 We are discussing the notion of dual basis okay let me start with an example, example where I show how a dual basis corresponding to a basis of R3 could be constructed okay let us consider the following vectors last vector is 2, 2, 0 you can verify that these 3 vectors form a basis of R3 what is the basis dual to this basis okay let us try to construct that simply using the definition of what the dual basis is. We are using B star okay to denote the dual basis so let me say B star has the vectors functionals F1, F2, F3 be the dual basis corresponding to the basis that we started with then we know that by definition this 9 equations must be satisfied by these 3 functionals F i of U j equals delta i j, i j vary from 1 to 3 this 9 equation must be satisfied by these functionals using these equations we will determine these functionals but you will see that this is where what we learnt for solving systems of equations will come in handy what are the equations that we need to solve let me write it in full and then write the compressed form and then use what we learnt earlier what this means is for to begin with F1 of U1 is 1, F1 of U2 is 0, F1 of U3 is 0 okay so I have these equations but okay before I write let me also use this observation let me also use this observation that I made the other day if you have F as a linear function on Rn then F as the representation F of X equal to A1 X1 plus A2 X2 etcetera An Xn okay let me use that representation by observing that F element of R4 star R3 star in this example dual space F belongs to this implies that F of X equals let us say alpha 1 X1 plus alpha 2 X2 plus alpha 3 X3 where as usual X is in R3 the notation for X is X1 X2 X3 okay so this was proved first immediately after giving the definition of F linear functional we will make use of this now so these F1, F2, F3 must satisfy these equations where each Fy will have certain coefficients okay so let me say for F1 I will use the coefficients alpha 1, alpha 2, alpha 3 that is F1 of X is alpha 1 X1 plus alpha 2 X2 plus alpha 3 X3 F2 is another linear functional beta 1 X1 plus beta 2 X2 plus beta 3 X3 so alpha beta and then let us say for F3 gamma 1, gamma 2, gamma 3 okay so what is the problem is to determine these 9 numbers for instance if I determine the first 3 numbers I know what F1 is F1 of X is alpha 1 X1 plus alpha 2 X2 plus alpha 3 X3 so I need to determine these 9 numbers there are 9 conditions 3 into 3 9 conditions coming from this so there is a unique solution okay okay let us then write down these equations in the light of these representations I have the following F i of Uj so I am going to write down F1 of U1 this is 1, F1 of U2 is 0, F1 of U3 is 0 I am taking first the functional F1, F1 goes with the 3 numbers alpha 1, alpha 2, alpha 3 okay so what is F1 of U1 on the let me write like this it is F1 U1 equals 1 gives me alpha 1 minus alpha 2 equals 1 do you agree F1 of U1, U1 is 1, 0 minus 1 what is it okay just to confirm the notation that I am using here is maybe I will go back and write it here this is a general functional for F1 what does this mean this means F1 of X is alpha 1 X1 plus alpha 2 X2 plus alpha 3 X3 for F2 it is beta 1 X1, F2 of X is beta 1 X1 plus beta 2 X2 plus beta 3 X3 similarly for F3 so I am going to determine F1 of U1, F1 of U2, F1 of U3 this is the definition of the dual basis I have just written down the first 3 equations so I get alpha 1 minus alpha 2 equals 1 that is the first equation F1 of U2 minus alpha 3 yes F1 of U2 just sum up alpha 1 plus alpha 2 plus alpha 3 that must be 0 F1 of U3 the first 2 2 alpha 1 plus alpha 2 is 0 that is alpha 1 plus alpha 2 is 0 okay let me write 2 alpha 1 plus 2 alpha 2 and then we can simplify it later this is the system that alpha alpha 1 alpha 2 alpha 3 must satisfy there is a similar system for beta 1 beta 2 beta 3 similar system for gamma 1 gamma 2 gamma 3 okay in order to solve the system what do we do this is like AX equal to B we need to we need to reduce A to the row reduced echelon form and probably A will be row equivalent to Y now the dual basis unique so in this case A will be row equivalent to Y in general it may not be but in this case a dual basis unique means the solution for this system must be unique AX equal to B has a unique solution square system if and only if A is invertible that is that gives rise to unique solution so this will this should be an invertible matrix okay instead of solving all these 3 systems one after the other we would just solve one system and substitute for the right hand sides see it is the same okay what happens to the next system for beta you will again get something like this see for beta I am determining F 2 for F 2 this is 0 F 2 of U 1 is 0 F 2 of U 2 is 1 F 2 of U 3 is 0 so that will give me beta 1 minus beta 3 equal to 0 beta 1 plus beta 2 plus beta 3 equals 1 this time 2 beta 1 plus 2 beta 2 equals 0 this must be the system that beta will solve and similarly for gamma the other system is gamma 1 minus gamma 3 equals 0 gamma 1 plus gamma 2 plus gamma 3 equals 0 this time finally the last equation 2 gamma 1 plus 2 gamma 2 equals 1 so it is essentially the same coefficient matrix A the right hand sides change so let us do it more efficiently than solving each of the systems individually so I will take the general right hand side reduce it to the row reduced echelon form and then substitute these vectors okay in other words I consider solving the following system consider so I write on the coefficient matrix 1 0 minus 1 1 1 1 2 2 0 just the row vectors that I have here this I want to solve this for a general right hand side this time I will take B 1 B 2 B 3 do elementary row operations reduces to row reduced echelon form where this part will be reduced to the identity matrix then instead of B 1 B 2 B 3 I will substitute these 3 right hand side vectors to get the solution okay so this I need to do elementary row operations from here I get 1 0 minus 1 B 1 will be kept minus this plus this 1 2 B 2 minus B 1 minus 2 times this plus this B 3 minus 2 B 1 please check calculations here okay next step I will keep this as the pivot row I need to make this 0 okay I will keep this as the pivot row so the next step is I get this matrix 0 1 2 B 2 minus B 1 this will be oh the first row is also kept as it is this will be minus 2 times this plus this minus 4 minus 2 minus 2 times this plus this B 3 minus 2 B 2 then I will divide throughout by minus 2 the next step then I make these 2 entries 0 so last row is kept as it is just add these 2 add these 2 2 B 2 okay let me write like this 2 B 1 plus 2 B 2 minus B 3 I need some space I am just adding these 2 rows to replace the first row then minus 2 times this plus this still have enough space okay minus 2 times this plus this this will go with a negative sign minus 2 okay what are the calculations minus 2 times this minus 2 B 2 plus B 3 minus 2 times this plus this plus B 2 minus B 1 is that okay minus 2 B 2 minus 2 that becomes plus B 3 yeah so that is B 3 minus B 2 minus B 1 that is this entry let us now it is clear this is the row row disc declaw this is the identity so the system as a unique solution as we expected let us now determine the solutions alpha 1 alpha 2 alpha 3 beta 1 beta 2 beta 3 gamma 1 gamma 2 gamma 6 for determining alpha 1 alpha 2 alpha 3 the right hand side is the vector 1 0 0 B 1 is 1 B 2 is 0 B 3 is 0 so what is F 1 then so let me write F 1 right away alpha 1 so that is the first vector B 1 is 1 so I get a 1 here so it is X 1 minus 1 this is 0 so F 1 of X is X 1 minus X 2 that is alpha 1 is 1 alpha 2 is minus 1 alpha 3 is 0 is it clear what I am doing this is for a general right hand side vector B to determine the solution corresponding to this system I will replace the right hand side vector by this vector this is B 1 this is B 2 this is B 3 just substitute there and then these are diagonal entries the first equation gives you X the first equation gives you alpha 1 equals this alpha 2 equals this alpha 3 equals this where B is 1 0 0 okay what happens to the second one second vector so B 2 so that is again a 1 X 1 minus 1 plus 1 is that okay I get a 1 here I get a minus 1 here I get a 1 again so that is F 2 F 3 corresponding to 0 0 1 0 0 1 minus 1 by 2 minus X 1 by 2 0 0 1 minus X 2 0 0 1 minus X 3 by 2 these are the 3 functionals now you can verify that these 3 functionals satisfy these 9 equations F i u j equals delta i j for instance look at F 1 of u 1 u 1 is 1 0 minus 1 so F 1 of u 1 is 1 F 2 of u 1 1 0 minus 1 X 1 plus X 3 1 and a minus 1 that is 0 F 2 of u 3 2 2 that is minus 1 plus F 3 of I am sorry I am calculating F 1 of u 1 F 1 of u 2 F 1 of u 2 is 1 minus 1 F 1 of u 3 is 2 minus 2 okay so just to verify so this is one method of doing it okay and we have done it little more efficiently than solving 3 systems okay this is an example where we have constructed a dual basis compare this with the example that I gave you last time I have given the dual basis first and then I have asked you to find the basis corresponding to which this is the dual basis okay that corresponded to the Lagrange interpolating polynomials for the nodal points T 1, T 2, T 3 okay let me move on I want to discuss the following problem a formula which somewhat reminds us of the rank nullity dimension theorem okay we will derive a formula and then look at two corollaries of this formula so what is the formula for I state this formula let me make the following observation suppose you have dimension of V is n okay so V is a finite dimensional vector space F is a linear functional let us say F is not the 0 functional then what is the rank of F what is the rank of F range of F see F is a function F is a function from V to R, R means R over R as a vector space what is the look at range of F range of F if F is not 0 is one dimensional so rank is 1 rank of F is 1 if F is not the 0 vector 0 functional then it means that rank of F rank of range of F is a subspace of R okay if it is not 0 then the subspace will be the whole of R and so rank of F equal to 1 if V is finite dimensional then by rank nullity dimension theorem it follows that the nullity of F equals n minus 1 okay rank plus nullity is the dimension of the domain space nullity of F is n minus 1 remember that nullity of F is what dimension of the null space of F that is n minus 1 null space of F is a subspace of V range of F is a subspace of the core domain null space of F is a subspace of V this subspace has dimension 1 less than the dimension of the whole space such subspaces are called hyperspaces any subspace of V of dimension n minus 1 with dimension n minus 1 with dimension n minus 1 is called a hyperspace dimension 1 less than the dimension of the original space that is called a hyperspace so what follows is that if F is a non-zero linear functional then null space is a hyperspace okay so F not equal to 0 implies null space of F is a hyperspace hyperspace of V the question is whether the converse is true this is what we will try to answer what is the converse I have a subspace of dimension n minus 1 is there a functional corresponding to this subspace that is is there a functional F such that null this subspace is null space of that function is the converse true what is the converse given a subspace of dimension n minus 1 is there a functional F such that null space of that functional F is equal to this subspace okay we will see that the a little more general question can be answered affirmatively okay in order to answer this question we need the notion of the annihilator so let me give this definition first the annihilator of a subset of a vector space to answer this question we need the notion of the annihilator of a subset so let me give this definition let us be a subset of V the annihilator of S let us denote it by S0 let us denote that by S0 is defined by S0 this is the set of all functionals set of all linear functions on V set of all linear functions on V that have the property that f of S equal to 0 for all S element of this set of all linear functions on V which take every element in S to the 0 number 0 real number so this is a subset of V star the annihilator is a subset of V star so it consists of linear functions it consists of certain linear functions now it does not matter what S is S0 is always a subspace S0 is a subspace of V star now that is easy to see because if f, g belong to S0 then I must show that alpha f plus beta g also belongs to S0 but if f, g belong to S0 then alpha f of S is 0 beta g of S is 0 their sum is 0 f is a linear function so this is a subspace so S0 is a subspace of V star our interest it is in determining the dimension of S0 given V is finite dimensional S0 is a subspace let us look at two extremes for this S0 if S is single term 0 what is S0 two extremes I said are you sure all functions all linear functions a linear functional must satisfy the condition that f of 0 is 0 if S is equal to single term 0 then S0 is a whole of V star if S is equal to V then S0 is the 0 functional S0 is a 0 functional okay let us now look at look at the dimension of a subspace and the dimension of the annihilator of that subspace they are related and this is a formula that will lead to an affirmative answer for this question for the converse question. So I want to prove the following theorem let V be finite dimensional so I will assume dimension of V is n W be a subspace this time not just a subspace of V then the following formula which I told you must remind you of the rank penalty dimension theorem holds look at dimension of W plus dimension of W0 this is the dimension of V okay now you will see the proof is also somewhat similar to that theorem okay remember this is not a straight forward result because this connects two numbers one for the vector space V the other one for the vector space V star okay so proof let us start with the basis for W let me say I have B equals U1 U2 etc UK let me emphasise that it is a basis for W this be a basis for W so W is K dimensional any basis for that matter any linearly independent subset can be extended to a basis of the whole space so I can extend this extend BW to a basis for the whole space V is finite dimensional so let me write BV for basis for V BV contains BK so it contains the vectors U1 U2 etc UK and also other vectors I will call them UK plus 1 etc U n U1 etc U n because dimension of V is n this is a basis for V I can construct there is a dual basis for this basis so let B star equal F1 F2 etc Fn be the dual basis for the basis BV this is the dual basis what we will show is that you kind of remove the first n functionals the remaining n minus K functionals we will show forms a basis for W not similar to rank probability dimension theorem the first vectors form a basis for the null space the remaining vectors form a basis for the range space okay so let us now look at FK plus 1 etc Fn we want to show that the claim is that these vectors form a these functionals form a basis for W not see we want to determine the dimension of W not we will be able to explicitly write down a basis for W not so this is what we will prove let me just list these functionals FK plus 1 FK plus 2 etc Fn these functionals form a basis for W not this is a claim suppose we have proved this claim then the result follows because the number of functionals here is n minus K there are n minus K functionals so if I proved that this is a basis then W not is n minus K dimensional W is K dimensional so K plus n minus K is n so it is enough if we prove this okay. Let us recall the following formula which was proved in the last lecture proof is as follows proof of the claim if I have any functional on V then this any linear functional on V then this linear functional is a linear combination of the dual basis functionals F1 etc Fn where the coefficients also can be given explicitly that is F is summation i equals 1 to n you look at F of U i and then multiply that with F i F is a linear combination of F1 F2 etc Fn because F1 F2 etc Fn form the dual basis so any linear functional can be written in terms of those in addition we also have information on the coefficients the coefficients are F U i okay if okay what is it that we want to show that these vectors form a basis for W not first of all are these are these functional in W not otherwise there is no sense are these functional in W not what do we need to show let us just keep this aside for a while I want to show that these functionals belong to W not that is these functionals must satisfy the property that they take the value 0 for all the points for all the vectors in W okay but look at this what we know is the dual basis satisfies this condition F y U j equals delta i j so if i is say I am looking at functionals from k plus 1 to n so if i is greater than or equal to k plus 1 and j is less than or equal to k there is no way these two can become equal so you will get the value 0 so if i is greater than or equal to k plus 1 and j less than or equal to k then F i of U j is 0 for all the i and j that satisfy these conditions F i U j has to be 0 in other words is that clear in other words F k plus 1 of U 1 F k plus 1 of U 2 etc F k plus 1 of U k they are all 0 F k plus 2 of U 1 etc F k plus 1 of U k they are all 0 so what is it that we have proved we have proved that these functionals take the you take okay you now take X and W see I want to show F is 0 on W then it follows that if I want to show F is 0 on W I must show that F of X equal to 0 for all X okay look at X belongs to W then X is a linear combination of those basis vectors so let me take some delta 1 U 1 etc plus delta n U n delta k U k now look at look at F i I am again considering i greater than or equal to k plus 1 I am again considering i greater than or equal to k plus 1 that is let us say F k plus 1 is a functional that we are using now look at F k plus 1 apply to this that will be delta 1 F i of U 1 etc delta k F i of U k but you observe that these scalar these superscripts 1 2 etc k they are always distinct from k plus 1 and so this has to be 0 and so what is it that I have shown whenever i is greater than or equal to k plus 1 and X belongs to W F i of X is 0 so in the first place these are functionals that belong to W not these are functionals that take 0 that take the value 0 for any vector in W okay now we need to show that this forms a basis we must show they are linearly independent and they span W not but linear independence is obvious because this is just a subset of the dual basis subset of a linearly independent set is independent so linear independence is easy we only need to show that span of these is equal to W not so we will show that W not is contained in span of these that is the last step F k plus 1 etc F n they are in W not so span of those vectors will again belong to W not there is nothing to see W not is a subspace it is only the other way around we need to show W not is contained in span of this okay okay let us now look at this representation I have written down if F belongs to V star then F equals i equals 1 to n F of U i F i W not is a subset of V star so W not consist of functionals W not consist of linear function let us remember so if F belongs to W not if F belongs to W not that is what I want to show is F belongs to W not implies F is a linear combination of these functions so let F belong to W not this means what then W not is a annihilator of W so any F will take the value 0 for anything in W so F of U 1 F of U 2 etc F of U k all these must be 0 because U 1 U 2 etc U k belong to W W not is a set of all functionals that take every vector in W to 0 in particular that must take the basis vectors to 0 so if F is in W not then these are 0 go back to this equation so F belongs to W not implies this representation has n terms but if F belongs to W not the first k terms are 0 so there are only n minus k terms summation i equals k plus 1 to n F of U i F i forget about these numbers this F is a linear combination of F k plus 1 etc F n which is what we wanted to prove this belongs to span of F k plus 1 etc F n okay so what we have done is we have shown that if F belongs to W not then F is in the span of this so that proves this part and so it follows that if you go back this claim has been proved and so it follows that W dimension W plus dimension W not is n okay W is k that is a dimension of W dimension W not is n minus k which we proved just now that is n that is a dimension of V okay let me stop.