vectors that is a non-trivial linear combination of fellows sitting inside S which results in 0. So, this must be true any set that contains 0 can never be linearly independent 3 claims that we have made we have not even sort of dirtied our hands by trying to prove them, but hopefully we have presented arguments that have convinced you that these 3 are true. So, that is the first important concept we have introduced today which is that of linear independence or linear dependence the complementary properties. So, once you understand one you understand the other any questions so far no S bar is a superset of S yeah S bar is contained in the vector space of course. Why not? So, S bar this is a vector space. So, S bar contains only fellows that are qualified as vectors within the vector space V. So, I mean why not? So, for example, let us say you have 1 x 2 x x squared alright and you have 1 let us say 1 x x squared that is it and 1 x 2 x x squared and let us say what should I write 5 x squared 5 x squared yeah should I say let me just add a random number of elements here 7 x 3 x squared 5 x squared yeah and let us say this is 2 x ok. So, already this is linearly dependent set see these are all coming from that example that I had given the vector space over R this is a linearly dependent set why because I can identically cook up just these 2 fellows here yeah. So, minus 2 times this vector plus 1 times this vector and the rest I can just put 0. So, that is a non-trivial combination that leads to the 0 polynomial therefore, this is a linearly dependent set now this one contains every fellow that is here. So, this is my S and this is my S bar now you see S bar must come from this as well that is all that I am saying now my vector space is this this is the V right that is all that I am claiming. So, therefore, if you cook up a super set of a linearly dependent set to begin with that super set also must be linearly dependent because otherwise it does not make sense we are talking about subsets from the same vector space you cannot compare apples and oranges right. So, this said I mean you can just add maybe 3 apples to this I do not even know how to take their linear combinations unless I have all the objects coming from the same vector space. So, 1 x 2 x 7 x 3 x squared 5 x squared and 5 apples that is not a vector space right yeah yeah yes yes both of them must be from the same vector space only then we can compare this property of linear independence and dependence and this subset relation and so on right. So, we will now introduce the second important concept for today's lecture which is that of generating. So, what is the generating set once again at the back of our minds we should always do well to remember that there is a vector space that we have in mind everything that comes does come from that same vector space nothing is outside of that structure okay. So, suppose G this is a subset or a set sitting inside the vector space V. So, this is the vector space this is a set okay let us you know subset is a given. So, you just read it as a set there is a set that is contained in the vector space V. So, every element in this set G belongs to the vector space V is what I mean by this right such that the span of G is equal to V okay that means by combining fellows that are contained inside this set G I can cook up every vector that lives inside this vector space V okay. So, if I have such a set right then G is a generating set for V do you think that the generating set needs to be unique once again think of the polynomial I love to give the example of the polynomial because that is the first like you know simple example that you can give and yet it is not something that looks like an n-touple at least not at first glance later on we shall see when we assign coordinates they are also like those n-touples of numbers but nonetheless the point being that once again you see 1 x x squared you are using these vectors you can generate every polynomial of degree 2 or less by the same token you can also take a 5 7 x 8 x 3 x squared plus 2 4 x squared this is also a generating set you can check right this is also a generating set for every polynomial of degree 2 or less okay. So, of course, there is no question of having a unique or degenerating set okay it is I can all at best be a generating set for a vector space yeah that is why we do not talk about the generating set for a given vector space once you come up with one generating set we are happy with that. So, these are examples yeah so these are generating set this is g 1 and this is g 2 of course the vector space we have in mind is again that same right so that is the idea of a generating set if by combining fellows inside a given set you can cook up every possible vector that lives inside the vector space it is a generating set for the aforesaid vector space right but now we are going to try and trim it. So, how do we trim this generating set we introduce the idea of a basis. So, now we have the ideas of linear independence and of a generating set and we combine them to define what is a basis for a vector space okay set B contain inside the vector space V is said to be a basis for V if it is a linearly independent generating set for V see none of those terms we have used now are not are undefined we have defined what is linear independence we have defined what a generating set is. So, based on our understanding of these two notions that is all we need in order to understand what a basis is right. So, let us look at a few examples just to make things clearer. So, example one let us say the one we understand very well Rn n tuples of real numbers over R of course and I put it to you that this bunch of vectors is a basis. How will you try and establish this well it has to have two properties you take any n tuple of numbers it should be representable as a linear combination of these fellows. So, without loss of generality choose alpha 1 alpha 2 alpha 3 till alpha n you will see that it can be written as alpha 1 times the first one plus alpha 2 times the second one plus alpha 3 times the third one plus so on and so forth till alpha n times the last one. So, it is definitely a generating set then the next thing to check is whether this is also linearly independent suppose it is not when you choose c 1 times the first one plus c 2 times the second one plus c 3 times the third one so on till c n times the last one equate that to the 0 vector in our n tuples that is all 0s. Now of course, entry wise you have to compare if two vectors are equal if two n tuples are equal then every 1 1 to 1 entry should be equal. So, therefore, c 1 is equal to 0 c 2 is equal to 0 and so on likewise all the c i's must be equal to 0 and therefore, nothing but the trivial linear combination of these fellows can lead to the 0 n tuple and therefore, this is also linearly independent. So, this is indeed a basis let us look at example 2 which is c n over c once again the same b is still a basis go ahead and check you can have n tuples of complex numbers and you can combine them linearly over the field of complex numbers. So, every one of those n tuples is a complex number. So, you are free to choose those coefficients instead of real they can now be complex. So, again by the same token it has to be linearly independent and generating set again checking generating set whether the fact that this is indeed a generating set for this vector space over c that is also readily evident the fact that this is linearly independent is also follows the same line of reasoning as presented before. But now if I tweak this a bit let us I should say n because this is probably a non-example let us say c n over r that is also a legitimate vector space right the field has changed the set v has remain the same. Now, do you think b now b is candidature is under question why is it still not linearly independent well of course, it is it is still linearly independent just go ahead and take linear combinations of fellows in b over the real field we have already seen as linearly independent. So, why does it fail the span? So, what do you need to do in order to actually cook up a generating set for because it is a generating part where this has failed not the linear independence part. So, can you fix this right? So, let me call that give that a new sort of a nomenclament notation like this and then I take the union of b and i b. Now, this does not remain a non-example anymore now it becomes an example of a basis yeah. So, see that is why the field is very important the vector space can only be studied and it is structured understood in entirety only once we know what field we are looking for right. So, same c n over c and c n over r the basis what remains a basis in this fails to remain a basis in this right. So, these are important notions of course you can go ahead and check that polynomial the vector space of polynomials of degree 2 or less you can go ahead and check that 1 x and x squared is also a basis for that yeah. The fact that it is a spanning or a generating set is obvious just whatever the coefficients of your polynomial just go ahead and put them as c 0 times 1 plus c 1 times x plus c 2 times x square and you have all possible values of c 0 c 1 c 2 covering all possible polynomials of degree 2 or less yeah. So, thus generating part is done the linear independence we already spoken of in general you can take polynomials of degree n or less and check that 1 x x squared x cubed till x to the n they do indeed provide you with a basis for polynomials of degree n or less right. So, that we have with that we have understood two the three fundamental notions and how they are tied together the idea of linear independence the idea of a generating set and now the idea of a basis these are very intimately connected with each other right and we have also seen a few examples I am sure you will find a lot more examples not just in the textbooks, but also in the problem sheets that we shall be subsequently sending out ok. Any questions thus far? Oh I B is basically I am just going to replace the once here with I that is the square root of minus 1 the imaginary number it is just a name that I am giving it is not something sacrosanct that you will find in textbooks it is just I am defining for my purpose of or my ease ok. So, every element in the set B being so called multiplied with the imaginary I am calling that set as I B it is just my nomenclature ok you can even say maybe it is non-standard, but that is the way it is like we have the argon plane we say R and J R in electrical engineering we use J for the square root of minus 1 right. So, J R is the imaginary axis. So, I am just saying B and I B right this is union yeah no, but these are sets these are not vector spaces B and I B are not vector spaces these are sets. So, we are we have only spoken about sums of subspaces remember not sums of sets right ok. Any further questions otherwise we will bring this module to a close and we will move over to the next ok yeah C n over R yeah all the all the scalar R's can come from the set R yeah yeah no, but I B is the set of vectors. So, members of B are first coming from C n. So, members of I B are also coming from C n. So, they are obviously n tuples of complex numbers. So, I is a legitimate complex number with specifically 0 real part and 1 imaginary part. So, those are definitely members of C n it is only the scalars that you are now restricted to choose from only the real numbers not complex numbers. So, just that restriction includes I mean entails that you will now have to choose twice the size of the basis twice the number of elements that you had in the original basis in the this basis yeah ok.