 So we are discussing the notions of dual basis dual spaces annihilators later we will discuss the double dual the transpose of a linear transformation okay last time we had proved the following result that dimension of W plus dimension W not equals N we are assuming that V is finite dimensional okay V is finite dimensional and W is a subspace of V then this holds where W not is an annihilator of W so I am just recalling the definition W not is the set of all set of all functionals set of all linear functionals on V that take each X and W to 0 this is W not called the annihilator of W and I was going to state two consequences the first one is the following let V be finite dimensional and W be a subspace of V of dimension K then W is the intersection of N minus K hyperspaces W is the intersection of N minus K hyperspaces of V what is a hyperspace of a vector space any subspace of dimension one less than the dimension of the space V that is called a hyperspace. So one of the consequences of really not this result but the proof of this result okay so let me only take the first few steps of the proof of the previous theorem from that we will derive we will derive the fact that this W is the intersection of N minus K hyperspaces okay so recall the proof the first few lines of the proof of the previous theorem dimension W plus dimension W not equal to N let us say W is W has this as a basis B equals U1, U2, etc UK this is a basis for W and what we did was to extend this to a basis for V I will call this BW BV, BV is U1, U2, etc UK, UK plus 1, etc UN the dimension of V is taken to be N this is a basis for and then what we did was to construct the dual basis for this let B star equal F1, F2, etc FN be the dual basis of BV then how are these functionals and these vectors related by definition we must have F i of Uj equals delta ij okay in this notation I will make the following claim so let us look at the following the claim that I am making here is that this subspace W is the set of all X and V such that F i of X equal to 0 for all i K plus 1 less than or equal to i less than or equal to n remember this W is a subspace of V of dimension K I am trying to describe this W in terms of the earlier functionals FK plus 1, FK plus 2, etc FN okay suppose let us say I have proved this claim then can you see that this is okay what is this again this is a set of all X such that FK plus 1 X equal to 0, FK plus 2 X equal to 0, etc FN plus 1 sorry FN X equal to 0 this is a set of all X that lie in the intersection null space of FK plus 1 null space of FK plus 2, etc null space of FN that is if I had proved this then it would follow the W is intersection null space FI I equals K plus 1 to n intersection of null spaces of linear functionals each linear functional has a property that the null space is one dimensional okay null space is n minus one dimension range is one dimensional null space is n minus one dimensional so each linear functional has a property that the null space is a hyperspace I have written this as a intersection actually n minus K sub K plus 1 to n, n minus intersection of n minus K hyperspaces okay so we need to only prove this is that clear suppose we prove this then it follows that W is a intersection of n minus K hyperspaces it is clear that it is intersection of n minus K subspaces but each subspace is null space of a linear functional each is a subspace of a null space each subspace is a null space of a nonzero linear functional null space of a nonzero linear functional is n minus one dimensional and so it follows that this W is intersection of n minus K hyperspaces so we need to only demonstrate this okay so let us prove this one way is obvious but I will prove both I have two sets A equals B I must show A contained in B and B contained in A okay let us start with so we need to prove this claim now so let me take X and W I must show that this X has a property that whenever I is greater than or equal to K plus 1 Fi of X is 0 that X element of W then for W I have taken this as a basis so this X can be written as alpha 1 U 1 plus alpha 2 U 2 etc plus alpha K U K look at look at I greater than or equal to K plus 1 and then Fi of X I greater than or equal to K plus 1 Fi of X remember I need to show that Fi of X is whenever I want to show left hand side contained right hand side I must show that if X belongs to W then FK plus 1 X is 0 FK plus 2 X is 0 etc Fn of X is 0 so I take I to be greater than or equal to K plus 1 I must show this is 0 okay but Fi of X Fi is linear so this is alpha 1 Fi of U 1 etc plus alpha K Fi of U K I have just used linearity of Fi okay but remember that this is a F1, F2 etc Fn is a dual basis and so whenever I is greater than or equal to K plus 1 each of this is 0 because look at these indices you get from U 1 etc up to U K so when I is equal to K plus 1 or more I is not equal to J so each term is 0 so this is 0 I is greater than or equal to K plus 1 so Fi for any U J when J is less than or equal to K is 0 that is the definition so Fi of X is 0 so what we have shown is that the left hand side W is contained in this subset so W is contained in set of all X and V such that Fi of X equal to 0 for all I satisfying we need to show the converse. Suppose that Fi of X is equal to 0 for all I I running from K plus 1 to N suppose X is a vector that belongs to the right hand side subset right hand side subspace I must show that left belongs X belongs to the left hand side W now X is an arbitrary element so I can write X using this basis so I let me use some other scalars X is a linear combination of U 1 etc U n so X is beta 1 U 1 plus beta 2 U 2 etc plus beta K U K plus beta K plus 1 U K plus 1 etc beta N U n U 1 U 2 etc U n is a basis for V and so I can write X in this manner Fi of X is 0 so I start with 0 that is Fi of X Fi of X Fi is linear beta 1 Fi of U 1 etc plus beta K Fi of U K plus etc okay let me write the next term also beta K plus 1 Fi of U K plus 1 plus etc plus beta N Fi of U n after applying Fi I get this for each I running from K plus 1 to N X must satisfy Fi of X is equal to 0 but look at what you have on the right hand side on the right hand side there is only one term that remains because Fi of U J is delta I J what is that term see I runs from K plus 1 to N I is fixed okay and so this is Fi of U I all other terms are 0 all other terms are 0 beta I Fi U I but Fi U I is 1 so this is beta I so what have we shown we have shown that if Fi X is equal to 0 then beta I is equal to 0 but what are the values that I can take I takes I runs from K plus 1 to N which means beta K plus 1 0 beta K plus 2 is 0 etc so what we have shown is that beta K plus 1 equals beta K plus 2 etc equals beta N equal to 0 so go back and look at the representation for X look at the representation for X X is beta 1 U 1 etc from K plus 1 the term onwards they are all 0 so X is just this this term is 0 the representation of X the scalars corresponding to B K plus 1 etc they are all 0 that is what we have shown because see this Fi X is equal to 0 for all I running from K plus 1 to N so these scalars beta I running from K plus 1 to N are 0 so X is a linear combination of U 1 etc UK but then that is U 1 U 2 etc UK is a basis for W so this X must belong to W so I have started with an arbitrary vector on the right hand side subset I have shown that that belongs to W so right hand side subspace is contained in the left hand side subspace so W is equal to this and hence the theorem corollary let me just write one more step to make the final part transparent let me write that here so what we have done is W is a set of all X in V such that Fi of X is 0 for all I running from K plus 1 to N now you see that this is the intersection of the null space of Fi I running from K plus 1 to N anything on the right hand side must be in the null space of Fi for each I running from K plus 1 to N so this is and there are N minus K this intersection has N minus K terms so there are N minus K subspaces each is a null space of a non-zero linear functional so each subspaces of dimension N minus 1 null space of Fi that is N minus 1 so I have written W as an intersection of N minus K hyperspaces okay that is the complete proof of this corollary there is another corollary which talks about the relationship of two subspaces and they are annihilators really so second corollary V is finite dimensional and W1, W2 are subspaces of V then W1 is equal to W2 if and only if they are annihilators are equal W1, 0 is W2, 0 two subspaces are equal if and only if they are annihilators coincide this again uses the proof of the previous theorem out of which one part is easy if S is equal to T then annihilator of S is equal to annihilator of T that is easy to see so W1 equals W2 implies W1, 0 equals W2, 0 annihilators must be the same okay it is a converse that is non-trivial here to prove the converse to prove the converse what we will do is assume that W1 is not equal to W2 so that W1, 0 is not equal to W2, 0 conversely let us suppose that W1 is not equal to W2 we will show that W1, 0 is not equal to W2, 0 this is really the converse because what is the meaning of this W1 not equal to W2 implies W1, 0 is not equal to W2, 0 this is the same as saying W1, 0 equals W2, 0 implies W1 equals W2 that is a converse if A implies B then not B implies not A that is what we are using the statement A implies statement B the negation of statement B implies negation of statement A so if we demonstrate W1 not equal to W2 implies this then it follows that if this statement is not true then negation of the statement is true so it follows that if this is not true that is W1, 0 is equal to W2, 0 implies W1 is equal to W2 which is really the converse okay so let us prove this. Now W1 is not equal to W2 they are subspaces so as sets they are not equal so one is not contained in the at least one of them is not contained in the other for the sake of using this notation let me let me assume that W2 is not contained in W1 okay I am saying without loss of generality so this is what I am going to prove I am going to prove that W1 not equal to W2 implies W1, 0 is not equal to W2, 0 without loss of generality let us assume that W2 is not contained in W1 okay we show that W1, 0 is not contained in W2, if W1 were not contained in W2 one could show that W2, 0 is not contained in W1, 0 by a similar argument so there is no loss of generality in assuming W2 is not contained in W1. Now I will have to go back to the previous notation for W1 let me take U1, U2 etc UK as a basis I am calling that BW1 this is a basis for W1 this can be extended to a basis for V okay as before I am assuming that V is n dimensional so there are n vectors here this basis has been extended to a basis for V, if W2 is not contained in W1 what it means is that there is a vector in W2 which is not in W1 okay is it clear then that there exists S greater than or equal to K plus 1 such that US belongs to W2 obviously US does not belong to W1 do you agree with this W2 is not contained in W1, W1 has U1, U2 etc UK as a basis so anything in W1 is spanned by these vectors now you take US where S is not 1 to K so S is greater than or equal to K plus 1 if all of these vectors belong to W1 then W1 is a whole of V in which case this cannot happen W2 is a subspace so there is at least one S for which US does not belong to W1 but those UIs that belong to W1 are indexed by 1 to K so if there exists a US that does not belong to W1 it must be corresponding to an index that is greater than or equal to K plus 1 so there exists US where S is there exists S greater than or equal to K plus 1 such that US does not sorry US does not obviously belong to W1 but it belongs to W2 okay now look at the functional FS that will do what we require here look at the functional FS I have now constructed dual basis let B star equals F1 F2 etc be the dual basis of V corresponding to the dual base corresponding to the dual basis BV corresponding to BV then as before FI of UJ equals delta IJ I am just calling your attention to FS look at FS FS of US we know by definition must be equal to 1 okay and that is not 0 it means FS cannot belong to see US belongs to W2 and FS is a functional that takes US to a non-zero value so FS cannot belong to W2 0 FS does not belong to W2 0 W2 0 is a set of all functionals that take all elements in W2 to 0 I have produced one element in W2 which is taken to a non-zero value by the functional FS so FS does not belong to W2 0 but obviously FS belongs to W1 why but FS of UI I will use J FS of UJ equal to 0 for all J such that 1 less than or equal to J less than or equal to K by the definition of the dual basis because J runs from 1 to K S is greater than or equal to K plus 1 so J can never be equal to S J runs from 1 to K S is greater than or equal to K plus 1 so S is never equal to J so this is 0 in other words FS of U1 FS of U2 etc they are all 0 which means FS belongs to W1 0 because anything in W1 0 is a linear combination of these and so if you apply the linear functional FS to that vector that will take the value 0 should elaborate okay let me do that quickly so if X belongs to W then X is some linear combination of these UK's so that FS of X is what I want FS of X is delta 1 FS of U1 etc delta K FS of UK each term is 0 so this is 0 that is FS belongs to W1 0 so remember that is what I wanted to show if W2 is not contained in W I wanted to show W1 0 not contained in W2 0 I have a functional that belongs to W1 0 I have a vector or a functional that belongs to W1 0 but that is not in W2 0 FS is not in W2 0 and so this holds and so the converse holds okay so W1 0 equals W2 0 implies W1 must be equal to W if the annihilators coincide then the corresponding subspaces only for subspaces subspaces must coincide by the way this formula does not hold if you take arbitrary subsets if W1 is a subset W2 is a subspace then W1 0 could be equal to W2 0 without W1 being equal to W2 the underlying subsets must be subspaces okay okay to consolidate let us work out 2 examples to consolidate the ideas of dual basis annihilators and then determining elements in a dual basis. So let us consider the following examples let us say I have first one determine the subspace W of let us say I take R4 determine the subspace W of R4 for which the functionals given below are the annihilators I am given let us say 3 annihilators I will call F1 of X X is an R4 so let us say I have X1 plus X2 minus X3 plus X4 F2 of X is X1 minus 2 X2 F3 of X is 3 X2 plus 2 X4 the question is you are given functionals these functionals annihilate a certain subspace what is that subspace remember a subspace just now we have seen W is equal to set of all X element of V such that Fi X equal to 0 for all I running from K plus 1 to N subspaces can be given by the set of annihilators of that subspace okay subspace can be given using the annihilating functionals also okay so we need to solve this problem. So what is the definition of W? W is a set of all X in R4 in this case such that F1 of X equals F2 of X equals F3 of X equals 0 now what you will see is that again it is elementary row operations write down these three equations these are homogeneous equations these three systems give me the following sorry there is only one system X1 plus X2 minus X3 plus X4 equals 0 X1 minus 2 X2 equals 0 3 X2 plus 2 X4 equals 0 elementary row operations so I have the matrix 1 1 minus 1 1 1 minus 2 0 3 0 2 I will apply elementary row operations to determine the set of all solutions I will keep this as a pivot and then straight away make this 0 okay let me do it like this I have 0 1 minus 1 by 3 1 by 3 I can divide this also by 1 0 1 0 divided by 3 2 by 3 taking the second next operation will be 2 okay multiply this by 3 so that was unnecessary I have 0 0 1 1 0 1 minus 1 by 3 1 by 3 1 0 minus 2 by 3 2 by 3 yeah obviously I am not doing it very efficiently but let us look at the solution so now I have a 3 by 4 system where this 3 by 3 part is the identity part so what follows is that remember that this came from this set there are 4 variables what this tells me is that I must fix the last variable X4 so let us say X4 is alpha then X1 is minus 4 by 3 alpha X2 is also minus 4 by 3 alpha X3 is alpha minus alpha there is a problem with the solution from this step to this step it is correct from here to here see I am getting this as the pivot row then 1 by 3 times this plus this so these entries this becomes 0 1 by 3 times this plus this 2 by 3 let me look at it once again up to this step is correct okay so I am keeping this as fixed okay then multiply this by 2 by 3 to cancel this 1 0 2 by 3 2 by 3 plus 2 by 3 4 by 3 that is okay so this entry is minus 2 by 3 alpha please check the calculations so what is W then W is one dimensional W is one dimensional because it is a any multiple of let me just say W is span of this vector let me multiply throughout by 3 X1 is minus 4 X2 is minus 2 so let us say 4 2 X3 is multiplying throughout by 3 and minus 1 I am multiplying by 3 I am multiplying by minus 3 okay 6 minus 3 minus 3 X1 minus 2 X2 3 X2 that is 6 minus 6 is 0 so this is the subspace W which is annihilated by these 3 functionals okay again it is only solving homogeneous systems one more problem where we will determine the annihilator second example find W0 if W is spanned by these vectors let us say U1 is let me take one more W is subspace spanned by these 4 vectors I am not claiming that these vectors form a basis for W, W is spanned by this set of vectors what is W0 annihilator of W okay remember that W is a subspace of R5 not the whole of R5 there are only 4 vectors here so W is the subspace of R5 I must find the functionals that generate W0 okay. So I will determine a dual basis for a basis corresponding to for a basis containing for a basis contained in this set these 4 may not be independent okay so let us what do we need to find we need to find W0 okay so let us look at the functional if F belongs to W0 what is the condition that F must satisfy okay but before that let us look at these 4 I will again write it in the matrix form 1 1 minus 1 minus 1 minus 1 minus 1 1 1 0 0 0 0 0 2 apply elementary row operations W is a subspace which is the row space of this matrix W is a subspace corresponding to the row space of this matrix okay row space does not change if we do elementary row operations so I want to do just probably one more one operation I not reduce it to the Rode-Steclone form I will just do one of these so minus this plus this I will keep this one also as it is and then observe the last one is 0 0 2 in the next step what could be done is this 2 could be made 0 so I will make that here itself this could be made plus 2 does not make a difference so what is clear is that the 4 vectors are not independent only 3 of them are independent dimension of W is 3 we need to determine W0 okay what we know is that any functional F in R5 any functional on R5 if F belongs to R5 star then F can be written as alpha 1 X1 plus alpha 2 X2 alpha 5 X5 any functional on Rn can be written as A1 X1 plus etc A and Xn so Fx is of this form I need to determine okay I do not know how many are there so I will just keep I need to determine F such that F of these 2 vectors I will call them V1 V2 V3 I will call these row vectors V1 V2 V3 I must determine all F that satisfy these 3 equations again homogeneous equations okay so let us write this it is almost in Rode-Steclone form so I want 1 1 minus 1 minus 1 1 into X1 X2 X3 X4 X5 equal to 0 I will have to interchange okay let us say I make I keep this as it is so from here I get this 0 I do not have to really do it but I get a 1 here this is not the Rode-Steclone form so let me not get into the details there from this I can write away tell you what these functions are see from this what follows is that remember I need to determine F so I need to determine the Phi unknowns here alpha 1 alpha 2 etc alpha 5 is that clear I need to determine W0 F belongs to W0 if F satisfies these equations these equations give me this into alpha 1 alpha 2 etc alpha 5 that is equal to 0 now what is clear is that from this alpha 5 is 0 is that okay alpha 5 is 0 okay let me write down second equation tells me alpha 5 is 0 third equation okay so from this I have 3 equations in Phi unknowns so I need to fix 2 of them which one do I fix it cannot be the third one it cannot be the first one and so it is a second and the fourth which you get by the Rode-Steclone matrix so let me say I fix alpha 2 equals alpha see this diagonal entry this is 1 2 this is 3 so these 3 these 2 will not be fixed alpha 5 is already 0 1 3 so 2 and 4 alpha 4 is beta then determine the others in particular the third equation gives me alpha 3 alpha 3 is alpha 5 minus alpha 4 minus beta the first one should give me alpha 1 this is gone alpha 4 plus alpha 3 beta plus alpha minus sorry alpha 5 is 0 alpha 4 plus alpha 3 that is 0 so alpha 1 plus alpha 2 is 0 so alpha 1 is minus alpha so can you see that the basis consists of just 2 functionals because there are just 2 variables alpha beta all the others are in terms of these let me summarize f belongs to W0 implies f is of the form this is what I have minus alpha x 1 alpha 2 is alpha plus alpha x 2 alpha 3 is minus beta alpha 4 is beta alpha 5 is 0 alpha 5 does not figure here now I can write alpha and beta are can take arbitrary values in particular alpha 0 beta 1 alpha 1 beta 0 so I can write this as alpha times x 2 minus x 1 plus beta times x 4 minus x 3 which means I can write this as alpha times f 1 of x plus beta times f 2 of x f 1 and f 2 are independent functionals f 1 of x is x 2 let us say minus x 1 plus x 2 f 2 of x is a second term x 4 minus x 5 sorry x 4 minus x 3 f 1 is this f 2 is this any f and W0 is a linear combination of these 2 these 2 are independent these 2 are independent because you can think of f 1 as the vector minus 1 0 sorry minus 1 1 all other entries 0 f 2 is 0 0 minus 1 1 0 so these 2 are independent okay finally W0 is span of f 1 f 2 so please verify the calculations essentially it is solving homogeneous equations even remember the first problem when we determine the dual basis that was solving homogeneous equations okay. Let me stop here next time let us discuss the notions of the double dual once we have done once we go from V to V star can we go from V star to V double star so this can be done and when we discuss the notion of double dual we will also consider the following question which we have not dealt with before what we know is that given a basis for V finite dimensional case given a basis for V there is a dual basis for V so there is a basis for V star where there is a natural correspondence between the basis for V that we started with and the basis for V star that we constructed. The other question is given a basis for V star is there a basis for V such that the basis for V star is the dual for the basis of V that we would construct the answer is yes for finite dimensional space the answer is yes for infinite dimensional space the answer is no infinite dimensional spaces will be discussed in function analysis for finite dimensional spaces we will show that the answer is yes okay this is one of the main results that we will prove in the next lecture okay let me stop.