 Now, let us look at it the other way. So, this is part 2. Suppose x is an extreme point, extreme point of p. Now, if x is an extreme point of p, we need to show that x can be written and we need to show that x takes this sort of form x1 till xm followed by n minus m0. So, suppose I arrange the coordinates of x in this sort of way, suppose I write x as x1 till xk followed by some n minus k0. Now, k could be n in which case there will be no 0s, that is also fine. So, this is every x can be written in this kind of form, this is without loss of generality. Now, if the columns of A corresponding x1 till xk, so that means these are a1 till ak, these columns, if these columns are linearly independent, then what can one say? If you have k linearly independent columns of A, what must be the case? Well, the only way that can happen for you to be able to find k linearly independent columns of A is that k should be less than equal to m. So, then if these columns are linearly independent, then it means that you have then k is less than equal to m. If k is less than equal, then what that means is therefore that x takes this sort of form x1 till xk, this is your first k. Then you have some 0s here, how many of these these are m minus k0s and then you have the remaining 0s, which is n minus m0s, correct? Because k is less than m, less than equal to m, it means that m is some you have first you have these k, then you have some you have some 0s here, m minus k0s and then some more 0s, right? So, all of these together, the number of 0s here is still n minus k, just as I had written before. But what is the meaning of this? What I can do is now notice that well, this is actually what I can do is I can pick up I can pick up m minus k columns from here such that together with the columns of columns of together with a1 to ak, they form a basis. See a1 to ak are linearly independent, so k has to be less than equal to m. From the remaining columns, I can pick up from the remaining columns of a, I can get I can choose m minus k, so that together they form a basis, right? And then so using, so we can find m minus k columns from ak plus 1 till an such that a1 to ak these columns are linearly independent. So, these are these should all be linearly independent in which case we can then say that this means that x can be written in this form, x1 till xm followed by z, x1 till xm followed by n minus m0 where the columns corresponding to these are linearly independent. So, if I wanted to show that x can be written in this form for that I if k if the columns that are I can write x as x1 till xk followed by n minus k0 and if the columns a1 till ak corresponding to x1 till xk if they are already linearly independent, all I need to do is pick m minus k columns from the remaining put that together that then makes creates for me a linearly independent and then that becomes my that then becomes a basic feasible solution. So, which means then that this is a x is a basic feasible solution. Now, I can do this if under this condition which is if the columns corresponding to x1 till xk which means they are the a1 to ak if they are linearly independent, if they are not linearly independent then I cannot say that k is even less than equal to n. If they are linearly if they are not linearly independent then k need not be less than m, k could be even greater than m. So, this logic will not work. So, then that gives you the other case. So, the second case is that if the columns a1 till ak are linearly dependent. So, if they are linearly dependent what does it mean? This means that I can find there exist say a w such that w in rk such that summation w i a i equals 0 going from 1 to k. So, if these are linearly dependent then I can find a linear combination of these such that that linear combination is 0. There exists a w not equal to 0 in rk such that I can if I take this weighted sum of these columns that that gives me 0. This is what it means to be linearly dependent. So, since they are linearly independent. Now, what I do is the following look I have x which is written like this x1 till xk followed by n minus k zeros what I will do to it is the following I will I add to it add this vector let us call this w hat w hat is comprised of w which is what I have earlier my the w above and a bunch of zeros I have w and a bunch of zeros right. So, now look at the vector x plus epsilon w hat what what what can you say about x plus epsilon w hat where epsilon is just some small positive quantity x plus epsilon w hat what kind of components it has if you look at the last n minus k components all the n minus k components are all 0 the last n minus k components are 0 the first k components are getting perturbed a little bit. So, you you have the components of x1 you have the x1 to xk but then on top of that you are adding a w epsilon times the w i right. So, it is being perturbed a little bit but I can find we can choose we can choose epsilon small enough so that x plus epsilon w hat remains greater than equal to 0 right. So, remember x x was so here we we started off saying that x is an extreme point so of p that means x is certainly feasible which means x these are all greater than equal to 0 these components the last m guys were all last n minus m were 0 right. So, now what I am saying is I can put a small enough perturbation I can put a put a small enough perturbation so that these are all greater than equal to 0 because what I have done the form of x I have assumed is that the first the all the non-zero ones are here and the remaining guys are all 0 right. So, so we can choose epsilon small enough so that this is greater than equal to x plus epsilon w hat is greater than equal to 0 not only that x minus epsilon w hat is all can should also be greater than equal to 0. So, fixing a w I am scaling by epsilon so I so x plus epsilon w hat is greater than equal to 0 x minus epsilon w hat is also greater than equal to 0 and moreover look at this x ax plus epsilon w hat what is this this is ax plus epsilon a w hat what is this quantity ax is is b and epsilon times w hat is what it is it is w it is the so all the all the the last n minus k are all 0 in w hat so what you are really doing is doing this sum here the sum written here summation w i a i and that is equal to 0 right. So, this term therefore is 0 so what you are left with is just b and similarly x minus epsilon w hat is also equal to b. Now what did we conclude as a result x plus epsilon w hat is greater than equal to 0 x minus epsilon w hat is greater than equal to 0 a times x plus epsilon w hat is great is equal to b a minus a into x minus epsilon w hat is also equal to b right what this means is if you look at these two points x plus epsilon w hat that belongs to p x minus epsilon w hat also belongs to b now if these two both belong to p but then x is actually nothing but half this plus half plus the other one x is half of x is simply the average the midpoint of these two right so what have we concluded we have concluded that if the columns if it so happens that the columns a 1 to a k are linearly dependent then you can find x 2 points in p x plus epsilon w hat n x minus epsilon w hat 2 point whose midpoint is x right which means what which means x cannot be an extreme point right. So, x cannot be an extreme point and that is a contradiction because we started off assuming x is an extreme point of p started off assuming x is an extreme point and we found that well if in this case a 1 to a k if they are linearly dependent then it cannot be an extreme point. So, the only possibility is that they are linearly independent and then if they are linearly independent we said that it is actually possible to look at view x as a bfs by padding additional columns from the remaining remaining n minus k column is this clear. So, what this what we have concluded as a result is the essentially is go back to the statement of the theorem we concluded just this that x is an extreme point of p if and only if x is a basic feasible solution. Now, these sort of characterization which basically take you from a geometric property to an algebraic property are extremely useful because computers can only do algebra they cannot do geometry right extreme point is a geometric property that has no formula associated with it bfs has a formula associated with it right. So, there is an algorithm you can now try to design around this. So, what so the most the most popular algorithm for solving linear programs actually just basically does this it just goes from one basic feasible point to another looking for trying to see if that is a solution all right that is what is called a simplex method we will come to that in later if possible. So, but let me tell you of couple of corollaries that follows in a straightforward manner from here yeah which part of the proof yeah. So, these were all assumed to be positive that was in the first case. So, going back here let me revise this see x can always be written in this form where you put all the non all the positive guys as x1 till xk and the rest is rest are 0 it could be as I said k is equal to n in which case there will be no 0 element all elements will be positive. But okay now if the if the yeah so if you have if the columns corresponding to x1 till xk are linearly independent then I have fewer then I have I actually have fewer columns than I need to make this a BFS. So, I have k of them they are linearly independent all I have to do is borrow m minus k additional columns from these guys from the ones that are multiplying these elements. So, you have these k then you have m minus k and then you have the last n minus m. So, all I am saying is I can find some I can find I can find m minus k columns from these remaining ones here such that together with these a1 till ak they form a linear they form a basis they form they should form they become m linearly independent columns no we have not used that here. So, x if x is an so if x is an we just use that yeah. So, we basically said well if we did not use that x is an extreme point we said that suppose it has these many if this is the form of it then then all I need to do is I need to find m minus k and if these a1 to ak are linearly independent then all I need to do is find these remaining additional columns and create that into a basic feasible solution. This what you are saying is right this observation will work whenever k is less than m whenever the a1 to ak are linearly independent this will work does not matter whether x is an extreme point or not. And that is not the only possibility the other possibility is that they are linearly dependent and that is where we need that x is an x is correct. So, what he is asking is is it possible that you can have a polyhedron is an extreme point of a polyhedron may not have some component 0 right that is a fair comment. But the reason why this theorem works is because this is not an arbitrary polyhedron anymore it is it is an LP in standard form it once you have a polyhedron written in this form ax equals b and x greater than equal to 0 then extreme point must have this feature. This is now this polyhedron now is in the non-negative orphaned of your n dimensional space and there is an affine set that cuts through that. So, the part of the affine set the affine set is a is this ax equals b that the part of it which lies in the non-negative orphaned that is your polyhedron such a poly you are we are characterizing the extreme points of such polyhedron. What you are saying is absolutely right if I wrote out an arbitrary polyhedron drew an arbitrary polyhedron on a page there is nothing there is nothing to say that it should have you know the extreme point must have some coordinate 0. But these kind of polyhedron do have that right. So, this is one of the also an unwritten advantage of converting an LP to a standard form that you get firstly you get two things one is that if the feasible you get a formula for the extreme point that is one thing the other thing is that if the feasible region is a non-MPO for free getting a guarantee that there is an extreme point. So, let me just write this out since we are on that topic. So, let me write this this is a corollary. So, if P which is x such that ax equals b x is greater than x greater than equal to 0 if this is non-MP then P has an extreme point why does it have an extreme point because it always has a BFS right. If it is non-MP it must have a BFS right because and once it has a BFS it has an extreme point and then combining with what we already know if there exists an optimal solution which means optimal values what this means optimal value is not minus infinity it is greater than minus infinity then there exists an extreme point that is an optimal solution. Here is a simple question for you how many extreme points can you have if I gave you a bunch of if I told you that I have a polyhedron in n dimension how many extreme points can it have is that a simple answer to this question say a polyhedron just in R2 how many extreme points can it have any number of extreme points right because you just you can you can just keep making you know polyhedron that looks or more or more like a circle and you can get as many extreme points as you like. So, just the dimension does not automatically say how many extreme points you should have you can have but the beauty of this the standard form is that now that you are in the standard form you can actually say put a bound on how many extreme points you can have how many would that be see every extreme point is a BFS and how do you generate a BFS you generate a BFS by looking at m linearly independent columns of A right how many such choices do you have for linearly independent columns of A you have you have n columns you want to choose m of them n choose m is the number of linearly possible choices not no more than that and each such choice will give you potentially 1 BFS or 1 extreme point right. So, the number of the number of extreme points or basic feasible solutions of this set x such that A x equals B x greater than equal to 0 this is the number of extreme points or basic feasible solutions no more than n choose n right where but remember yeah where A is r m cross n and full row rank. So, what this this is another thing that the that that putting an LP in standard form what it gives you is that well it tells you that at the most you need to search over these many points this is the number of points that your solution lies in one of one of these if it has a solution it lies in one of these.