 See remember in the last lecture we were confronted with this question given a basis B star for V star is there a basis B for V such that this B star is a dual for that basis B okay. The answer is yes and we will prove that today we need the notion of the double dual. So what is a double dual? The double dual is just V double star okay over R let us say real vector space then the double dual of V denoted by V double star is defined by V double star is V star star okay that is a double dual. It is the vector space of all functionals on V star it is a vector space of all linear functionals on V star that is a double dual. Now for we will now establish a correspondence between V and V double star as follows. So the objective first is to establish a correspondence. Let me take X in V I will define a functional I will define L X of F to be equals L of L X of F is F X for every F in V star I am defining a functional on V star so I am defining an element in V double star provided of course I show that L X is linear. So first let me show that L X is linear let me show that L X is linear then it would follow that this L X of see this L X acts on V star if I had shown L X is linear it follows that L X belongs to V double star okay. Let us prove that first that L X is linear to improve L X is linear I must show that L X of alpha F plus G equals alpha L F plus L G. So let us look at L X of alpha F plus G F and G come from V star F and G come from V star consider L F L X of alpha F plus G by definition this is alpha F plus G of X but this is addition of function as we know is point wise. So this is alpha F X plus G of X alpha F X alpha times F X that is alpha L X F plus G X is L X G so it follows that L X is linear is that okay. So L X is linear that is L X belongs to V double star remember that it the domain of L X is R see L X of F is F X F is a linear functional so F of X is a real number so L X is a function from V star to R L X is a function from V star to R so it is a functional we have shown it is a linear function okay. So for every X in V there is an L X now can it happen that this L X is 0 do we have plenty of these L X that will be answered by the next theorem but let us quickly observe that if X is not equal to 0 then L X cannot be 0. If X is not 0 we will show that L X cannot be the 0 functional okay X is not equal to 0 look at if L X is a 0 functional then what it means is that L X of F equals 0 for all F a function is 0 if it takes the value 0 for every X in the domain so if L X is a 0 functional then L X F is 0 for all F in V star but L X F is by definition F X F X equals 0 for all F in V star can we conclude from this that X is 0 how yes okay so there are at least two ways one is a little long winding the other one is to simply identify that since this is true for all F linear function one could take the identity function identity linear functional identity linear function identity of X is X so this means X equal to 0 contradiction to what we started with so L X cannot be 0. So X not equal to 0 implies L X is not equal to 0 that is if X is not 0 there is at least one functional F such that L X F is not 0 okay we want to show there are plenty of functions okay this correspondence between X in V and L X in V double star we will make it formal and prove the following theorem our concern is over finite dimensional vector spaces. Let V be a finite dimensional vector space define the function phi from V to V double star by means of this correspondence by Phi of X equals L X Phi of X equals L X see what we now know now is that this L X indeed belongs to V double star and the definition of L X for any fixed X is that L X of F for F in V star is equal to F of X. So this is a this takes the vector X in V to an element in V double star okay then what is the natural conclusion that one would like to draw Phi is an isomorphism let us state this and prove then Phi is an isomorphism then Phi is an isomorphism it is this result that will give a proof of the fact that for every basis for every basis B star in V star there is a basis B for V such that this B star is a dual basis of that basis B at least for finite dimensional spaces okay. So let us look at the proof of this theorem I must show that Phi is an isomorphism is the claim I must show that Phi is linear I must show Phi is linear Phi is 1 to 1 Phi is on 2 then it follows that Phi is an isomorphism Phi is linear that is the first thing let me consider Phi of Alpha X plus Y I must show that this is Alpha Phi of X plus Phi of Y so look at Phi of Alpha X plus Y this by definition is L of this L of Alpha X plus Y I want to know what L of Alpha X plus Y is so what I will do is look at the following L of Alpha X plus Y that is known if I know the action of this on F in V star L of Alpha X plus Y of F this by definition is F of Alpha X plus Y so remember F is in V star L is a functional on V star okay so this is a number F is in V star this is a functional on V star V double star this is a functional on V star this is an element in V double star okay F is linear Alpha F X plus F of Y this is Alpha F X I have L X F plus F G is L X F Y L Y F Alpha F X plus F Y because F is linear and then I go back to the definition of L subscript X L X F plus this so I can write this as Alpha L X plus L Y acting on F okay so now I know what L of Alpha X plus Y is L of Alpha X plus Y is precisely Alpha L X plus L Y so I will use that here that is if T of X equals S of X then T equals S, T of X equals F of X T equals X T equals S so these functions are the same so L of Alpha X plus Y is Alpha L X plus L Y I again go back Alpha L X is on the right Phi X plus L Y is Phi Y so I have shown that Phi is linear Phi of Alpha X plus Y is Alpha Phi of X plus Phi of Y so Phi is linear I want to show first that Phi is injective so I know it is linear so injectivity I will show null space consists of just single term 0 so I start with Phi of X equal to 0 I will show X is equal to 0 suppose Phi of X equal to 0 then L X equal to 0 but L X is a functional on B star this means if a function is identically 0 it means it is action on any element in the codominus 0 so L X of F is equal to 0 for all F in V star definition L X of F on the other hand is F X, F X equal to 0 for all F in V star as before we have considered that here F X equal to 0 for all F in V star in place X is 0 so X is equal to 0 so I started with Phi X equal to 0 I have shown that X is equal to 0 so it follows that Phi is injective it follows that Phi is injective to prove that Phi is surjective I will make use of the rank nullity dimension theorem if I have a linear transformation on a vector space V then I know that and it is finite dimension then I know that it is 1 1 if and only if it is on 2 okay Phi is a function from V to V double star what is the dimension of V double star the dimension of V star is a dimension of V if V is finite dimension dimension of V star is a dimension of V that is a dimension of the dual space is the same as a dimension of the original space the dimension of V double star should be the dimension of V star because I am taking the same dual V star is finite dimension I am using so dimension of V is equal to dimension of V double star dimension of V is equal to dimension of V double star so and I have shown that this is a Phi this is a linear injective map so it must be by rank nullity dimension theorem this must be surjective also Phi is surjective also so Phi is an isomorphism okay that is the proof so Phi is an isomorphism which is what we want to prove okay so let us now look at two consequences. The first corollary of this theorem let V be finite dimensional and L belong to V double star so let me say L be a linear functional on V star then there exists a unique F in V star such that I am sorry there exists a unique V is finite dimensional L is a linear functional V star then there exists a unique X in V such that L of F equals F of X for all F in V star this is an immediate consequence of the previous theorem linear functional on V star so L belongs to V double star okay linear functional on V star so it is an element of V double star okay proof L is a linear functional on V star so L belongs to V double star the previous theorem exhibits an isomorphism explicit isomorphism Phi from V to V double star by Phi of X equals L of X it means that for any L and V double star there exists an X such that Phi of X equals L of X we have established that this Phi is an isomorphism now look at what you have on the right this function is on to so for any L and V double star there is an X such that Phi Phi of X is equal to that L okay so by the previous theorem there exists there exists X element of V such that Phi of X equals L by the previous theorem Phi is on to that is what I am using here so anything in V double star has a preimage in V so that is my X but since Phi is an isomorphism this X must be unique and so X is unique because Phi is an isomorphism the other thing is Phi X the definition is L X so this L is equal to L X the form of the L is also known this is equal to L X for some unique vector X in V so this L X equals L L is equal to L X so L is equal to L X so if you look at the action of L on F it is the same as the action of L X on F but L X on F I know that is the evaluation functional that is F X. So what I have shown is L F is equal to F X that X is unique so it is a almost a one line proof right from the previous theorem okay so remember that we have made use of the factor Phi is on to here Phi is on to so for any L in V double star there is a preimage this preimage must be unique because Phi is an isomorphism Phi is injective and so there exists an X such that this happens but then what is the definition of Phi Phi of X is L X so this L is equal to L X write down what this definition L X means we have written it is an evaluation functional so the L that I started with in V double star must satisfy this equation okay this is one consequence the other consequence is what I the other consequence is what I mentioned right in the beginning this correspondence between a basis for V star and a basis for V in fact there exists a unique basis there exists a unique basis I can include that also given a basis V star of V star this theorem states that if V is finite dimensional then there is a basis V of V such that this V star is a dual of the basis B okay this is again a consequence of the previous theorem really the previous corollary. So proof is as follows I will start with the basis B star I can write down the elements explicitly B star is let us say F 1, F 2, etc F n let B star be a basis V star V star is finite dimensional I have a basis corresponding to this there is a dual basis let B double star equal to corresponding to any basis in a finite dimensional vector space there is a dual basis so let B double star equal oh please remember that this B double star should not be this is just a notation okay see we will later identify V double star with V so this is just please remember this is just a notation maybe I can okay I will just leave it as it is see this is I am writing down this is the notation for a basis in V double star basis for V double star. So B double star equals let me say L 1, L 2, etc L n I know that the number of elements in this must be the same as a number of elements in B star let this be the dual basis corresponding to B sorry corresponding to B star then I know how these are related then L i of F j this is delta i j F i of U j equals delta i j U i's are replaced by F i's U j's are replaced by F j's F's are replaced by L now look at each L i each L i is in V double star this is a basis of this is a basis of V double star okay this is a basis of V double star each is an element in V double star by the previous corollary each element has a pre-image okay for every i there exists unique U i that is what the previous theorems there exists unique U i such that unique U i in V let us get back to the previous result for every L in V double star there exists a unique X such that Phi of X equals L for every i there exists U i in V such that L X equals L tell me if this is alright I will use U i L U i equals L i instead of L in the previous corollary I have L i instead of L X I have L U i for every i I can do this take L 1 for instance take i equal to 1 for instance I am looking at L 1 L 1 is in V double star there is a unique pre-image I am calling that U 1 how are these two related L is equal to L X is the previous theorem L 1 equal to L U 1 I do this for all i L i equals L U i for all i this is okay collect these U i's it is natural to expect these U i's form a basis for V we will only prove independence then it follows that these vectors form a basis okay so I will collect U i's is the construction clear let us set B as U 1 U 2 etc U n then we show so the claim is that this B is a basis for V such that B star is dual to this these two things we have to prove this B is a basis and the fact that the B star that we started with is dual to this okay it is is it clear that it is enough to show that this is independent linearly independent it is linearly independent the number of elements is n the dimension of V is n so it must be a basis so we will show independence of this linear independence of this so you start with a linear combination equate that to 0 suppose alpha 1 U 1 plus alpha 2 U 2 etc is 0 we must show that each scalar is 0 if this is a 0 vector I apply for any I apply a linear functional f f of 0 0 apply f apply f f is in V star take any linear functional on V and then apply f to this so I get alpha 1 f of U 1 plus alpha 2 f of U 2 etc alpha n f of U n this is 0 but f of U 1 you go back to this f of U 1 is L 1 of f agreed f of U 1 is L 1 of f say I am using I am using this really use this L 1 of f is evaluation of f at U 1 right this equation tells us that L I of f is f at U I because L U I is a evaluation function L x is f x L x of f is f x so L I of f is L U I of f let me write down L also something that we have used already also L I of f is L U I of f because each L and V double star for each L and V double star there is a unique x L U I of on the other hand is f at U I L x of f by definition is f at x so f U I can be replaced in the above left hand side some by L I of f so alpha 1 L 1 f plus alpha 2 L 2 f etc alpha n L n f is equal to 0 remember you need to apply carefully the appropriate argument is this okay now this is true for I started with an arbitrary f so this is true for all f this is true for all f and V star do I have the required conclusion I have something like T x equal to 0 for all x so T must be the 0 map let me elaborate so what this means is that alpha 1 L 1 plus alpha 2 L 2 etc alpha n L n this acting on f this is 0 for all f in V star that is what this means alpha 1 L 1 f plus alpha 2 L 2 f etc this is equal to 0 for f and V star means this because what is the definition of this is alpha 1 L 1 f plus alpha 2 L 2 f etc okay so I have a function which for every point in the domain gives 0 so this function must be 0 but this is now an equation involving the basis vectors basis functionals since L 1 etc L n are linearly independent it follows that each scalar is 0 okay and so we wanted to establish the linear independence of these vectors u n etc u n I started with this combination I have shown that each scalar is 0 so B is a basis for V why is B star dual to this we must verify that f i of u j equals delta i j okay also f i of u j equals f i of u j go back to this equation go back to this equation f i of u j okay let me write once again what I am using also li equals lu i this is what I want to use that is these are elements in V double star these are identical elements in V double star so their action on f must be the same okay that is li of f is equal to lu i of f but lu i of f is f of u i for every f in V star okay now consider f i of u j consider f i of u j f i of u j is f i of u j instead of i I have j l u j of f i be careful with the indices here I want f i of u j forget f i for the moment call it just f I want f of u j f of u j is l u j of f l u j of f but instead of f I have f i so it is l u j f i is that okay see I want to show that B star is dual to this I must show that f i u j equals delta i j so I am computing f i u j f i u j the definition comes from this f of u j is l u j of f but I want f i of u j so that is l u j of f i but l u j of f i is equal to you can go back to that li f j equals delta I could have restricted my attention to li but does not matter li f j is delta i j so this is equal to delta i j l u j okay maybe one more step right l j f i l j f i delta i j equals delta j i delta i j equals delta j okay so this is delta i j this is actually delta j i but delta j i is delta i j identity matrix is a symmetric matrix is it okay then f i u j equals delta i j so the basis B star that we started with is in fact dual to the basis u 1 u 2 etc u n that we have constructed is it clear so for every basis for every dual basis rather for every basis of V star there exists a basis B of V such that the basis that we started with is the dual basis for B okay okay now there is one final section on the double annihilator I will prove maybe one or two results on the double annihilator and one consequence now before I discuss the double annihilator say I want to discuss something like S double not I want to know whether this is equal to S any subset of V first of all does it make sense first of all this is a subset of V S double annihilator is a subset of V double star whether this make sense we have to see for that you go back and look at this functional isomorphism phi from V to V double star defined by phi of X equals L X see this establish an isomorphism a 1 to 1 correspondence between V and V double star so what we do here after is given any element in V double star we will identify that with a unique element in V this identification will enable us to think of V double star as being the same as V this is identification we will identify V double star with V what is the basis for this identification that is this isomorphism any L in V double star we know can be mapped to a unique X in V so any L in V double star there is a unique X so we can think of V double star as being the same as V in the rest of the discussion that enables us to talk about S double annihilator being equal to S now for any subset this is not true we need to look at subspaces so for subspaces we will prove this equation okay that is the first result. So what I want to do is to prove this theorem that if you have S is a subset of a finite dimensional vector space V then the double annihilator of S is equal to span of S the double annihilator of S is equal to span of S the reason why the right hand side cannot be an arbitrary subset is that it must be a subspace see when you take the annihilator it is always a subspace so the double annihilator is a subspace we will show that this subspace is same as span of S the proof is as follows okay so remember that these two are subsets of different spaces but there is an identification that is possible by means of this isomorphism it is with this identification in the mind that we will prove that these two subspaces are the same okay. Let me set W as span of S then this is the same as this is what we will show we must show that double annihilator is W the proof will make use of the fact that dimension of W plus dimension annihilator of W is equal to dimension of V which we proved last time whenever W is a subspace dimension W plus dimension W not is equal to dimension of V the annihilator of W the dimension of the annihilator of W satisfies this I can do this for the double annihilator that happens in V star dimension of the annihilator of W plus dimension of the double annihilator of W that must be equal to dimension of V star this is happening in V star this happens in V star and so this is equal to dimension of V star all these equations make sense V is finite dimensional so V star is finite dimensional cancel the right hand sides right hand sides are the same so this minus this this minus this so I am cancel I am subtracting one from the other it follows dimension W is dimension double annihilator okay and yeah so these subspaces are the same dimension I want to show that W is equal to the double annihilator is it enough if I show that one subspace is contained in the other then it will follow that subspaces are the same okay so we will show that W is contained in the double annihilator it will then follow that W is equal to W double annihilator again this must be understood in the sense of the identification W is a subset of V W double annihilator a subset of V double star but we must use the identification coming from the isomorphism Phi okay let me write down W not W not by definition is a set of all functionals that take the value 0 for all vectors in W so it is a set of all F in V star such that let us say F of X is equal to 0 for all X in W this is W not the first annihilator what is a double annihilator the double annihilator is a set of all L in V double star set of all L in V double star such that see W annihilator is a subspace it is L in V star such that L of F is equal to 0 for all F in W not this is a double annihilator I have just used the same definition as the previous one notation is different previously W is in V so it is a vector here these are linear functionals okay now write this in their expanded form identify with this L is in V double star so there is an identification with a vector in V use that identification we will be able to show that W is contained in the double annihilator okay this is the set of all let us say L in V double star such that set of all L X in I know that each L in V double star is of the form L X each element is of the form L X Phi is an isomorphism so it is set of all L X in V double star such that 0 equals L X of F that is F X by definition this for all F all that I have done is to remember that any L in V double star is of the form L X but this L X I told you can be identified with X so I will use X itself set of all X in V double star I am identifying that with V so set of all X in V such that I will just use the last equation F of X equals 0 for all F in W this is the double annihilator set of all X in V such that F X equal to 0 for all F in W star so I have made use of the identification L X can be identified with X and the rest is just use this I have removed L X now F X equal to 0 for all F in W not okay there is only one last step what is that last step let X element of W then for all F in W not F of X equals 0 by definition that is X belongs to W perpendicular W perpendicular we now know is a set of all X that are sent to 0 by any linear functional that is in W not right is a set of all X that is sent to 0 by any linear functional in W not if X is in W then for any linear functional F in W not F X is 0 so if X is in W then X must be in W double annihilator so W is contained in double annihilator two subspaces having the same dimension with one contained in the other means they are the same okay this is another consequence when you do the double annihilator you could identify that with the subspace that you started with okay so let me stop here in the next lecture I will discuss the notion of the dual linear transformation of a linear transformation and then derive two natural results one result that we proved earlier that the row rank is equal to the column rank this was this was proved using linear equations homogeneous equations then row reduced echelon forms etc we will give a conceptual proof this time okay without calculations a conceptual proof of the fact that the row rank of a linear transformation is equal to row rank of a matrix is the column rank of the matrix and also how are the matrices of a linear transformation and it is transpose related if T is a linear transformation A is the matrix of the linear transformation relative to a basis then what is the matrix of the linear transformation T transpose relative to two particular bases we will observe that it is the transpose of the matrix of the linear transformation that we started okay so these notions we these results we will prove next time.