 So, I will I will state the separating hyper plane theorem that we need. So, the theorem is the separating hyper plane theorem that C subset of R n be a convex set and let y be a point in the exterior of C bar. C bar is simply the closure of C. I am allowing the set to be not closed, but let us take the closure of the set and then consider a point y that lies outside it. So, if you have you have a set C which probably not closed, but then you take its closure that closure will the closure is also a convex set and then you take a point y take a point y that lies outside the outside the closure. Then the claim is there exists a vector a not equal to 0 such that a transpose y is less than I take the infimum of a transpose x as x ranges over C. So, there exists a vector a not equal to 0 such that a transpose y. So, this is a vector in R n such that a transpose y is less than the infimum of a transpose x as x ranges over C. Now, how does this define a separating hyper plane? So, for that the normal to the hyper plane is the vector a is the normal of the hyper plane. Where is my b? Where is for a hyper plane I need to define I need to talk of a as well as b a scalar b. Yes, so I can I have that these two values are one the left one is strictly less than the right one. So, I can always fit in a take a value b that lies in between them. So, this is a value b that is greater than a transpose y, but at the same time less than the infimum of a transpose x over x and C. So, look at a consider a value b that lies in between then then clearly and then look at the hyper plane h defined by such that a transpose x equal to b. Then in that case y lies on one side of the hyper plane and the entire set C lies on the other side. Why does the entire set C lie on the other side? Because we know that infimum of a transpose x x and C is greater than b, which means that a transpose x is greater than b for all x and C. So, the entire entire set C lies on one side of the hyper plane and on the other and and as far as y is concerned, we know a transpose y is less than b. So, it lies on the other side. So, this theorem basically guarantees for you implicitly a separating hyper plane. It is giving you that it is giving you the slope as well as the implicitly through this is also letting you pick an intercept. You can pick any intercept ranging from anywhere in between here that is that is perfectly ok. So, now using this we will be we will be proving the strong duality theorem. So, now what are we trying what are we going to prove? Let us go back to the let us go back to the statement of the theorem we want to prove this red statement here. The statement says that if either primal or dual has a finite optimal value then so does the other and these values are equal. The other theorem other statement which is that if either primal or dual is unbounded then the other must be infeasible that is trivial we just we argued that directly by looking at weak duality. So, now we are going to look at the strong duality theorem. So, suppose the optimal value of the primal is finite and equal to say some z 0 let me just denote that by z 0 that is the optimal value of the fine of the primal. Now, I am going to define this set this set C as follows it is the set of r comma w such that r is equal to T z 0 minus C transpose x and w is equal to T b minus A x as and allow x to be greater than equal to 0 and T to be greater than equal to 0. So, C is this set of r comma w that satisfy this. So, it is the set of r comma w that can be expressed in the following way it can r can be expressed as T z 0 minus C transpose x for x greater than equal to 0 and T greater than equal to 0 and w can be expressed as T b minus A x for x greater than equal to 0 and T greater than equal to 0 for the same x and T greater than equal to 0 your r can be expressed in this way and w can be expressed in this way this is this is the set C. Now, what is the what kind of a set is this set C if you look at this set C how can you tell me what is its dimension how many in what space does it lie what is the dimension of r and what is the dimension of w. So, r here is r a scalar or a vector yes r is the scalar r is T z 0 minus C transpose x. T here is has to be a scalar otherwise I will not be able to add it to C transpose x right C transpose x is a scalar so and z 0 is another scalar so T is also a scalar. So, T is a scalar so T is a scalar x is a vector x is a vector or your original vector this is in the original space so x is a vector in r n. So, r therefore, is a scalar what about w what is the dimension of w yeah w has as the length of w is as many as the rows of a right. So, w is in r m so C itself is a subset of r m plus 1 alright. Now, so now let us ask what type of what kind of a set is C. So, suppose r comma w belongs to C and I give you a lambda greater than equal to 0 then what can you say about lambda w comma lambda lambda r comma lambda w. So, if r comma w belong to C that means r it just means that there exists x greater than equal to 0 T greater than equal to 0 such that r is equal to T z 0 minus C transpose x and w is equal to T b my T b minus A x. Now, if I multiply both sides by lambda in both equations then what would I get I would get lambda r equals lambda T my z 0 minus C transpose lambda x and lambda w as lambda T into b minus A into lambda x lambda is a scalar right. So, what this means is if I have if lambda if r comma w belongs to C and lambda is some scalar greater than equal to 0 then lambda r comma lambda w can be are also in C and the corresponding value of T and x is just lambda times the earlier s while T and the lambda times the earlier. So, by so you said so these are also greater than equal to 0. So, which means lambda r comma lambda w belong to C. Now, if a set C is of this kind where if it is if r comma w belongs to C then lambda r lambda w belongs to C for lambda greater than equal to 0 then what kind of a set is this it is a cone. So, C is a cone it is also very easy to show that C is actually a closed cone. So, in fact, C is a cone and C is closed C is a closed cone what about the convexity of C is C convex it is convex you see C is defined using a bunch of linear equations and inequalities right. So, it is just r comma w that satisfies some linear equation you can think of it this way you can think of a set comprising where the variables are it is you can think of a set S like this which is r w x and T such that r is equal to T z 0 minus C transpose x w is equal to T b minus A x and x is greater than equal to 0 and T greater than equal to 0. If you look at a set S like this the projection of this set on to the r w on the r w axis is my set C right. So, if I take the shadow of this set on the r w space that that set is actually C is projection of S on the r comma w on the r w space and what kind of a set is S well S is actually a polyhedron it is just a bunch of linear it is it is it is a vector that satisfies some linear inequalities right it is a set of vectors that satisfy linear inequalities. So, S is a S is itself a polyhedron. So, the shadow of a polyhedron has to be a polyhedron right. So, this is a projection. So, therefore, C is a polyhedron and hence convex. So, many proofs in optimization begin with this one question first what kind of a set is this right you you encounter a set you ask what kind of a set is this right and more you can say about the set more you will be able to say about the problem ok alright. So, now, what have we concluded well that C is a cone C is convex closed and C is also convex alright. So, what we have is a closed convex cone alright. Now, let us look at this point let us say a point like this say a point 1 comma 0 is this 1 comma 0 what I mean by that is. So, 1 here is a scalar is a scalar and 0 is the vector this is a vector this is a point in r m plus 1 right. So, this vector is a 0 vector in r m. Now, this point can you does this question is does this does 1 comma 0 lie in C is this 1 comma 0 out in C right. So, let us check. So, suppose it does suppose it does then then means that there exists t greater than equal to 0 and x greater than equal to 0 such that your 1 which is in the place. So, your r equals 1 and w equals 0 now. So, r that means 1 is t z 0 minus C transpose x and w equal to 0 means that 0 is equal to t b minus A x right. Now, here again so if 1 comma 0 belongs to C then it means that you can find t greater than equal to 0 and x greater than equal to 0 that meet that satisfy these equations. Now, let us take 2 cases. So, case 1 case A suppose is that here the t that you got is positive if the t that you got is positive if you if the t that you got is positive then I can divide throughout by t and I get that I get that 1 by t minus z. So, I get let me write this again I get that C transpose x by t is equal to z 0 minus 1 by t and the second equation if I divide throughout by t I get I get also that x A into x by t is equal to b. Now, I already have x greater than equal to 0 t greater than equal to 0 this means that x by t is greater than equal to 0. Now, what has what have I got now as a result in this case. In case A if t is positive then I have concluded that these equations must hold but what are these equations this is saying x by t satisfies A into x by t equals b and x by t greater than equal to 0 and the. So, x by t is therefore feasible x by t is feasible for the primal but then what is the what is the value of x by t it is z 0 minus 1 by t that means it is value is strictly less than z 0. So, it is a feasible value feasible point for the primal whose value is less than less than z 0 and what was z 0 here z 0 we had said that z 0 is the optimal value of the prime. So, what this means is if t is greater than 0 we have effectively constructed a point whose value is even better than the optimal value that is a contradiction. So, x by t is feasible for the primal and has objective value strictly less than z 0 which is contradiction. So, case A therefore is impossible means you cannot have t to be strictly qualitative ok. Now, let us go to the other case case B case B would be since t is greater than equal to 0 the only case remaining is t equal to 0 now if t has if t is equal to 0 which means that these boxed equations here these boxed equations must are hold with t equal to 0 right. So, what do I get if I put t equal to 0 I get C transpose x equal to minus 1 right and I get A x equals 0 right. So, this implies C transpose x equals minus 1 and A x equals 0. Now, what does this mean? So, remember x x and t x and t ok x and t are simply you simply have x greater than equal to 0 and t greater than equal to 0 the x is here is not necessarily feasible for the primal it is just some vector of length n we are not saying A x equals b here we constructed an x by t which became feasible right. Now, if A x equals 0 and C transpose x equals minus 1 what do we get what is what is this saying if A x equals 0 it means that x is in the null space of A right. So, you have found a direction such that it if I go in that null in that direction by A times that direction we will not change right. So, what this means is so if I have say any feasible value. So, if I have say an x tilde such that A x tilde equals b and x tilde is greater than equal to 0 then x plus x sorry x tilde plus x would also satisfy this because x tilde plus x would give me still would still be A x tilde and x tilde plus x would also be greater than equal to 0. Now, what thought what does this remember what we we used to we had we had a name for this a direction along which you can keep going but without leaving the set this is a recession direction right. So, what this means is your x here is a recession direction right x is also remember greater than equal to 0. So, it is greater than equal to 0 it satisfies A x equal to 0. So, x is means x is a recession direction or a ray of the set right it is a direction along which you can keep traveling and not lose. But then every time I travel one unit of distance along x look at what is happening to my objective I keep I keep get incurring my C transpose x goes down by minus 1 goes down by 1 right. So, if I take this point x tilde and look at C and add an x to it well x plus x tilde is still in the set but what about C transpose x plus x tilde C transpose x plus x tilde is now C transpose x tilde minus 1 right. So, I can keep go. So, and now if I if I so this is because I took a unit step in the direction x if I took a step of length lambda in the direction x I would still be within the set and my objective value would have dipped by lambda greater than equal to 0 right eventually I can go along I can make my lambda as large as I want and that will drive my optimal value as low as I want right. So, this is again a contradiction contradicts what contradicts the assumption that the primal has a finite optimal value right. So, this optimal value of the primal is minus infinity which is a contradiction. Contradiction since we assume that the primal must have a has a finite optimal value which means again case B is also not possible right. So, what have we got so far we concluded case A is not possible because in case A you get something better than the optimal which means you are contradicting optimality in case B we are getting something that that the optimal value is minus infinity which the hence we are of we are contradicting that the optimal solution exist right it is that it is finite alright. So, in summary what we have shown is after all this what we have basically shown is this which I let me put this here we ask we began through this analysis asking does the point 1 comma 0 belong to C and what have we shown we have considered both cases and shown that no it cannot belong to C right. So, so the point what we have effectively shown is the 1 comma 0 does not lie in C this is what we have shown.