 See in the last lecture we had discussed the properties of unitary operators on a finite dimensional inner product space. In today's lecture I will discuss the normal operator the case of normal operators and prove what is called as the spectral theorem for normal operators on a finite dimensional inner product space. We must compare this spectral theorem for a normal operator with the one that I proved a couple of lectures ago for the silver joint operator okay. We will make this comparison just before the statement or perhaps after the statement of the spectral theorem for a normal operator. Let us look at some properties that we will require for a normal operator. So first of all what is a normal operator? Let me we have seen this before let me quickly recall the definition and prove some properties of normal operators. So definition let V be an inner product space a finite dimensional inner product space. Let T be a linear operator on V be a normal operator I am sorry I want to define a normal operator. So let me say I want to define a normal operator an operator T element of L V is called normal if it is called normal if T T star equals T star T, T star at the adjoint operator of T okay that is the definition we have seen this before let us look at a couple of examples one for the real inner product space and another one for the complex inner product space. Just a couple of examples we will define the linear transformations through matrices. So the first one let us take A actually A theta to be the rotation operator we have come across this operator before. So I am writing down the matrix first and then I will define the operator through this matrix A theta is cos theta sin theta minus sin theta cos theta for theta lying in 0 to pi define a linear transformation I will call it T theta from R 2 to R 2 by the formula that T theta of X is equals A theta X X in R 2 for a fixed theta then you can verify that T theta is a normal operator that is see this is a real inner product space so instead of star we consider the transpose T. So this equation T T star equal to T star T in this case is T T transpose equals T transpose T that is T is a normal operator okay so this is an example of a normal operator on a real inner product space. Let me give an example for a complex inner product space example 2 again I take a matrix and then define the linear transformation through the matrix say 1 I and 1 define let us call it S perhaps S from C 2 to C 2 by S of X equals A X X belongs to C then I leave it as an exercise for you to verify that S is normal that is S S star equals S star S if I remember right it turns out to be identity so these are this is a unitary operator this is an orthogonal operator okay so these are some of the examples of normal operators. Let us now look at some properties a couple of properties really for a normal operator that will be required in the rest of the discussion so the first property is the following this first property that I am going to write down should remind you of a result for a self joint operator okay so the first one is let me call it as a theorem let T be a normal operator on a finite dimensional inner product space we do not really need the space to be finite dimensional but still I will assume that V is finite dimensional let X be an eigenvector of T corresponding to the eigenvalue corresponding to the eigenvalue lambda of T then we will show that X is an eigenvector and X is an eigenvector of T star corresponding to the eigenvalue lambda bar corresponding to the eigenvalue lambda bar we will require this result in a later discussion okay so let us prove this result as I told you you must make a comparison of this result with the case of a self-adjoint operators in the case of self-adjoint operators we have shown that the eigenvalues of a self-adjoint operator are real numbers we have also shown that eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator are orthogonal okay let me include the second part also eigenvectors so we have a so all the time saying is that we have a similar result for a normal operator also. So eigenvectors corresponding to distinct eigenvalues of T are not just linearly independent they are orthogonal also are orthogonal okay so let us see the proof of this result T is given to be a normal operator let us now define an operator U by the formula T minus lambda I lambda is any complex number in particular we take lambda to be the number the eigenvalue that we started for example so take U to be this then it is easy to see that U is normal I leave that part for you to verify then U U star equals U star U okay so let us now look at so U is normal let us now look at norm of U star x the whole square norm of U star x the whole square this is by definition inner product U star x with itself I use a property of the adjoint operator this operator U star when it comes to this argument it goes as U so this is x U U star x I will now use the equation U U star equals U star U so I can write this as x comma U star U x again I bring this to this argument it will come as U so that is U x inner product of U x with itself which is now norm of U x square so in particular what this means is that norm of U star x equal to 0 if and only if norm of U x equal to 0 now look at U star what is U star U is T minus lambda I then U star is T star minus lambda bar I so 0 equals norm of I am writing T star minus lambda bar I of x if and only if norm of T minus lambda I x equals 0 so I have just written U star here and U here from this it follows that if lambda okay look at this equation if lambda is an eigenvalue x is a corresponding eigenvector for the transformation T for the norm of operator T then it follows that x is an eigenvector the same x is an eigenvector for T star corresponding to the eigenvalue lambda bar okay now this is what we wanted to show that is the first part. Let us show the second part second part is to show that eigenvectors corresponding to distinct eigenvalues of a normal matrix are orthogonal again this part of the proof should remind you of what we did for the case of a self adjoint operator so let me go through this quickly. The second part let T x equals lambda x and T y equals mu y where I know that mu and lambda are distinct I must show that x is orthogonal to y so I must show that inner product x y is 0 so let us start with consider lambda times inner product x y I can bring this lambda to the first argument lambda x comma y and then substitute lambda x equal to T x this is T x comma y apply the definition of the adjoint that is x T star y now use the previous result y is an eigenvalue sorry y is an eigenvector corresponding to the eigenvalue mu of the operator T then we have seen that it must follow that y is an eigenvector corresponding to the eigenvalue mu bar for the operator T star that is T star y equals mu bar y T star y equals mu bar y so this is the using the first part and mu bar is in the second argument it comes out with the conjugate and so conjugate double conjugate gives me mu times x y now the result follows if lambda is not equal to mu then inner product x y must be 0 and so x and y are orthogonal okay so x is perpendicular to y that is what we wanted to show. Let us go back and see every unitary operator is a normal operator okay but a unitary operator the entries at least when you treat the columns or rows of a unitary operator they are very much tractable in the following sense if u is a unitary operator then without writing on the board let me just explain this if u is a unitary operator then it satisfies u u star equals u star u equals identity okay what this means is the following just let us look at the equation u u star equal to identity it means that if you look at the inner product of the first row of u into the inner product of the first column of u star okay that is the inner product of the first row of u with itself then by looking at the right hand side it follows that this is equal to 1 whereas the inner product of the first row with the inner product of the second column of u star third column of u star etc the last column of u star these inner products are all 0 this is the same as saying that the inner product of the first row with the second row the third row etc with the nth row that inner product is 0 is that clear so every row of the matrix u if u is unitary has the property that every row if you take the dot product of the row with itself then it is a vector of norm 1 whereas the dot product of the vector with every other row is 0 this means what the rows of a unitary matrix form an orthonormal basis for Cn a similar argument can be given by looking at u star u equal to identity to make the following statement that the columns of a unitary matrix form an orthonormal basis for Cn a similar statement can be given for the case of an orthogonal matrix that is a real unitary operator so for an orthogonal matrix the rows and columns form an orthonormal basis for Rn for instance but the case of a normal operator is much more general the product is not equal to identity not it not so it need not be invertible a normal operator need not be invertible so the a statement about the entries of a normal operator is not easy to make but we will look at a particular case that is the essence of the next theorem that is if you know that an operator a normal operator t has the property that the matrix of t relative to a particular orthonormal basis is triangular that is either upper triangular or lower triangular then we can say something it follows there is a matrix must be a diagonal matrix okay so this is what we will show so let me make this statement precise so this is what I want to prove I will state this as a theorem let t belong to Lb where v is a finite dimensional inner product space suppose that there exists an orthonormal basis I will call it script b as before orthonormal basis b of v such that the matrix of t relative to this basis equals a is let us say upper triangular till now I have not made any assumption on t, t is just any operator with the property that there exists an orthonormal basis with a extra property that the matrix of t relative to the orthonormal basis is upper triangular or lower triangular it does not matter then t is then we can characterize normality of t necessary sufficient condition for t to be normal is that a is diagonal okay then t is normal if and only if a is diagonal as I told you this say something less about the matrix a that is you need to make some assumption on the matrix of t relative to orthonormal basis but nevertheless this will be useful for us in proving the spectral theorem for a normal operator okay okay let us see the proof of this so proof let us perhaps prove the easy part here suppose that a is diagonal by the way all the spaces are complex spaces suppose that a is diagonal then any two diagonal operators matrix is commute so I can say this much that a a star equals a star a this is always true any two see way is diagonal a star is also diagonal so this happens this means what thus what is a a is the matrix of t relative to b and a star is the matrix of a star is a star is a conjugate of the matrix of t relative to b but then we know we have proved this before that t b star is t star b so I have this equation and on the right hand side I will write a star a that is t star b into the matrix of t relative to b have this equation and there is a property that we have proved for the product you can replace that by the composition composition I will not use the circles I will simply say that t t star relative to b is t star t relative to b now this is a calculation that we have done a number of times before what follows from this is that the operators must be the same that is action of the operators so I have two operators let us say s and t the action of s on a basis is equal to the action of a t on that basis then we have seen that this defines as uniquely so from this it follows that t t star equals t star t so is it clear so t is normal if a is diagonal then we have shown that t is normal let us prove the converse part that is not straight forward the converse part is to show the converse part is to show that if t is normal then a is diagonal we will do we will consider one column of the matrix a at a time and show that all elements except the principal diagonal element must be 0 okay. So let t be normal what is given I am just reiterating and I am calling that as a matrix a right what is given is the following a equal to the matrix of t relative to b is upper triangular so let me write the matrix the form of the matrix it is upper triangular so it is something like this a 1 1 a 1 2 etc a 1 n this entry must be 0 a 2 2 a 2 3 etc a 2 n this entry must be 0 this must be 0 etc let me write the last row 0 etc 0 this is a n n. So, this is a form of the matrix a if t is normal I must now show that all these entries are also 0 this is the principal diagonal this is the principal diagonal I must we must show that all these entries are 0 okay so we will do it one row at a time okay now let us look at the relationship between the entries of a and the matrix of t relative to b in other words I will just use the formula for this equation so this says t of u j equals summation i equals 1 to n a i j u i this is true for all j 1 less than or equal to j less than or equal to n. Now this equation we have encountered a number of times earlier if a is the matrix of t relative to b then this is what this equation must hold where see we are making the standard assumption that what are these u j's u j's form a base that is u 1 u 2 etc u n this is an orthonormal base where b is an orthonormal basis for v this is standard terminology okay let us now look at let us now look at t u 1 that is j equals 1 we have t u 1 to be so j equals 1 so it is a i 1 u 1 so t u 1 is let me just write this a 11 u 1 plus a 21 u 2 plus etc plus a n 1 u n but see a 21 a 23 etc a 21 a 31 etc a n 1 they are all 0 so this is just a 11 u 1 okay t u 1 equals a 11 u 1 I will appeal to one of the properties that we have proved just now by the way what this means is that u 1 is an eigenvector corresponding to the eigenvalue a 11 for the operator t what follows is that u 1 must also be an eigenvector for the conjugate a 11 bar corresponding to t star so let me write that equation. So let me say that a 11 bar u 1 equals t star of u 1 and then look at t star of u 1 what is the formula for t star of u 1 similar to what I wrote down for t u j I will write down a similar equation I will write down an equation for t star u j and then look at the particular cases also t star of u j is summation i equals 1 to n this time it is a j i bar t star u j it is a j i bar u i so where I have made use of the fact that the matrix of the conjugate of the matrix of t relative to b is the matrix of the conjugate t star relative to the same basis b okay. So I have this again 1 less than or equal to j less than or equal to n in particular look at j equals 1 and so let me go back to this equation a 11 bar u 1 equals t star u 1 which can now be computed as j is 1 so it is a 11 bar u 1 plus a 12 bar u 2 plus etcetera plus a 1 n is that okay j is 1 and i varies from 1 to n so a 1 n bar u n. Now a 11 bar u 1 that can be cancelled look at what remains what remains is a 12 bar u 2 etcetera plus a 1 bar u 1 equal to 0 now u 1 u 2 etcetera u n form an orthonormal basis in particular they are linearly independent so subset is also linearly independent so u 2 etcetera u n is linearly independent subset so this is an equation like let us say alpha to u 2 etcetera plus alpha n u n equal to 0 so each of the scalars must be 0 so what follows is that a 12 bar so see the conjugates are 0 so the complex number also must be 0 so a 12 equals a 13 etcetera equals a 1 n all these must be 0. These are precisely the first row entries except the first entry so the first row entries have been shown to be 0 what we have shown is that a 12 a 13 etcetera a 1 n they are all 0 in particular okay in particular see we will appeal to induction but we need to make one observation in particular a 12 is 0 so if you look at the formula for t u j when j equals to what happens the formula for t u j when j equals to gives me a 22 u 2 and that is like this equation I am sorry yeah that is like this equation t u 1 equals a 11 u 1 so once I have a guarantee for that then we can appeal to induction that is I will say that in particular what follows is that t u 2 equals a 22 u 2 because a 12 is 0 and so I am looking at the second column all entries below a 22 are already 0 because it is an upper triangular matrix now what follows is that u 2 is an eigenvector corresponding to the eigenvalue a 22 for the operator t so u 22 will be an eigenvector corresponding to the eigenvalue a 22 bar for the operator t star I proceed as before there are only finitely many columns so it follows that a is diagonal okay so I will simply say by induction it follows that a is diagonal okay now this is an important intermediate result in order for us to prove the spectral theorem for a normal operator okay now before going to the spectral theorem we will prove what is called as a Schur triangularization theorem okay Schur triangularization theorem so let me write this theorem this is called the Schur triangularization theorem Schur triangularization the statement is a following let V be a finite dimensional complex in a product space and t be any operator now this result is true for any operator there are no conditions on t normality there is no such condition so what is the statement finite dimensional complex in a product space t is a linear operator on V then it says that you can do a little less than diagonalization then there exists an orthonormal basis script V of the vector space V such that the matrix of t relative to B is upper triangular okay for any operator t on a complex inner product space what is important to observe is that the Eigen values belong to the underlying field because C is an algebraically closed field okay so let us understand that there is no counterpart of this result for the case of a real inner product space however if you know that t is an operator on a real inner product space with the property that all the Eigen values of the operator t are real then we can write down a similar statement okay so let us understand the importance of the fact that this is a underlying field is a complex field it is an algebraically closed field okay we will show that is upper triangular the proof will be by induction okay proof by induction on the dimension let us call this n on the dimension let us call this n the case n equal to 1 there is nothing n equal to 1 it is trivial that is when n equal to 1 what it means is that we want to show that there is an orthonormal basis such that tp is upper triangular there is just one entry so 1 cross 1 matrix now what is that matrix so this is where we are using the fact this is a complex inner product space there is an Eigen value for the operator t there is an Eigen value for the operator t in other words in other words let me say that there exists x not equal to 0 such that such that tx equals lambda x for some complex number lambda we can call the x1 as x by norm x the Euclidean norm x is well defined then it follows that tx1 equals lambda x1 with norm x1 being 1 you take the basis consisting of that just the single vector then the matrix of t relative to b is the number lambda which is trivially upper triangular let us assume that the result is true for finite dimensional inner product spaces of dimension n minus 1 and then prove it for spaces of dimension n so in that the induction hypothesis is made we will prove this result for all finite dimensional inner product spaces of dimension n so let us take dimension of v to be n let us now start with okay suppose dimension v is n see we made a statement that the operator t has an Eigen value a similar statement holds because the underlying field is complex a similar statement can be made for t star so let me exploit that let x not equal to 0 be such that this time t star x equals lambda x for some complex number lambda existence of an Eigen value so there is a by definition in on 0 vector x satisfying this equation let us now define w to be span of this vector x and take its perpendicular so w is span x perpendicular then what follows is that then observe that x belongs to w perpendicular okay x belongs to w perpendicular so what it means is that t star of w perpendicular is contained in w perpendicular that the space w perpendicular is invariant under t star because x is an Eigen vector okay so one dimensional w perpendicular is a one dimensional subspace span by the single vector x okay now t star w perpendicular is contained w perpendicular we know that this is the same as saying that t of w is contained in w this is one of the results that we proved earlier if t star is if w perpendicular is invariant under t star then w perpendicular perpendicular must be invariant under t star that is w is invariant under t so t w contained in so it makes sense to so s from w to w defined by so I define a new operator s this is defined as the restriction of s to the restriction of s I am sorry the restriction of t to w this is well defined so this operator is well defined s from w to w is well defined see all that I need to know is that if you take an element w and w whether s of w belongs to w okay but by the definition s of w is t of that little w but t of w is contained in w so this s is well defined this is another construction that we have used when we discussed the self adjoint operator case so s is well defined what is the dimension of w what is the dimension of w dimension of w is n minus 1 see it is orthogonal complement of a single vector so dimension w is n minus 1 I have an operator s I have an operator s on a finite dimensional neoproduct space of dimension n minus 1 so the induction hypothesis is applicable that is there exists an orthonormal basis so let me say there exists an orthonormal basis let me write down the elements as u1 u2 etc may be x1 x2 etc x n minus 1 right this space is of dimension n minus 1 orthonormal basis of w such that such that the matrix such that the matrix of s relative to this x1 etc xn minus 1 is upper triangular okay so I have this by the induction hypothesis let us just write this in full and see what it says okay but before that let us I will make use of this and then define a new vector xn as the first vector x that we started with divided by its norm the vector x that we started with the norm of norm 1 so I will take that divided by its norm and call it xn then it is clear by the construction that the vector x1 x2 etc xn minus 1 which have been obtained before and the vector xn that we have defined newly through this vector x is an orthonormal basis. This is an orthonormal basis for v that is we are writing v as w plus w perpendicular and then for w this is a basis orthonormal basis for w perpendicular x2 etc xn minus 1 is an orthonormal basis so come we know that when you combine these two it gives rise to orthonormal basis for v all that we need to do is to look at the matrix of t relative to this basis script b which means what you need to look you take t apply it on x1 write it as a linear combination of x1 etc xn minus 1 xn etc finally take xn look at the image of t look at the image of xn under t write it as a linear combination of x1 etc xn now t x1 x2 etc xn minus 1 belong to w and the action of t is like the action of s and for s I have an orthonormal basis I am sorry the form of the matrix of s relative to x1 etc xn minus 1 is upper triangular okay so let us make use of that. So all that I am saying is that the matrix of t relative to this basis b is of the form let us say a11 it is upper triangular for the first a12 etc a1 n minus 1 then it is upper triangular this entry is 0 a2 to etc a2 n minus 1 etc all these entries are 0 a n a n minus 1 n minus 1 and then look at what happens so this purely comes from the matrix of s relative to x1 etc xn minus 1 what is t of xn t of xn see xn belongs to w I am sorry xn belongs to w perpendicular and so I have I make use of this invariance of space thing so what follows is that the last entry will be 0 0 etc lambda okay just check these steps now it is now clear that this is a required form and so the matrix so if t is normal I am sorry if t is an operator on a finite dimensional complex in a product space then the matrix of t relative to b is upper triangular this is the so called true triangularization theorem let us now move on to prove the spectral theorem for a normal operator spectral theorem for a normal operator let t be a normal operator on a finite dimensional complex in a product space v then there exists a basis script b of v such that the matrix of t relative to b is diagonal that is every diagonal you can say this is the same as saying that this is a matrix version of the statement that if you have a complex normal matrix on if you have a complex normal matrix then it can be diagonalized by a unitary matrix there is also another way of stating this which is that if t is a normal operator on a complex finite dimensional in a product space v then there is a basis orthonormal basis b of v which has a property that each vector in the basis is an eigenvector for t we have seen this before we have seen the statement before so let me not emphasize okay how does the proof go it makes you suffer the previous two results the true triangularization theorem and the fact that if the matrix of a linear transformation corresponding to an orthonormal basis is triangular then the operator is normal if and only if the matrix is diagonal okay so let me say by Schur's theorem Schur's theorem is applicable for any operator by Schur's theorem the matrix of I will simply say there exists a basis such that such that the matrix of t relative to b let us call it as a is upper triangle this upper triangle so that is Schur's theorem now a is upper triangular appeal to one of the results the earlier result that we proved since t is normal it follows that a is diagonal which is what we wanted to show that is there exists an orthonormal basis b of the vector finite dimensional vector space in a product space v with the property that the matrix of t relative to this basis is a diagonal matrix okay now you must compare this statement with the statement for a self adjoint operator see I will leave it as an exercise for you to write down the matrix versions the matrix version is something that I told you immediately after writing down this theorem so that is anyway there now you must compare this with the spectral theorem for a self adjoint operator where the underlying inner product space was not assumed to be complex okay to summarize this is what we have if you have a self adjoint operator on a finite dimensional inner product space it does not matter whether it is a real inner product space or a complex inner product space it always has the property that there is an orthonormal basis for the space v which has the which also satisfies additional condition that each vector of the orthonormal basis is a eigenvector for the operator t for the case of complex for the case of normal operators I have already made this I have already made this statement that if you have a normal if you have a normal operator on a real space then it is not necessarily diagonalizable that is if t satisfies t into t transpose equal to t transpose into t then it does not necessarily follow that t is diagonalizable the example that was given was that of the rotation operator which we also see which we have also seen in the beginning of today's lecture but if it is a complex inner product space and t is a normal operator then there is an orthonormal basis with the property that each vector is an eigenvector let me stop.