 We have seen that the adjoint exists for any operator over a finite dimensional inner product space. Let us look at some properties of this adjoint remember the correlation between similarities between the adjoint and the operation of complex conjugation okay I told you that they are similar the following result will tell you why this is so. So I have the following theorem which lists the properties of the adjoint operator that V be a finite dimensional inner product space T, U the linear operators then the following is hold first one is if you look at T plus U, T plus U star this is T star plus U star property 2 look at alpha times T star this is alpha bar T star property 3 look at T U star that is U star T star and property 4 is that this operation of taking the adjoint is an operation of period 2 you do it 2 times you get the same operator T proof is to just verify the basic equation for the first one you look at T plus U X, Y T and U are linear so and this addition it is also a linear operation that I can split this as T X, Y plus U X, Y that is X, T star Y plus X, U star Y this is possible because T and U have adjoint it is a finite dimensional inner product space this can be written as X, T star Y plus U star Y that is T star plus U star of Y now the definition of adjoint is if you have L X Y equals X, M Y then M is equal to L star L X Y equals X, M X, M Y then M is equal to L star that is T star plus U star is T plus T whole star that is the first one okay. Second one similar second one I will leave the thing is for the second it is a conjugate it will come with the conjugate let me prove the third one so second is straight forward third is also straight forward but I will prove it in any case you consider what is the order in which I take T U X, Y T U X, Y is T this is a composition okay so that is T of U X, T star Y then this is U X, T star Y that is the definition of T star this goes here goes with the conjugate applied once again X, U star, T star Y when I write U star, T star it is the composition of these linear maps so what is the meaning of this X, X but this left hand side is just to emphasize the left hand side is X, T U star Y and so 3 holds we have shown that T U star is U star T star this is called the reverse order law for the adjoint operation similar to the reverse order law for the usual transpose transpose of linear transformations that we studied earlier similar to the operation of taking transpose for matrices similar to the operation of taking conjugate transpose for complex matrices so that is property 3, property 4 you look at T X, Y I want to show T double star right so let me start with T double star this time T double star X equals Y this is by definition when it goes here 1 star is removed 1 star actually appears so the problem becomes complicated so what I will do is I will start with T star X I will start with T star X, Y this is X, T Y which is what I want to prove that is X, T star star Y but the left hand side is X, T left hand side is X, T Y right hand side is X, something Y okay we do not know but okay this much we know left hand side is Y, T star X conjugate okay this is T star X, Y Y, T star X conjugate which is the conjugate of T Y, X which is the same as X, T that is the left hand side okay so that needed explanation so X, T Y is X, T double star Y so from the set this is true for all X and Y follows that T star star is T okay so there is to list some properties. Now you will see that the reason why this adjoint operation is compared to the operation of complex conjugation if you have Z1, Z2 two complex numbers then Z1 plus Z2 bar Z1 bar plus Z2 bar Z1 okay look at this constant times Z is that constant times bar into Z bar and look at this this is Z1, Z2 bar their complex multiplication is commutative so you do not have to be you do not have to impose this route okay this is just T star U star there if you want and when you do it twice Z bar bar is Z okay. What is also to be compared for an operator is the following for an operator you can think of a real part and then imaginary part using the adjoint operation so let me just illustrate that and then move on to unitary operators. So if T is an operator on a finite dimensional inner product space V then I can write T as U1 plus I U2 where U1 is T plus T star by 2 and U2 is T minus T star by 2 I okay you can verify this now U1 and U2 are unique because T star is unique and you can also observe that U1 has the property that U1 is its own adjoint it is a self-adjoint operator what about U2? U2 is that self-adjoint it is see this is called Hermitian so what is the thing with U2? It is Q Hermitian so U2 star is minus U2 just check these calculations so there is a kind of a U1 star is U1 is what I want to say U2 star is minus U2 U1 is self-adjoint so this is just to tell you that any operator on a possibly complex finite dimensional inner product space can be written as a real part and an imaginary part now U1 is self-adjoint so that is a kind of a real part we talk about why is it the real part for instance a complex number is real if and only if Z bar is equal to Z, Z bar is equal to Z so what is real about this the thing that is in the mind when one says that this is a kind of a real part is the following any self-adjoint operator satisfies the following in particular this U1 satisfies the following U1 x, x this is real for all x element of V so that is what we have in our mind when we say that this is a kind of a real part U1 x, x is real for all x in V whenever U1 star is U1 in fact the converse is also true the converse is also true first U1 star equal to U1 implies U1 x, x see this is a kind of a quadratic form U1 x, x is a quadratic form so if U1 is self-adjoint then this is real that is straight forward the converse will use polarization identity I am going to leave that as an exercise I will discuss polarization identity a little later but the converse is also true T is self-adjoint if and only if T x, x is real okay why are self-adjoint operators important this is a self-adjoint operator the real part U1 is a self-adjoint operator self-adjoint operators are important for one single reason which is that they there exists an if T is a self-adjoint operator on a finite dimensional in a product space V then there exists an orthonormal basis of V each of whose vector is an eigenvector for T okay so this is saying this is the same as saying that T can be diagonalized by means of a unitary matrix I will explain these terms later but this is just a kind of a preload any self-adjoint operator on a finite dimensional in a product space V has a property that there is a basis of there is an orthonormal basis of V each of whose vectors is an eigenvector for T okay this is another very important property that a self-adjoint operator satisfies. So this is just to kind of relate the notion of the adjoint with the conjugate operation for a complex number let us look at this further look at subclasses of operators which satisfy certain other properties with regard to the adjoint okay one of them is a unitary operator but before you discussing unitary operators I want to discuss the notion of isomorphism between inner product space okay using the operation of star I want to discuss notion of a unitary operator but before that isomorphism between inner product spaces let V be an inner product space not necessarily finite dimensional and T be a linear operator on V then T is said to preserve the inner product said to be inner product preserving if T satisfies the following inner product of T x with T y is the same as inner product x with this is true for all x y in V T is said to preserve the inner product of this equation holds okay this leads to the definition of an isomorphism remember isomorphism between vector spaces is a 1 1 it is a linear injective surjective map linear bijection for inner product spaces the extra requirement is that it must be inner product preserved okay. An isomorphism isomorphism means same structure right morphism is structure iso is the same in addition to this being a bijective linear map it has got to do with vectors inner product spaces V, V and W for instance so it must preserve the inner product also that relation should also come an isomorphism I will say now between an isomorphism on an inner product space an isomorphism on an inner product space is a linear bijective inner part preserving map is a linear bijective inner product preserving map linear bijective map which preserves inner product before I go to characterizing isomorphism between inner product spaces let me tell you one property that an inner product preserving map must satisfy I will give examples a little later but there is one property that this inner product preserving linear transformation must satisfy in relation to the norm there is a relationship with the inner product this should lead to an relationship between norms that is the following I will state this as a lemma let P be an inner product preserving linear map linear operator on V, V is an inner product space not necessarily finite dimension any inner product this is definition also you will see for a general inner product space then norm of Tx equals norm of x for all x in V we say in this case that it is an isometric norm of Tx equals norm x for all x in V the converse is also true the converse also holds so if T is an isometry linear then we can show that it will preserve the inner products remember that this norm is a norm induced by that inner product okay whenever we talk about a norm and inner product space in this course this is a norm induced by that inner product okay one way the proof is easy if it preserves inner product then it preserves a norm is easy so necessity look at norm Tx square this is inner product Tx with Tx that is what I told you that this is a norm induced by this inner product but T is the inner product preserving so this is inner product x, x that is norm x square you take the positive square root you get these are non-negative numbers you get T is norm preserving okay so norm Tx is norm x for the converse you need a little work which is actually to first prove the polarization identities. So for the converse sufficiency we first observe but we can quickly see the proof also observe that the polarization identities hold so what is this one for real one for the complex case V is a real inner product space in this case the polarization identity is inner product x, y equals 1 by 2 sorry 1 by 4 norm x plus y square 1 by 4 norm x minus y square when it is real for a complex inner product space it is a little more complicated V is a complex inner product space then the polarization identity takes this form you can guess the next two terms I by 4 norm x plus I y the whole square minus I by 4 norm x minus I y the whole square these hold for all x, y out of which let me prove see what you can do is in order to prove this look at the right hand side expand by the inner product you will get the left hand side for example norm x plus y square is inner product x plus y comma x plus y that is norm x square plus 2 times real part of x comma y plus norm y square when you subtract the 2 times real part will go sorry it goes to the minus sign for the second term so that is the only term which will remain 2 plus 2 4 that 4 will get cancelled with this simple verification proof so I will not get into the proof of this that is really straight forward using either let us I will prove it for the case when V is real that if it preserves the norm then it must preserve the inner product okay complex case is similar so let us take the case when V is real I want to show that inner product T x T y is x comma y so let me start with inner product T x T y this by definition is 1 by 4 norm of T x plus T y the whole square minus 1 by 4 norm of T x minus T y the whole square I am assuming I am proving the sufficiency part I am assuming that T preserves the norm I will show that T preserves the inner product this is 1 by 4 norm of T of x plus y T is linear minus 1 by 4 norm of T x minus y use the fact that T preserves the norm so this first term is 1 by 4 norm of x plus y square that is norm of T x equals norm of x so norm of T x square is norm of x square minus 1 by 4 norm of x minus y square the real polarization identity tells you that this is inner product x comma y so T preserves the inner product okay so if it is inner product preserving then it must be norm preserving and conversely I have proved it for the real case complex case is similar so let us now look at necessary sufficient conditions for a linear transformation to be an isomorphism between inner product spaces by the way I do not have to write 2 inner product spaces V and W are said to be isomorphic inner product space set by isomorphic if there is an isomorphism between the inner product space isomorphism is not just a vector space isomorphism it must also preserve the inner products okay then this relation is an equivalence relation V is isomorphic to itself by means of the identity map V isomorphic to W means W is isomorphic to V by means of the inverse transformation inverse exists because T is bijective inverse is linear because T is linear inverse preserves inner product because T preserves inner product that takes a little observation but it can be done and again if V is isomorphic to W W isomorphic to Z then V is isomorphic to Z through the composition map okay so that is quickly it is an equivalence relation isomorphism between vector spaces is an equivalence relation but let us look at the following result which gives isomorphism in by means of other conditions so I have the following result you must compare this theorem with a corresponding result that we proved for vector spaces finite dimensional vector spaces where we characterized isomorphisms here we have finite dimensional inner product spaces let V and W be inner products finite dimensional inner product spaces over the same field such that dimension of V is equal to dimension of W and I am assuming that V is a finite dimensional inner product space so V and W are finite dimensional inner product spaces with the same dimension for a linear map T from V into W the following conditions are equal the following conditions are equal okay so same dimension the following conditions are equal the first condition is quality T preserves inner product just this much T preserves inner product dimension V is dimension W condition to T is an isomorphism to emphasize it is an inner product space isomorphism second is apparently stronger than the first one but under in this framework these are the same these are equal in conditions so T is an isomorphism where I am just emphasizing that it must be an inner product space isomorphism condition C for every orthonormal basis of V let me list the elements for every orthonormal basis okay just write down the elements U1, U2 etc U n for every orthonormal basis this of V it holds that T U1 etc T U n is an orthonormal basis of W now this you must compare this I told you with the vector space isomorphism it will not be an orthonormal basis it will be just a basis okay you include orthonormality because there is an inner product here so it holds that T U1 T U2 etc T U n so I am given that the dimensions are the same it holds that this is an orthonormal basis of W for every orthonormal basis we may call this a image this is an orthonormal basis apparently weaker is condition D which says for some orthonormal basis V1 okay I will use the same U1 U2 etc U n of V T U1 T U2 etc T U n is an orthonormal basis condition D looks weaker than condition C but in this framework they are the same the conditions are equivalent okay let us prove this again as before we will prove A implies B B implies C C implies D D implies A implies B T preserves in a product so you can give the proof tell me T preserves in a product I must show that it is a nice inner product space isomorphism so I need to show that it is bijective that is all how does that follow the conditions are the same as to be used but how do you prove it is bijective that is the only thing left you need to show T is bijective okay null space is 0 so how do you show that null space is 0 preserves inner product okay you can use that norm T x equal to 0 implies x norm x equal to 0 just use that that is the first part only to show T is bijective okay sufficient to show you must show that T is bijective but sufficient to show that T is injective sufficient to show that T is injective because the dimensions are the same by rank nullity dimension theorem it follows that it is injective and only if it is surjective T is injective but then we know that but since T preserves inner product we have that T preserves norm so norm of T x equals norm x for all x in v so T x equal to 0 implies norm x equal to 0 it is implies x equal to 0 and so null space of T is single term 0 and so T is injective so T is bijective so B holds it is in fact invertible that is what B says B implies C for every orthonormal basis we must show that this image is also an orthonormal basis of W what is given is that T is an inner product space isomorphism must show that okay let me just write down given that orthonormal basis U1 etc U n I must show that T U1 T U2 etc T U n is an orthonormal of W but why is it a basis for W? Why do not you just give me one word one word T is an isomorphism over T is an isomorphism B implies C T is an isomorphism so this must be a basis we want to show it is extra it is an orthonormal basis see T is an isomorphism so this is a U1 U2 etc U n is a basis implies T U1 T U etc T U n is a basis to begin with so first of all this is basis first this is a basis I need to show it is orthonormal but just use the factor it preserves inner product so just look at T Ui, T Uj I must show that this is equal to delta ij then it follows it is an orthonormal basis but this is by definition Ui, Uj because it preserves inner product Ui, Uj I started with this as an orthonormal basis so this is delta ij the chronicle delta so this proves that this set of vectors T U n etc T U n this set of vectors T U n etc T U n forms an orthonormal basis of W that is B implies C, C implies D there is no need for the proof if it holds for every orthonormal basis it must also hold for some orthonormal basis so only thing is D implies A so C implies D there is nothing trivial D implies A for some orthonormal basis U1 etc this is an orthonormal basis if this is satisfied then I must show that T preserves inner product okay let me just write down the condition D there exists an orthonormal basis such that T U1, T U2 etc T U n this is an orthonormal basis I know this I must show that T preserves the inner product Muttukumar you wanted to say something C is stronger than D so C implies D is straight forward okay so I am just proving D implies A okay how do I go about proving this I must show that inner product T x, T y equals inner product x, y for all x, y okay so we have to take a close to that let us take x as a linear combination of C I have I will take x, y in V I must show that inner product x, y equals inner product T x, T y they are in V they can be written in terms of these and I also know what the coefficients must be I will call the coefficients x1 etc x and y1 etc y n because they correspond to x and y so I have the following representation x equals x1 U1 plus x2 U2 etc plus x and U n I know what these coefficients must be y will be y1 U1 plus y2 U2 etc see we need to use the linearity of T so you write x and y in terms of these basis vectors what is the inner product of x with y let us calculate that let me use a summation notation here this is summation what do I use i, i equals 1 to n x i Ui this is summation j equals 1 to n y i Ui sorry y j Uj I need to calculate this so can I just write down now we have seen this several times i equals 1 to n x i j equals 1 to n y j inner product Ui Uj, yj bar okay but you see what is going on we started with an orthonormal basis the outer sum I mean the inner summation is for j and for the inner summation i is fixed so when j takes the value i it is 1 all other terms are 0 j takes the value i this becomes y i bar so this is just summation i equals 1 to n x i y i bar something that is similar to C n for instance something that this is precisely the inner product in C n with respect to the standard orthonormal basis okay let us calculate T x T y similarly consider T x let me write down T x T x will be a summation i equals 1 to n x i T of Ui and T y is summation j equals 1 to n y j T of Uj I take the inner product compute show that it is the same as this so inner product T x T y is again I will reduce the calculations i equals 1 to n x i j equals 1 to n y j bar inner product T Ui T Uj again the same thing T Ui T Uj that is part of an orthonormal basis when j takes the value i it is 1 all other terms are 0 because this is an orthonormal basis so this is again i equal to 1 to n x i y i bar so A holds so T preserves inner product if you know that T is a linear transformation which takes some orthonormal basis to some orthonormal basis of V to an orthonormal basis of W dimension W equals dimension of V okay an easy corollary I am going to leave that as an exercise for you let V, W be finite dimensional inner product spaces finite dimensional inner product spaces then V is isomorphic to W where it is isomorphism of inner product spaces between inner product space then V is isomorphic to W if and only if dimension V equals dimension W I will not write down the proof I want you to tell me how it goes there are two parts suppose V is isomorphic to W as inner product spaces then dimension V equals dimension W how does it follow see there exists an inner product space isomorphism from V into W but that must be in the first place a vector space isomorphism but we know vector spaces when they are isomorphic and they are finite dimension the dimension must coincide one way converse suppose dimension V equals dimension W finite dimensional spaces I must show that V is isomorphic to W that is all I must exhibit an isomorphism so take an orthonormal basis of V take an orthonormal basis of W orthonormal basis of V we can call it V 1 etc V n for W W 1 etc W n define a map T which takes V i to W i then you can show that it is linear bijective preserves inner product okay so this is an easy corollary of the previous result before I stop let us look at two examples one over finite dimensional spaces another one over infinite dimension spaces inner product preserving maps product preserving linear maps okay I told you two examples the first one is I will take V to be R 3 with the usual inner product W to be the subspace of the space of all 3 by 3 real matrices that are skew symmetric that are skew symmetric space of all real skew symmetric matrices of order 3 I want you tell me the dimension of W what do you expect the dimension of W to be first I want to give example of inner product space isomorphisms so what do you expect the dimension of W to be 3 okay can you can you show it is 3 how A transpose equals A see a matrix of order 3 square matrix of order 3 has 9 independent elements 9 independent ways of filling the elements the entries of A A transpose equal to minus A make sure that diagonals are 0 so 3 conditions are gone I need 6 independent apparently 6 independent conditions can I reduce it to 3 skew symmetric transpose is minus A so just fix up either the upper part or the lower part the other part is determined so dimension W is 3 okay I define a map T from V to W by T of the vector X take a vector X how do I fill it up as a matrix in such a way that I get a skew symmetric matrix the definition is it must be skew symmetric the diagonal entries must be 0 so 0 0 0 this is minus X 3 this is minus X 2 this is X 1 this is negative X 3 minus X 1 X 2 okay then it is an easy exercise to see that T is linear dimension of V is the same as dimension of W so in order to show that it is a invertible bijective it is enough if I show null spaces singleton 0 can you see that null space of T must be singleton 0 so T is injective dimension of same T is bijective I only need to show that T is an inner product space till now I have only vector spaces I will end R 3 with the usual inner product end out W with the following inner product for 2 matrices in W the inner product is defined as 1 by 2 times trace of A B transpose this is not new this is not new it is just trace of B star A if you want the extra thing is 1 by 2 you will realize why it is why that constant 1 by 2 appears there given 2 matrices A B and W okay I will leave it as an exercise for you to verify that inner product X 5 over R 3 the usual inner products T X comma T 1 so please verify this is just elementary calculation you calculate A B transpose look at the trace it will be 2 times the usual inner product that is the reason why this 1 by 2 is taken to cancel out the 2 okay finite dimensional case this is an inner product space isomorphism just an example in the infinite dimensional case it is inner product preserving but not an isomorphism example 2 I will take V to be C 0 1 real value real value to continuous functions over the interval 0 1 with this inner product it will be 0 to 1 F of T G of T together with a weight T square okay so I will include a weight T square any positive weight can be included so this is an inner product on V W is the same space C 0 1 real value to continuous function over 0 1 with the usual inner product same space with the usual inner product define the mapping T from V to W by T of F continuous function F at T so T F at T is T times F of T then you can show that this T is linear it is inner product preserving this T will come 2 times first argument second argument so this T square comes out okay so you can check that this T is inner product preserving, inner product preserving linear map it is an inner product preserving linear map is T invertible T F equal to 0 does it imply F equal to 0 of course T F equal to 0 of course implies F equal to 0 T is injective T is not bijective T is not bijective because take the constant function 1 take the constant function 1 and W the question is does there exist a function small f such that T times F of T equal to 1 does there exist a continuous function F such that F of T equals 1 by T so T is injective but not bijective please check this so T is an inner product preserving linear map injective but not bijective okay. See I am emphasizing this example because see remember we have shown that if T is if dimension V equals dimension W and if T is inner product preserving then it is invertible okay that does not happen here because we are over infinite dimension spaces that is the purpose of this example the result that we proved earlier holds for finite dimension spaces dimension V equals dimension W finite it is a same space W V W equals V same space inner products are different okay so this is injective inner product preserving but not bijective but not onto bijective okay let me stop see I want a continuous function F of T such that T F of T equal to 0 for all T the only way that can happen is F is 0.