 In the next few lectures I want to discuss a notion of linear functionals, duals, double dual and the transpose okay may be about 3 or 4 lectures. So let me write down the topics linear functionals, dual spaces, the concept of a dual space, the double dual and the notion of the transpose of a linear transformation okay. So let me start with linear functionals, a linear functional is a special case of a linear transformation when the codomind space is the underlying field okay. So definition of a linear functional, a linear map f from V to R where V is a real vector space is called a linear functional, it is called a linear functional. So what is important is the word functional, linearity we have encountered before throughout the course we will encounter this notion of linearity, we will in particular look at what are called linear functionals. So a linear functional is a linear map from V to the underlying field in most of the cases for us the underlying field will be R, so this is a real linear functional okay. Let us look at some examples okay by the way what does it mean, linear functional f means we always use T, T from V to W for a functional I will use the word f the first letter of functional, so what is the meaning of this, this means f of alpha x plus y is alpha f x plus y where x y come from V and alpha comes from R, only to emphasize what a linear functional is f of 1, what you must remember is that on the right I have addition of real numbers okay this is addition of real numbers, the addition on the left is happening in V okay. Let us look at examples, the first example is I can take any f n and define a functional over f where f is a field real complex etc but again I look at a specific case I will define f from R into R as follows, f from R into R any vector must be mapped to a real number, I will define that as f of x equals a 1 x 1 plus a 2 x 2 plus etc plus a n x n where I am given the numbers a 1 a 2 etc a n, x is an R n I am given the numbers a 1 a 2 etc a n. This is obviously a linear functional that can be verified immediately, in fact this is from f x equals a x where a is just the row vector, row vector consisting of the elements a 1 a 2 etc a n, so row vector into the column vector x 1 etc x n, this is a functional, this is a linear functional. What is important is to see that any linear functional on R n can be given by this representation, so we that gives a complete understanding of linear functional over R n in fact over any f n. So let me prove that quickly, this is a linear functional alright but if you take any other linear functional that must also be of this form then we have completely understood what linear functionals over R n R, okay. So let me prove that quickly, in fact any linear functional on R n is of the form above, let me prove that quickly, now in order to prove this let me make the following observation, observe that f of E j, if E j is the j standard basis vector then f of E j equals A j, is that clear? This is my f, f of E j, see this x is in R n what is f of E j, f of E j must be the vector, E j is the vector whose jth coordinate is 1 all other entries are 0, so it is only A j on the right, so f of E j equal to A j let us make use of this. So what is it that I need to prove? Given f, okay let me say given g, given g a linear functional on R n I am given a linear functional on R n I want to show that g has this representation, I want to show that g has a representation, so I will just make use of this that is given g a linear functional on R n I have g of E j to be some number, g of E j some real number I will call that alpha j, so I will call the number alpha j as g of E j this is a real number I know, okay then I want to show that g of x is of this form, I want to show that g of x is of this form this time it will be alpha 1 x 1 etc alpha n x n that is the claim. So let us consider, okay what is the claim? Claim is g of x equals alpha 1 x 1 plus alpha 2 x 2 etc alpha n x n, okay that is easy to see let us do that quickly. Proof of this g of x, x is in R n I use a standard basis g of summation okay let us say x can be written as x 1 E 1 plus x 2 E 2 plus etc plus x n E n this is a unique representation of any vector x in R n in terms of the standard basis vectors E 1 E 2 etc E n, g is linear so this is x 1 g of E 1 plus x 2 g of E 2 etc plus x n g of E n now I will take these numbers and replace g E j by these numbers. So this is alpha 1 x 1 plus alpha 2 x 2 plus etc plus alpha n x n so I have proved what I wanted g of x for me is alpha 1 x 1 plus alpha 2 etc alpha n x n so g of x has this form, okay. So linear functionals on f n are completely determined I know how they look like, okay. Let us look at other examples this is my first example let me look at other examples one of the important examples of a linear functional is the so called trace functional let me define f from R n cross n to R f of a equals trace of a, f of a equals trace of a what is the trace of square matrix it is the sum of the diagonal entries. So this is equals sum of the diagonals of a we will usually denote the entries of a by a i j and so this is equal to a 1 1 plus a 2 2 etc plus a n n summation i equals 1 to n a i i that is the trace functional it is easy to see that the trace functional is a linear functional on R n cross n second example third example let us look at for the third example we have so third example is a special case of what we call as the evaluation map. Let us say I have the space of all functions v equals the space of all functions over R, space of all functions over R functions of a single variable that is what it means functions over R I will define f okay let me use some other notation this time define L from v to R by L T of p, L T equals p of t this p belongs to v this is called the evaluation functional I take a fixed t and then define now and define L T like this t is a real variable see p is a function so t is a real variable p belongs to v, v is the space of all functions over R so t is a real variable I fixed t let us say t equal to 0 and then I define L 0 as L 0 of p is p at 0 the value of p at 0 this is called the evaluation map you can easily show that this is a linear functional L T is a linear functional for each t see this evaluation functional will be discussed once again when I give an example of a dual basis that is the reason why I wanted to include this example also. Then let me give you one more example the last but not certainly the least linear functional is the following for me v this time will be C 0 1 let us say C 0 1 this time is the space of all real valued continuous functions over the interval 0 1 C 0 1 is the space of all real valued continuous functions over the interval 0 1 I define f from v to R by f of let us say f function p that will be integral 0 to 1 p of t dt integral 0 to 1 p of t dt remember that p is a continuous function so this is a Riemann integral this exists f of p is integral 0 to 1 p of t dt this exists and so this is a this is a number it is a definite integral so this is a number you can easily verify from the properties of Riemann integral as this is a linear map so this is a linear function okay f is a linear function on the space of continuous real valued functions okay. Now let us look at the collection of all linear functionals let us look at the collection of all linear functions let me use a notation v star let v star equals the set of all linear functionals on v set of all linear functionals on v now remember that if v and w are vector spaces then the set of all linear transformations t from v into w set of all linear transformation forms a vector space and we have also determined the dimension of the subspace when v and w are finite dimensional okay. So the set of all linear functionals is a vector space because it is a particular case of a linear transformation the core domain space is this underlying field so the set of all linear functionals is a vector space so v star is a vector space and what is the dimension of v star if v has dimension n v star is a real vector space vector space over the same underlying field if dimension of v equals n then what is the dimension of v star dimension of the vector space l v w is m times n where m is a dimension of v okay n is a dimension of v m is a dimension of w what is the dimension of r over r 1 any field over itself is one dimensional so the dimension of v star is n same as dimension of v okay so we have a certain information on v star v star has the same dimension as v for finite dimensional spaces okay for finite dimensional spaces v star has the same dimension as v the question is can I write down a basis for v star can I write down a basis for v star in such a way that this in some sense corresponds to a basis for v that I start with that is the question okay can I write down a basis for v star in such a way that there is some natural correspondence between the basis for v star that I write down and the basis for v that I start with but before that let me tell you this v star is called the dual space this v star is called the dual vector space of v it is a vector space that is dual to v dual vector space of v set of all linear functionals the space of all linear functionals okay so this is the question that we would like to address the answer is yes given a basis of v there is a natural basis that one could associate with v star that basis will be called the dual basis okay that was one of the topics that I had written down so how to determine what is a dual basis how to determine the dual basis so we would like to discuss the notion of a dual basis the notion of a basis dual to a basis of v a basis of v star dual to a basis of v okay we have in fact the following theorem that will define the dual basis let v be finite dimensional and I will write down a basis of v explicitly u1 u2 etc u n this is let this be an ordered basis ordered basis of v then there exists a basis I will call this basis b star script b star and I will call the elements of b star as f 1 f 2 etc f will stand for linear functionals f 1 f 2 etc f n see what I know for sure is that the number of elements in b star must be the same as a number of elements of b because if v is finite dimensional because we know that dimension of v star is equal to dimension of v then there exists a basis b star this has certain properties well the most important property that we are interested in is the following then there exists a basis b star of v star such that f i of u j is equal to delta i j this basis f 1 etc f n is related to the basis u 1 u 2 etc u n by means of these equations remember these are n square equations these are n square equation delta i j is a chronicle delta it takes n square values i and j vary from 1 to n so these are n square equations we also have further properties you take any other linear functional any linear functional on v that is f belongs to v star then I can write down explicitly f of x in terms of these basis vectors f 1 f 2 etc see this f must suppose that we have proved that this is the basis of v star then any f n v star is a linear combination of these vectors these functionals so any f can be written as something like alpha 1 f 1 plus alpha 2 f 2 etc plus alpha n f n okay in the case when we have a dual basis the numbers alpha 1 alpha 2 etc can also be written down immediately f x is nothing but f of u 1 x 1 plus f of u 2 x 2 plus etc plus f of u n x n this is the formula for f any x can in turn be written in terms of these functionals any x and v can in turn be written in terms of these functionals as follows okay I want this representation x equals again come back and look at this x is in v v has this as a basis and so v is so x is a linear combination of these vectors what are the coefficients what are the coefficients are f of u 1 u 1 plus f of u 2 u 2 I need to just check this one see then x can be written as I want to write f 1 of x yeah what I want is this f 1 of x see these numbers depend on x f 1 x u 1 plus f 2 x u 2 etc f n x this is the explicit representation of any vector x in terms of the dual basis vectors this is what I wanted to say any x is a linear combination of u 1 u 2 etc u n in the case of dual basis once you know the dual basis you can give the coefficients of x you can give the coefficients here u 1 u 2 etc u n explicitly in terms of the dual basis vectors f 1 etc f n okay so we will prove all these representations this is the representation for any functional on v any vector x and v can be written in this manner foremost is that f and u f i's and u j's are related by means of these n square equations okay so let us prove this yes sir any x is a linear combination of u 1 etc u n the question is what are the coefficients we are claiming that the coefficients can be given in terms of the dual basis vectors dual basis functionals okay let us prove this now okay let us start with u 1 u 2 etc u n this is a basis of v and the first step I will take these n numbers 1 0 0 etc 0 see this cons of n elements okay the first entry is 1 all the other entries are 0 n minus 1 0 1 1 okay this is a set of vectors in the vector space r this is a set of this is a basis for the vector space v so these are basis vectors for v this is any set of vectors in the codomain vector space I know that there is a unique linear transformation that takes u 1 to 1 u 2 to 0 etc u n to 0 this result we have seen given u 1 u 2 etc u n a linear transformation defined determined completely by its action on a basis and I know that given a set of vectors u 1 u 2 etc u n basis vectors w 1 etc w n absolutely no conditions on these vectors w n etc w n in w absolutely no conditions there is a unique linear transformation that takes u j to w j okay so for this consider this there exists a functional linear function I want to emphasize there exists a unique linear functional in fact let me write it again there exists there exists a unique linear functional unique is important I have not stated that in the theorem but this is a unique basis this basis is unique given the basis u 1 u 2 etc u n there exists a unique linear functional I will linear functional call it f to begin with unique linear functional f such that such that f of u 1 is 0 and all other vectors are mapped to 0 yes there exists a unique linear functional f such that f u 1 is 1 all the other vectors are mapped to 0 I can do this for so what I do next is take this basis and then take the numbers 0 1 0 0 0 etc I can keep doing this up to n steps now for each step I have a unique linear functional so what I will do is now index these by f 1 f 2 etc okay so I will say that there is a unique linear functional f 1 such that f 1 u 1 is 1 f 1 u 2 f 1 u 3 etc f 1 u n is 0 in general I have there exists a unique linear functional f 1 f 2 etc f n with the property that f 1 of okay f i of u j equals delta i j which is the first part of the theorem that is f 1 of u 1 is 1 f 1 of all other vectors is 0 f 2 of u 1 f 2 of u 2 is 1 f 2 of all other vectors is 0 etc so unless i is equal to j I would not get 1 when i is not equal to j all of them are 0 when i is not equal to j all of them are 0 i is equal to j I get 1 i equal to j corresponds to f 1 u 1 equal to 1 f 2 u 2 equal to 1 etc f n u n equals 1 so this proves the existence in fact the unique existence of unique set of linear functionals that satisfy these n square equations okay that is the first part we need to verify the other two representations okay let f belong to V star then I have just now proved that by the way what about linear independence of these functionals I have not proved I have simply said these functional satisfy those n square equations why is it a basis by the way are these are these functionals distinct are these functionals in the first place different they are different because f 1 takes a value 1 for u 1 f 2 takes a value 0 for u 1 etc so by definition these are different that is there is at least 1 vector for which the values that f 1 f 2 etc f n take are different so these are these are distinct to begin with what about linear independence of them suppose I prove linear independence then it follows that they must form a basis because dimension of V star is n linear independence of these functionals that is also easy to see maybe I will leave that as an exercise okay linear independence of these functionals I am going to leave it as an exercise simple straight forward so this forms a basis V star is a basis which is related to the basis B that we started with by means of these n square equations okay we also need to show that the other representations hold representation for f and representation for any vector x okay so this f is in V star f is a linear functional we know that f can be written as a linear combination of the dual basis vectors I have not yet defined dual basis vectors f is equal to let us say something like gamma 1 f 1 plus gamma 2 f 2 etc gamma n f n this is because f 1 etc f n these functionals form a basis for V star okay what is f of u j f of u j is gamma j it is gamma 1 f of f 1 of u j plus gamma 2 f 2 of u j etc gamma j f j of u j plus etc plus gamma n f n of u j so that is all gamma j f j of u j all of the terms are 0 that is just gamma j so do I have the representation then of this f x equals look at f x f of x equals f of this representation is rather incomplete so I have got to go back what do I mean by this what I mean by this is that see when I write down this I must know what x 1 x 2 etc x n are what is implicit is that x 1 x 2 etc x n are the coefficients of x in terms of when I write x in terms of u 1 u 2 etc u n I mean that the coefficients of x will be x 1 x 2 etc x n okay in other words f x I will write it as f of x 1 u 1 plus x 2 u 2 plus etc x n u n that is I am assuming that the representation of x in terms of the basis vectors go with goes with these coefficients x 1 x 2 etc x n okay so now this is x 1 f of u 1 etc x n f of u n f of u 1 is gamma 1 I think I have what I wanted right away no there is no do I have what I wanted right away yes what was the need for this there is no need for this this comes straight away okay the representation of f the action of f on any vector x this is the representation that is what I have written down there okay this may be I will use in the next part is it clear this is the representation that is if I want to know the action of f f is any linear functional I want to know the action of f on x then it is enough if I know the action of f on the basis vectors what I must know is the action yeah what I must know of course is what is the representation of x in terms of u 1 u 2 etc u n what are those coefficients okay the last one x has this representation I will go back and look at this representation finally x is my x 1 u 1 etc x n u n so that f of x f is linear f of x is x 1 f of u 1 plus x 2 f of u 2 etc I have determined is that okay what I really want is f 1 of x f 1 of x is x 1 f 1 u 1 etc x n f 1 u n now here f 1 u 1 is 1 so this is x 1 okay what is f 2 of x f 2 of x is u 2 sorry f 2 of x is x 2 etc f n of x is x n I go back and substitute I get that representation okay so I go back to this equation x equals x 1 u 1 plus x 2 u 2 etc x n u n x 1 for me I have just now determined is f 1 x u 1 plus x 2 is f 2 x u 2 etc x n f n sorry just f n x u 1 f 1 x u 1 plus f 2 x u 2 etc plus f n x u this is the representation of any vector x in terms of the dual basis vectors dual basis functionals that is what I do is given a basis I determine the dual basis and then compute these numbers f 1 x f 2 x etc f n x I know x then those are the coefficients corresponding to the representation of x in terms of u 1 u 2 etc u n okay. So let us observe once again the fact that this the elements of B star is unique given B okay this B star is called the dual basis B star is okay let me write it fully B star is f 1 etc f f 1 etc f n B star is unique given given B this is the basis of V this is the basis of V star and they are related by means of those n square formulas this B star is called the dual basis of V star we have determined a basis of V star in terms of a basis of V that we started with let us look at an example let us look at V as P 2 of R the real vector space of all polynomials of degree less than or equal to 2 this has dimension 3 the dual basis V star will also have dimension 3 I will determine 3 dual basis vectors corresponding to the usual basis of P 2 1 T T square I am sorry that is not what I will do I will determine a dual basis and then try to determine what is the basis for which this dual basis is the dual in this example I will determine a basis for V star and then determine which is the basis of V for which this is the dual basis okay. So for this we need this evaluation functionals so let me define let us take 3 numbers T 1, T 2, T 3 as distinct I have 3 distinct numbers let me define 3 functionals L 1 of P, L 2 of P, L 3 of P, Li of P is P at P i given a polynomial of degree less than or equal to 2 I will compute the value of this polynomial at T 1 that is my functional L 1 of P L 2 and L 3 are determined similarly I have 3 functionals here each is linear L 1, L 2, L 3 are linear functionals on V my claim is that this forms a basis of V star they form a basis of V star V star is the space of all functionals space of all linear functions on V I am claiming that these 3 functionals linear functionals form a basis of V star it is enough if I prove that these are linearly independent okay so let me prove that these are linearly independent suppose that the 0 functional is a linear combination of these 3 okay let us say alpha 1 L 1 plus alpha 2 L 2 plus alpha 3 L 3 I must show that alpha 1, alpha 2, alpha 3 are 0 okay what this means is that these are functionals so what this means is that the action of this functional on Li polynomial of degree less than or equal to 2 gives a value 0 alpha 1 L 1 of P plus alpha 2 L 2 of P plus alpha 3 L 3 of P this must be the 0 number okay this must be the 0 number here this is 0 is a linear functional 0 functional I will look at specific choice of P 3 specific choices of P take the case when P of T is 1 okay P of T is 1 constant polynomial so what does this equation give me in that case this is 1 so alpha 1 plus alpha 2 plus alpha 3 equals 0 take the case P of T equals T P of T equals T then what is alpha 1 of T alpha 1 of T alpha 1 L 1 of T T 1 do I get alpha 1 T 1 plus alpha 2 T 1 plus alpha 3 sorry T 2 alpha 3 T 3 to be 0 L 1 of T is T at T 1 that is T 1 okay I get this for the next polynomial T square I must get alpha 1 T 1 square alpha 2 T 2 square plus alpha 3 T 3 square that is 0 these are the 3 equations that must be satisfied by the numbers alpha 1 alpha 2 alpha 3 okay I get a homogeneous system I get a homogeneous system the first row is 1 second row T 1 T 2 3 T 3 third row T 1 square T 2 square T 3 square alpha 1 alpha 2 alpha 3 T 1 T 2 T 3 are distinct numbers it can be shown that this matrix is invertible you can do elementary row operations on this and reduce this to the identity matrix see this is called the so called van der moen matrix okay if T 1 T 2 T 3 are distinct then this matrix is invertible homogeneous system with an invertible coefficient matrix has 0 as a solution so alpha 1 alpha 2 alpha 3 are 0 L 1 L 2 L 3 are linear independent okay they form a basis for V star because V star has dimension 3 dimension of V star is 3 these form a basis I will leave okay I will just I will probably answer this question what is the basis of V what are the 3 polynomials what are the 3 polynomials which form a basis of V which give you this dual basis what are the 3 polynomials which form a basis of V which for which this is the dual basis that can be determined by looking at the defining equations of the dual basis F i of u j equals delta i j the answer is the so called Lagrange interpolating polynomials you will study this in numerical analysis the polynomials in V corresponding to which this is the dual basis L 1 L 2 L 3 are the Lagrange interpolating polynomials at T 1 T 2 T 3 okay so please try to determine these 3 polynomials for which this L 1 L 2 L 3 is the dual basis so let me stop here.