 Looking at linear functionals leading to the notion of the adjoint of an operator, okay so let me recall this result that we proved last time. Let V be a finite dimensional in a product space and F be a linear function on V then there exists a unique vector Y in V such that F of X is nothing but the inner product of X with this fixed vector Y for every X in V. This Y will of course depend on F because we have seen the formula for Y, okay. The formula for Y is summation J equals 1 to N F of UJ bar UJ so it depends on F and this UJ comes from a basis is an orthonormal basis, okay. Using this we will now look at the notion of the adjoint of an operator. So let us first prove this result. Let V be a finite dimensional inner product space and T be a linear operator. T is a linear transformation on a finite dimensional inner product space then there exists a linear operator T star on V so we will call this operator T star. There is a linear operator T star on V such that the following holds. Remember we are calling it T star so it does depend on T and it in particular satisfies the following, it is linear of course it satisfies the following property such that if you look at the inner product of T X with Y this is the inner product of X with T star Y this must be true for all X Y in V. So what this result claims is that there is on a finite dimensional inner product space V there is this operator this is unique so let me mention that also there exists a unique linear operator which we will call T star. This T star we will call it as the adjoint of the operator T that we will do a little later. First to prove this we will appeal to the previous theorem so in order to apply the previous theorem we need to look at a functional so the proof is as follows. Let Y belong to V be fixed I will tell you what T star Y is okay then I need to see that it is unique T star Y is unique and then the T star satisfies this and finally T star is linear okay define the following operator define Phi Y it depends on Y Phi Y from V to R or C within brackets R define this operator by Phi Y of X equals I will now invoke the operator T use the operator T in the definition of this functional Phi Y Y is fixed this X varies so this is an inner product Phi Y of X is a number real or a complex number now T appears here so you can show that Phi is linear Phi Y is linear then Phi Y is linear Phi Y of Alpha Alpha X plus beta Z you can show that it is equal to Alpha Phi Y of X plus beta Phi Y of Z so Phi Y is linear it is a linear functional I appeal to the previous theorem by the previous theorem there exists a unique I will call it Z by the previous result there exists a unique Z in V such that such that the action of Phi Y on any vector X is the inner product of the vector X with Z what is important is to realize again that it is unique this uniqueness will imply the uniqueness of this operator T star in fact okay so we started with Y we started with Y in V we have got a unique vector Z that satisfies this equation for all X so we will associate this Z to Y under T star so the definition is Z T star Y equals Z now you can do this for every you can do this for every fixed Y that is given of Y I will define this linear functional and then I know how to get this unique Z then for that while I will associate this Z so this is since this right hand side is unique for this Y this is well defined in the first place we need to verify it is linear but before that let us verify that this T star satisfies this equation that is straight forward also if you look at T X comma Y I will start with X with T star Y X with T star Y is by definition T star Y is Z so that is X with Z but X with Z is here that is Phi Y of X and Phi Y of X by definition is T X Y this is true first for all X in V and then I vary Y then this is true for all Y in V okay first I fix Y and then this is true for all X then I vary Y appealing to that Z which comes for this particular Y so this equation holds X comma T star Y is equal to T X comma Y for all X and Y we need to show T star is linear so that is also not difficult so we need to show T star is linear so let us consider I have used the variables X Y and Z okay let me use X will remain a variable I want to show T star is linear let us say T star alpha Y plus W now I know that T star satisfies this equation I will make use of that so I must so let us look at this I will show that this is equal to inner product X with alpha T star Y plus T star W this is true for all X so T star is linear so consider this by definition T star satisfies this condition that X with T star Y must be T X with Y so this is equal to inner product T X with alpha Y plus W that can be written as T X comma alpha Y plus T X comma W now here I will take alpha bar outside I will do two steps I will take alpha bar outside and then T X comma Y is what I will get that I will write as X comma T star Y the next term is X comma T star W now I will take this alpha bar inside so this gives me X comma alpha T star Y plus X comma T star W this X can be taken common X alpha T star Y plus T star W this is true for all X okay and so I started with X with T star of something I show that is X of alpha T star of Y plus T star W this is true for all X so it means that T star of alpha X plus W sorry alpha Y plus W I have shown that this is equal to alpha T star Y plus T star W C or again we have made use of the fact that if inner product of X with Z equal to 0 for all X then Z must be 0 X Z plus Y Z which term what where is your doubt this one it is an inner product no inner product of X comma Y plus W is inner product of X comma Y plus inner product of X comma W you can use so you can use that whenever there is no scalar okay your question is what happens why is this your question why is this true start with the left hand side left hand side is Y plus W X conjugate now it is you know that is linear with respect to the first one Y X plus W X conjugate then Y X conjugate plus Y X conjugate plus W X conjugate that is X Y plus X W if I have not mentioned it please take this as the explanation see the inner product is linear with respect to the first variable and with respect to the second variable it is additive it is conjugate linear with respect to the second variable but it is additive with respect to the second way that is something we are we have been using probably without mentioning okay. So for we have shown that this T star is of course well defined and it is linear why is it unique because Z is unique that is if there is okay I leave that part to you if there is some other operator S that satisfies the equation inner product T X with Y is equal to X comma S Y if there is some other operator S such that inner product T X Y equals X comma S Y for all X and Y then S is equal to T star okay you can show that S is equal to T star so T star is unique so I will just mention that you fill up the last line also T star is unique I have told you how to prove it but in any case that is not difficult so T star is unique so let us go to the definition of the adjoint using this result let V be an inner product space and T be linear if there exists an operator a linear operator T star on V such that inner product T X with Y equals X comma T star Y if this is satisfied for all X Y and V then T star is called if there exists again a unique linear operator then T star is called the adjoint the adjoint of the operator P the adjoint operator so what we have proved now is that if V is finite dimensional then any linear operator has the adjoint okay so remember that this T star is defined by means of this equation so it depends on the inner product T star of course depends on T it will also depend on the inner product because it must satisfy this equation okay okay we look at some examples of adjoint operators how to compute the adjoint infinite case the adjoint may not exist you can look at I will just give a loose answer you can look at the differentiation operator on the space of continuous functions you can show that this does not have the adjoint if time permits I will discuss this also if it is infinite dimensional adjoint may not exist for a linear operator okay you will discuss this in your functional analysis course infinite dimensions we will discuss in a functional analysis course okay look at some examples but before that let us settle 1 or 2 easy questions that would arise naturally for instance given a inner product space given a finite dimensional inner product space along with an ordered orthonormal basis I write down the matrix of T relative to that ordered orthonormal basis what will be the order what will be the matrix representation of T star what do we expect the answer to be it cannot be the same it cannot be the same unless T is self adjoint okay but before that before that let us let us address this question okay recall that when we looked at the notion of why an orthonormal basis is better than an ordinary basis in the representation of vectors that is if you look at the representation of a vector with respect to a basis X equal to let us say the basis is U 1 U 2 etc U n just an ordinary basis then X is alpha 1 U 1 etc alpha n U n where alpha 1 etc alpha n one needs to compute by solving a system of equations whereas if U 1 U 2 etc U n where an orthonormal basis you do not have to compute these numbers they are just inner product of X with U 1 X with U 2 etc so computation of the coefficients that becomes easier similarly remember the matrix representation of a linear transmission that we have discussed so let me discuss recall this and then go to the inner product space how is the matrix of a linear operator relative to a basis written so this is also something we can quickly recall I will use U 1 U 2 etc for the orthonormal basis let us take B to be say W 1 W 2 etc W n this is an ordinary basis this is an ordered basis for a vector space V okay there is no inner product space now for a vector space V T is also linear that is given T is linear then how do we write down the matrix of T relative to this basis B we looked at a more general notion of how we write down the matrix of T relative to two basis if that if the vector space see we discuss V to W if W equals V then it is convenient to deal with just one basis that is this so what is this? This is a matrix of T relative to B I will call that A that is given by the image of the first vector I am calling it W 1 write down T W 1 this is a vector in V that can be written in terms of B collect the coefficients that is the first column that is this etc T W j this is the jth column T W n this is the nth column this is how you write down the matrix of A relative to this basis okay this means what is the formula then connecting the entries of A and the left hand side images T of W j for instance do you remember that formula what is the formula all you have to do is just look at the left hand side take the jth column the jth column of A what are those entries jth column means the second entry varies is that sorry the first entry varies is it okay tell me if this is clear the formula for A ij is what I want this is I want the formula for T W j I want the formula for T W j so what is that in terms of the matrix A in terms of the entries of A do you remember this j summation i equals 1 to n A ij W i is that okay that is because if you write T W j is the jth column does this give the jth column the first entry is what varies so this gives the jth column okay this is the formula for the matrix of a linear transformation relative to a basis and how do you compute these A ij again A ij are compute it is a jth column so you go to the jth column that is you look at W j image of T under W j image of W j under T relative to this basis you write a representation that is amounting to solving a system of equations okay so in general if you have an ordinary ordered basis these coefficients are obtained by solving equations but if it is an orthonormal basis then you would expect the computation to be easier so let us first prove that let V be finite dimensional inner product space I expected at least one of you to answer this that V be a finite dimensional inner product space and B equals U1 U2 etc U n be an orthonormal basis of V let T be a linear operator okay you remember this notation L of V is a set of all linear transformations from V into V let T be a linear operator and A be the matrix of T relative to this orthonormal basis P then can you make a guess that is not correct but let me ask you the question what is the formula for A ij what is the formula for A ij yes it is an inner product yes inner product of what with what no now the basis orthonormal basis U1 etc U1 ordered orthonormal basis it is an ordered orthonormal basis I have not written but this is an ordered orthonormal basis I want the formula for A ij what is the formula for if I write X as a linear combination of an orthonormal basis in terms of an orthonormal basis okay so what is the inner product here there is no X it is the other way around T Uj with Ui and this is the formula okay which means what you simply take one inner product look at the image of Uj under T take the inner product of that with Ui then that gives the entry A ij this is the formula for A ij as before the entries the coefficients or the so called coordinates of a vector X relative to an orthonormal basis are obtained by simply taking the inner product of that vector with each of those orthonormal basis vectors you get this okay this is analogous to that result now to prove this just use this formula proof I want to look at T Uj T Uj so I will use this formula instead of Wj Wi it will be Uj Ui so T Uj is summation okay I am using I here so I will use a K here K equal to I am just rewriting this formula I am rewriting this formula relative to this basis summation K equal to 1 to n Akj Uk Akj Uk okay that is this formula this is T Uj all I have to do is look at inner product T Uj with Ui T Uj with Ui let us not do all these just K equals 1 to n Akj that comes out inner product this is the first term so this is Uk Ui K is a running index K is the summation index when see this I is fixed when I when I write this formula this I is fixed for me so when K is running index when K takes a value I this is 1 all other entries are 0 because an orthonormal basis when K takes a value I it is A ij all other terms are 0 which is what we wanted to prove T Uj Ui is A ij so this is a formula connecting this is a formula for computing the entries of the matrix A that is obtained by these n square inner products okay remember this formula connecting again this problem go back to the vector space case not the inner product space vector space case we define the so called transpose of a linear transformation the transpose was shown to be unique so the question is if A is the matrix of a linear operator T relative to a basis B what is a matrix of T transpose as a linear transformation we have seen that it is a transpose of the matrix of T okay and entirely similar results holds for result holds for the adjoint operator that is very easy so you can think of this star as a kind of a generalization of the transpose operation. So I will state that as lemma Aura theorem I will be cryptic I will follow what the notation that we have used before I will simply say that my matrix A is the matrix of the transformation T relative to B then the matrix of T star okay let I will call B as the matrix of T star relative to the orthonormal basis B then the conclusion is B star equals A remember the star of a linear transformation on a inner product space has been defined just today okay but what I am writing down here is the complex conjugate of matrices see this A and B are matrices that we know the complex conjugate of matrix we know you interchange rows and columns after taking the conjugate transpose after taking the conjugate right this matrix A has complex entries take the complex conjugates of that of the entries of A then take the transpose you get A star so this is how B star is constructed. So the matrices of T and T star relative to this particular basis relative to any basis that you start with provided you stick to that basis for T star they are related by this formula so proof again you have to simply use the formula for the entries of B star and A okay so okay let us start with let us use the previous one let me write down B i j what is the formula for B i j B i j B is defined through this B i j by definition must be T star u j u i is that okay T star u j u i but this is u j T u i because that is how the adjoint is defined but this is T u i, u j the conjugate of that but T u i, u j go back to the previous result is A i j A i j bar A j i bar yes this first and then this A j i bar that is what we wanted to prove that is B equals A star complex conjugation we know you do it once more you get the same matrix see this formula is correct right B i j next step how is the adjoint defined we are not using that see T star star is T is correct but we are not using that I have not proved it so I am not using it the same thing can be used like what the argument that I gave for the second argument that can be given here no but where have I used that I do not understand see T star u j u i this T star okay see you can take the conjugate and then go back to this you want that explanation okay let us you have objection then what you want is an explanation for this right so I can write this as u i T star u j conjugate okay then it is a conjugate of T u i, u j but that is the same as u j, T u i sorry the proof of ends here right T u i u j, T u i u j will be A j i going along with the bar agreed is that okay B i j i j that is T star and then it is u i with T star u j conjugate keep the conjugate then I use the adjoint operation T u i u j but T i u i u j is A j i that goes along with the bar so probably a more direct proof if you want and so as I mentioned before B star equals A I have shown that B is equal to A star but this star operation is of order is of period 2 you do it once more you get the same matrix, any problem again? This proves B equal to A star but you do the star once again because it is a matrix now then I get this so I still do not prove that for a linear transformation T, T star star is T I am not using that I am using it for a matrix I am using what it tells me for matrices for matrices you can verify that by just writing down the entries right okay. So let us look at some examples both you cannot expect this to happen it is only with respect to the same basis that the relationship holds in fact if you look at just an ordinary basis which is not an orthonormal basis then this relationship does not hold if it is not an orthonormal basis just an ordinary basis then this relationship is not true it is a more complicated formula how A and B are related if it is just an ordinary basis will not be the same as this that relationship is different it is more complicated but what is more meaningful and what is more applicable is writing it related to the same basis okay. I want to discuss examples let me take to one finite dimension the other one infinite dimension of transposes how do you compute the transpose of certain operators let us take V as C n x n with the trace in a product with let us say if I have AB in V then this is trace of B star A with respect to the trace in a product this was given as one of the examples of an inner product space define the operator L M define the operator L M from V to V for a fixed M in V by this formula L M of A is M times A this is a left multiplication that is why this L is given left multiplication by M A is the variable M is fixed A is a variable left multiplication by M what is the adjoint of this operator what is the adjoint of this operator let us calculate by the way this is linear that is an easy exercise L M is linear that is left to you I want to calculate the adjoint to calculate the adjoint I must appeal to the formula the first principle definition I want to look at L M A, B I want to write it as say I want to compute T star so I must write T x, y as x, T star y so I must write this as A, something of B that something will give me the T star L M star so this is what this is by L M A I will apply the definition it is M A B that is trace of B star the second one so this is trace of B star M A multiplication is matrix multiplication associated by I still insist that I write this bracket appeal to this fact the trace of the product that is trace of let us say C times D is trace of D times C I will keep applying that trace of till I am satisfied trace of M A B star that is the same as trace of M times A B star right B star M A is trace M A B star I will write this as trace M A B star so I want to take A outside okay no I will not write like this I want to keep A to the left I will write this as again the same thing A B star M which is trace of A M star B whole star this is A into B star M B star M is this whole star again I am dealing with this reverse order law for matrices not for transformations which I have not yet proved so I get this but this is same as now I have taken A outside and star of something so remember that the formula for the inner product then is given by A M star B okay do you agree this can be written as A, M star B is L M star of B left multiplication of M star with B L M star of B and so what does it mean if you look at L M star that is L M star this is true for all A B so the conjugate of L M is L of the conjugate with respect to M star M is what I started with left multiplication the conjugate is left multiplication with the conjugate of that matrix okay this is for the finite dimensional case let us look at an infinite dimensional example let us consider the space of all polynomials with complex entries complex coefficients space of all polynomials with complex coefficients the inner product is the usual inner product integral 0 to 1 f of t g bar f g bar now what is g t bar g is a complex polynomial the variable is real coefficients are complex the variable is real coefficients are complex so g t bar if g equal to let us say a 1 a 0 plus a 1 t etc a n t to the n where a 0 a 1 etc are complex numbers g t bar will be a 0 bar plus a 1 bar t plus a 2 bar t square etc plus a n bar t to the n okay so I take the complex conjugate of the coefficients that is my definition so with that definition I give this inner product this is an inner product so V with this you can show is an inner product space I will look at the following operator again something similar to the previous one left multiplication by a fixed polynomial define let us say L or t t from V to V by t of g is f with g for a fixed f element of V so I can actually call it t f of g take a fixed f and g fixed f and V take a fixed polynomial simply multiply given any g simply multiply on the left on the right does not matter polynomial multiplication is commutative unlike the previous one what is t f this is t f what is t star f that is the question what do you expect the answer to be what is f star for a function for a matrix M M star is known what is f star for a function it should be t f bar we will can show that t f t star is t f bar okay okay let us do that quickly again first principle use the first principle definition I want the inner product of t f g, h I want to calculate this I must write it as g, something and then that something will give me t star so this is by definition t f g h by definition t f g is f g f g h for f g h the definition is integral 0 to 1 f of t g of t h t bar I must kind of take out g but it is clear how we should do that 0 to 1 see this is matrix sorry this is polynomial multiplication so it is commutative I can take g t outside and then combine the other two and put them under a bar I want f to be without bar so f t bar h must go with this so do I get what I want double bar gives me f single bar gives me h bar I can call this L of t bar if you want and then write this this is integral 0 to 1 g of t L of t bar dt but that is by definition inner product of g with L which is g with L is f bar h which is I observe g t f bar of h so what I have proved is that t f g with h is g, t f bar h for all f g for all f g h for all g h I am sorry f is fixed g h are the variable polynomials so this means t f star is t f bar okay so this example tells you that star behaves like conjugation adjoint behaves like complex conjugate that is what I mean the adjoint operation star behaves like complex conjugate taking the bar it also behaves like complex conjugate see when I say complex conjugate it is z is a complex number z bar is what I am calling complex conjugate maybe just conjugate the previous example also has the conjugate L m star is m star m star already has conjugate okay so star the adjoint operation kind of behaves like the conjugation that you do for a complex number okay we will explore this a little further and see how some of the properties of the adjoint are similar to the properties of the complex conjugate okay so let me stop here