All right. Today, we'll talk about tonrmal basis and the role that they play in the study of inner product vector spaces.  So let me recall the setup which we discussed last time. We have vector space V Over F or F is either the field of real numbers or the field of complex numbers.  And we have fixed an inner product on v, which is a map from V cross V to F satisfying the axioms introduced last time. Then we have the definition of or normal list of elements of v. List one, et EM. I want to recall that I'm following here the terminology from the book. But sometimes people use a different term set one M. There is a subtle point which is that normally when we say a set, it is implied that we do not have not fixed a particular ordering of this set. But here, ordering is important. So sometimes people say ordered set, but in the book that we use the term that is employed in this case is a term list. I personally prefer the term set, but it is ambiguous because of the ordering. So that's why I'm following the terminology from the book. List one M obviously it's ordered by the indices is called orthonormal. I vectors are pairwise tgonal to each other, EI zero for not equal to g, and the norm of each of these vectors, which I recall is a square root of E with itself is equal to one. This condition, by the way, this condition is that Ei is a unit vector. Okay. Uni vector is one whose norm is equal to one. Now obviously, if we relax this condition and only require this, then we will refer to this list as togonal rtormal means that in addition, we also require each of these vectors to be a unit vector. Okay. Now, what are these lists good for? The following mas will show us. So first of all, so properties of tnrmal lists, SLM One is that we can easily calculate the norm. So suppose this one N is an tonmal then the norm of any linear combination, squared is just the sum of squares of the coefficients. More precisely the absolute values of the coefficients. And the proof is simple just this expression simply means that we take a one one up to am M with itself inner product. Now we use linearity and we get them A bar, E J. We use linearity for both for the first and the second argument. Keeping in mind that we pull out a scalar from the second argument, we should replace it by its complex conjugate. This does not affect what happens in the case of vector field over the real numbers, but for complex numbers, we have to do that. We get the summation independently over one. Now this, we can actually write using the Cronicer symbol, which we introduced before. We see that only diagonal terms survive, the corresponding inner product, I I is equal to one, what's left is just A bar one to n, and that's exactly what's written here. That's very simple, very straightforward. The second statement is that that B nortrmalst is linearly independent. So let's prove this. We will use part A, and so we will write again, A one, one, a linear combination. So if this is equal to zero, this is a zero for A 1:00 A.M. A. Then by part A, we can find that this part is going to be zero, and so the norm of zero vector is zero. The left hand side of this formula is zero and then it's equal to a one squared plus a squared, absolute value squared. Now each of this is greater than qual to zero. By definition of the absolute value, square root of any number real or complex is a non negative real number. If you have a sum of no negative real numbers, which is equal to zero, it means that each of them is equal to zero. For all I, and that means that each Ai is equal to zero. We see that if you start with an equation like this, linear dependence equation, we end up with a statement that all coefficients have are equal to zero. And that's exactly the property of linear independence. Independence of this one M, right? That's what the statement that they're linear independent means that if a linear combination of this vectors is equal to zero, then no coefficients must be zero. We have obtained it by applying formula A. Okay. So ask me if it's not clear. So now, so far, we are considering some collections of orthonormal orthonormal vectors. But the best case scenario is when we actually have a basis, which is orthonormal. So And so now that's two, suppose one EM EN, let's say, rtonrmal list, which is also basis. Of v. Then naturally then must be equal to the dimension of v. In this case, we will call this list an orthonormal basis, obviously, Then we have the following. For any vector in V, we can calculate explicitly in a very straightforward fashion, the coefficients of the expansion of v in terms of elements of this basis. In general, we know that in general, If E one, E N is any basis of. Not necessarily to normal. Then any vector in V can be written as a linear combination of the elements of this basis, where AI elements of our ground field, and they are uniquely defined, uniquely determined by the vector V. In general, it's a rather complicated question to find those AIs, and the only practical way to do it unless one n is a standard basis in say RN or cn Then vectors themselves are given as colons and then the numbers, a sub can be read off asi appearing in the colon. But if you have a more complicated basis, in general, it's rather difficult to find the coficients. You have to actually solve the system of system of inhomogeneous equation equations with n variables. But if the set, this list, this basis is normal, then we are in luck. We can actually calculate those coefficients in a very straightforward fashion by simply taking the inner product of v With EI. You see. It's a very nice formula. In general, let me actually emphasize them coefficients. And so I'll put them in yellow here to kind highlight them. Okay? In general, this AI, As hard to find. But if this basis is ormal Then it's easy. Namely, this AI is simply equal to v. We can simply take the inner product of v with each I. This is going to be a number and going to be an element of F, and we're claiming that that's exactly the coefficient in the expansion of v with respect to this basis. That's the first statement. The second statement is that we can also easily find the norm of v by simply taking the sum of the squares of the absolute values of these numbers. Okay. So let's prove it. We simply take to prove it let's observe the following fact. Okay. Which is actually very useful, not only in this context, but in general. Suppose we have two vectors in V. Call them v. Such that v j is equal to v j for all from one to n. Okay. Actually here referred to the basis vectors, one N. But in general, actually it would be true also, even if the basis is not artnrmal just any basis. Then it follows that v is equal to, you see. You can test whether two vectors are equal by simply taking their inner products, all possible inner products vectors in a given basis. If you find that the values are the same, the vectors are the same. How Let me explain. If this is so, then we can take this to the left hand side and use the linearity, so we can write v minus w j is equal to zero for all J. We take this to the left hand side, and then the difference is in a product of the difference. But if so, then it follows that v minus w with any vector is equal to zero. Okay. Simply because every vector can be written as a linear combination. Of elements of the basis. And then we use linearity well, it's a semi linearity because when we have a scalar, we pull out it with a bar, but it's not going to change anything because what's left will be zero. We find that V in a product with U is zero for all U. But last time we proved that this implies that VW is zero. The only vector which is tag to everybody in the vector space is a zero vector. Remember the argument was a bit of looks like cheating because you say, if it's argon to everything, it's tog to itself. But in a pro is square root of u, and one of the axioms is that square root of the norm of the vector is equal to zero only if the vector is zero. This implies last time that v minus w is zero, but this is equivalent to saying v is equal to do. That's why to show that the left hand side is equal to the right hand side, it's sufficient to simply take the inner products of both of them with all elements of our basis. So for us, this will be v. Well, it is already V, and this will be W of this fact. And so what we need to do is simply take the inner product of both sides and ascertain that they are equal to each other. So on one side, we will have V E J, and we're checking whether it's true. We want to share, we want to find out, right? And then here it will be the sum of V E, E, and then J. Then by linearity, this is the same as the sum i from one to n i, j. But this is a chronicer symbol. Only one term survives and that term is J. Indeed, the left hand side is equal to the right hand side. That's true for all J. We are exactly in the situation of what I call the fact here that we have checked that the left hand side of this equation. The right hand side of this equation has the same inner product with E J for average j. And that implies that the equation holds. Okay? Yes. Okay That's called linearity, according to the first factor. From here to here. All of these are equal to zero. If I is not equal to. That's what it means to be delta crane delta out of this whole sum, there is only one term which is non zero, when i is equal to j and that term is this. Because if I is not equal to J zero, disappears and if is equal to j by our assumption, that's the norm squared, which is what we get this. That's part A. Then part B follows from A. And part A of Lemma one. One, and Lemma two, A. So then we get done. You see how nice. First of all, we can write every vector easily in terms of the normal basis and we can calculate the norm. In terms of the coefficients. So the question is, but how do we know that normal basis actually exists in a given vector space. For dot products, we know because it's given by such a simple formula and we can just take the standard basis of N or C for that matter, and get an normal basis. This is where we come to the Gram Schmidt procedure, which I started explaining last time. In fact, not only we can prove that nonrmal basis exists, but we can do more. We can actually give an algorithm, a very nice algorithm, which gives us a lot of control, which converts any given basis into orthonormal basis. So to show that rtonrmal basis exists. In any vector space within our product, we prove the following theorem, which is called Gram Schmidt procedure. It really is an algorithm. The easily program was very first thing I program when I was a student I remember, calculate finding orthogonal basis in a particular setting of kind of similar to what we discussed last time where you have functions and the inner product polynomial functions and the inner product is given by the product of the integral of the product of the of the polynomials. So suppose we have here's how it goes. We start out with a basis. We know that finite dimensional vector space exists in a finite dimensional vector space. Finite dimensional vector space. If you have an infant Densal rector space, the existence of a basis becomes a bit of a headache because you have to use axiom of choice or some such thing, so we get into some hot water storia. For in denial rector space, we have shown that it's possible to construct it. In fact, every vector can be extended to a basis. We start with one But in fact, we'll do more, let's say y one, y m. We'll do it even for linear independent subsets, and then when it saturates the set becomes a basis, then we will get a basis. Suppose this is a linear B linearly independent subset. Of v. Now we're going to define set x one is equal to y one. After that, one is x one divided by x one. The next is x two, which is y two minus y two, one. One. That's x two. But then we normalize it, then we say two is x two divided by. We continue this way. So at the at the K step, we have x k is y k minus i from one to k minus one k. And then we said we said K is x k divided by an alarm. And so on. And finally, x m is minus the sum minus one and m divided by the norm. So then the claim is that then this list one to EM is an not normal list. Okay. So the proof obviously will be by induction on K, just verify that they satisfy the properties. So the k, k equal one, it's clear, for K equal one, we start with y one, and then we go to one, which is just actually because x one is equal to y one, we might as well just write it as y one divided. Note that this is a unit rector If you have any vector, call it y one, which is non zero. Observe that I have to justify the possibility of dividing by this because what if this number is zero? But this number cannot be zero because if it were zero, it would mean that y one is zero, but then it cannot be part of a linear independent set. Note that y one is non zero since it belongs to a linear independent set, belong list. By our assumption, y one is included in a list, y one to y m, which is assumed to be linear independent. If you have a zero vector cannot be a member of a linear independent set or list. That's why this is justified. This is not zero, and we can divide by it. As a result, we obtain a vector which is a unit vector because if you calculate its norm, you will see that it's equal to one because the norm of the numerator will cancel in the denominator. Okay. So therefore, for K one, we indeed get anormal list of one element. A Atonrmal list consists of one element is just a unit actor. There's nothing more to it. Now suppose we proved it. Suppose we have proved That one up to K minus one is tonrmal Let's show that if we adjoin K obtained by this formula, where refers to the vector to the left of it is also going to be tonrmal Okay. For that, we simply need to calculate. So So notice that not only we have shown that one up to K minus one is an normal set, but can we also see that the span of one K minus one is equal to the span of one K minus one. This actually is very easy to see from the formulas because each formula is obtained by x up to scalar, each j is equal to x j, and it's equal to y j minus linear combination of the previous one. You can prove it by induction two. In fact, I should have said a more neater way to formulate it would be that this is not normal set and the span of the first k of them is equal to the span of the first. K from one to. And so then in my inductive assumption, I should also verify my base of induction. I should verify that the span of y one is equal to the span of one, which is obvious because they are proportional to each other. That gives me the first step of not only the statement that the set is tnormal but also that the span of one up to k is equal to the span of y one to y k in the case one k is equal to one. Therefore, we proved already that tm and the span of one k minus one is equal to the span of y one, y k minus one. We can include that in the inductive assumption. Now, because since the span is equal to the span of y, and here I take y k minus this. I see that K it follows that y k is not in the span of E one, E K. Because if it wore, it would be in the span of y one to y k minus one, and that would mean that the set is linear dependent. K is not in the span, and when I construct x k, I take y k minus something which is a linear combination of y one to y k minus one. Let's show that it's non zero vector. X k is non zero. So it's norm is non zero, and therefore, k, which is x k divided by the norm is well defined. It's legitimate to divide. The case step is well defined, and all I need to show is that it satisfies the same property. That is to say one up to k is span is equal to the span of y one to y k, and the set is tormal to show that it's normal, Is normal. It's enough to just check so normal. We need to show that um, k j is zero for all j from one to k minus one. Because we already know that K with itself is one because we have constructed K as a unit vector by taking vector x k, which is non zero and dividing it by its norm. Every vector obtained in this way is a unit vector, so we are in good shape. We only need to show this. But this follows right away from this formula. Since E K is obtained by dividing this vector by its norm. This is equivalent to showing that x k j is zero for j from one to k minus one. Let's actually do it right here. We have to take the inner product of this Let me just write it a little bit more. K a E J. What are we going to get here? We are going to get this in a product with E J. Bilinearity. This will give me first of all, this guy will couple with E J. K E J minus. Then there will be minus the sum of IE J for all Is for one to k minus one, right? But by the same argument as before, all of those terms will disappear, except the one where they match. The only term will disappear when i is equal to j, when i is equal to J, I substitute J for, I get J J, but I have already shown by induction that these are unit vectors. So it's going to be one and then going to be K, J. I get this minus itself, which is zero. Okay. Very straightforward calculation. If you if by any chance, you're not following it because you know, maybe my writing is not, you know, clean enough. Just try to do it yourself. It's just follow your pen will follow itself. It's absolutely clear. You find that the tormal also, guess what? We also get this that the span of one K is the span of y one, y k, y? Because because K is an inter combination of the previous ones. And clearly that we find that one span is equal to another. The procedure actually allows you to gradually obtain those vectors at each step, when you construct K, you're only involving the Y is from one to k. You're not involving YK plus one, k plus two and one. You have a good control that you know that the span of the first few of the original basis vectors will be the same as the span of the new basis vectors which are to normal. Any questions? Okay. That's very cool. Now I want to give you an example. And to show you how powerful this method tools are. This example goes under the umbrella fur analysis. For analysis is a very powerful method which arises in the study of harmonics. So originally the study of harmonics. And so I like to, you know, the analogy, think about the sound of a symphony. And so if you just measure the sound, so you'll get some function. And you know that this function is going to be superposition of the sounds coming from different nodes of different instruments. And each node is more or less given by sine or cosine function kind of trigonometric function with frequencies which are actually not random, but they are in the usual Western music chromatic scale. The ratios between the frequency satisfy some special proportions. For instance, if you go one octave up, the frequency will double and so on. And then the question becomes, automatically if you're given a function, how do you write it as a near combination of functions of the form sine x, where n is an integer, say. That's Question of harmonic analysis. Now what does it mean to write? The idea is that actually you can introduce an inner product and space of functions so that those harmonics are actually form an tonrmal basis. Then you can write things as a combination by using the previous result where we could find recall this formula from m one part A, that you can write a given B. So if you have an antonrmal basis, one N, you can write an tonrmal basis. You can write the sum of AI, I because it's at normal, as I just explained, you can find those, those ffient easily as just the inner product of v and i. In the case of inner product space where you have an normal basis, you can easily decompose any given vector as a linear combination, and now we apply it in the case of harmonics Except I will consider a complexified version of it, where instead of my harmonics normally harmonics will be Something like sine x, and cosine x or is in. Now, this does not capture, strictly speaking, an arbitrary sound of an arbitrary syphon, only if they play C of all the octos because it's only then that they will differ by an integer. But this is a kind of idealized version of what you could apply in mogenal setting mogenal setting, these numbers would be some powers of two, two to the power, which is a fraction with the denominator 12 in a standard chromatic, Western scale. Anyway, so instead of sine and cosine to simplify formulas and to accelerate this example, we'll consider Instead, we use the formula, the earlier formula, which is that e c the power x is equal to cosine x plus i times sine x, which you're probably familiar with. This way, we can actually replace the familiar technmetric functions by the exponential functions of this form. This function are going to have complex complex values, not real values. We will consider as our vector space H, I'll call it H. The space of continuous functions. From the interval 0-2 Pi. In fact, what we are doing, in fact, the proper way to think about it is that were cans funtion is not on an interval, but on the circle. Zero to two Pi is actually the angle coordinate on the circle. With values encompass numbers. It will be function f of t, I'll call a generic element of t will be our argument. Here we're going to play a little bit loose with infinite dimensional vector spaces. But that's because the payoff will be really amazing if we assume that all the formulas that we are used to in the final denial case actually work out in the infant Denial case, and they do in this case, it's not difficult to prove. But keep in mind that we are strictly speaking a little bit out of the scope that way I consider. Normal. What I want to say is that I want to I'm going to produce a collection of. I'm going to produce an list in this space of functions. So here's our noormal list. It will be labeled by integers. I will denote the F N. But before I explain or normal, I have to define the inner product. Before I define inner product, it doesn't make sense to say that something is normal or even arthogonal. The inner product in this age will be very similar to This is going to be over the complex numbers. In space over complex numbers. So it's a complex vector space, obviously. This is a vector space. Verse. Naturally, my inner product is going to take values and complex numbers and will satisfy the axions. Here's how it's defined. If I have two functions, f and g, by definition, the inner product is 1/2 Pi integral 0-2 Pi f of t g of t bar d t. But this is very similar to what I described last time as one of my examples. The only thing is that there is this factor 1/2 Pi is just for convenience. You will see it will normalize the list that I'm producing now. Here's my list. F N is equal to n t is an integer. Integer means all positive numbers zero, and all negative numbers. You see? These are harmonics in this world. If you only keep the real or imaginary part, you will get cosines and signs with x or, I see a typo. Okay. And since I'm I'm correcting it, I might as well make it closer to what we're doing. Okay. So first, let me show you that this is tonormal Indeed, okay? So let me calculate F. F N. By this formula is 1/2 Pi interval 0-2 Pi e t times remember have two complex conjugation. But t is a real number. T is tax value in the interval 0-2 Pi. Complex conjugation of is negative. That's why it's going to be negative. Maybe I emphasize it was yellow. It's because of the complex conjugation for the second function. That's my second function t d t. Everybody on board with that? Now, this is e to the i m minus n t d t. I will separated into two cases. For the first case is when m is not equal to n. In this case, I want to show that this is in a product is equal to zero. In this case, I have this integral, and then the anti derivative of this because this is non zero is going to be one divided by minus e minus t. Is this anti derivative? Right? Because if you take the derive tip of this, this factor will come down and we'll cancel this and I will get this. I get one divided by i m minus n e minus t from two Pi to zero. But the value in both cases are equal to one because e k times two Pi is equal to one for any integer. Here, k integer is m minus. This is a two Pi is equal to one and a zero's bvioly also equal to one. And so the result is zero, right? They are tognal to each other. Great. Now I want to calculate the inner product of a pen with a pen. So, in this case, it's like I substitute here is equal to n, so this will become one. I have 1/2 Pi, integral of dT. Guess what? That's one. Because the ing is equal to two Pi and I divide by two Pi. Now we can appreciate why I took I chose 1/2 Pi as my factor. That's my overall factor. Indeed, this is normal. In fact, in fact, Okay. I defined it here as continuous functions, but there is a slightly bigger space, which is obtained by completing it by adjoining limits of convergent sequences of continuous functions, which is called Hilbert space. Hilbert space, which is an notation L two is called L two of the circle is very close to what's called L two of one where one is zero to two Pi with the two end points identified, this is an example of what's called a Hilbert space. Which is a setting for inner product spaces, which applies equally well to find infant Denial vector spaces. It's exactly the bread and batter of quantum mechanics. Hopefully, at the end of next last lecture, I will speak in more detail about applications of quantum mechanics. Then in this setting, in this setting, and that's part of a subject called functional analysis, which is basically an extension of what we're doing here to the case of infant Denial vector spaces. And in fact, there is a there is a rigorous sense in which this collection is a basis. Is a basis. I is an t normal basis. Of L two. But in particular, in particular, it means that all the formula that we are used to for writing an element as linear combination applies here too. So every function can be written as a sum of fans called G indicate. For energy in this space. For any particular for every element in our H, which is continuous functions. This is called F series. The whole thing. This is the whole thing is called F series, and each coefficient is called fficient after French mathematician, Furies No furies. Coefficient. If you ever encounter coefficient now, you know what it is. It's just the inner product of your function with one of the harmonics, one of the elements of dart normal basis in your space of functions. You see, this can be made very explicit. And so I want to use this to prove a very important formula which was proved by Leonard earlier in 17 34. About the sum of inverse squares of natural numbers. One plus 1/4 plus 1/9 plus 1/16, and so on. Turns out to be equal to Pi squared root over six, which is an absolutely astounding result that the sum of some rational numbers is on the nose, not approximately, but on the nose, and the limit is Pi square root over six. You recover Pi, which has to do with the circle. But you'll see why because it has to do with the circle because of the way we're defining things. In some sense it demystifies the formula. Bear with me. I will use the Mundane function for this. I'm going to do for analysis for most dane function on the interval 0-2 Pi, my function f of g of t equal t, just linear function. I'm going to calculate. What I'm going to do is I'm going to calculate the norms. Remember Lemma two, Part B actually told us that the norm squared is the sum of v i squared. In this case, it is going to be function GFN n. So we prove this that the norm squared can be calculated as the sum of the squares of the fficient because this is antonrmal This was a part B of m two. Let's calculate take g of t equal t. That's an element of h and calculate it square of its norm. That's going to be 1/2 Pi. It's just with itself. And this is I'm using this in a product. In this case, because it takes real values because t is zero to two Pi. We don't need complex conjation doesn't do anything, so it's going to be an integral of t squared. We are back to single variable calculus. Of course we know how to do this. You get two Pi squared, 1/2 Pi times one third two Pi squared, right? Which is to cube, which is four Pi squared over three. If I got my math right. Yes. That's the left hand side. Now the right hand side, I have to calculate gn, and that's 1/2 Pi integral 0-2 Pi t times e minus and t. Now I have to put minus because of the cos conjugation for the second function. Well, I switch GMF in that form. I have to take the FN, remember, is this. I kind of I and it's complex conjugate will have a minus. This, you have to do by parts. Since I'm running out of time, I'll just leave it for you to do. It's just a simple thing in a single variable calculus. The result is minus two cases, the case when n is equal to zero and n is not equal to zero. If n is not equal to zero, it's minus one divided by n when n is equal to zero, it's Pi. Okay. So let me collect now. Now I want to apply it here. Yes. Let me collect the terms. So on the left hand side, I have four pi squared over three. Because of this calculation. That's this. On the right hand side, I see that when I have to take the absolute value. But absolute value means that I have and negative i. Let's just calculate. Let me just put it. What is one divided by minus one N. It's one divided by minus n times this complex conjugate, which is N, then that's going to be minus one divided by i squared squared, but i squared is minus one, so it cancels this one, so it's one over and squared. It's the same for positive and negative. Instead of taking the sum over all non zero n, I will take the sum over positive n and double it. It's going to be two times the sum from one to infinity, one divided by n squared. Plus I have to take the square of the zero term, which is pi squared. Now, I take it to this side, I get Pi squared over three and then divide by two, and I get bi squared over six is equal to the sum one over n squared. Okay. So that was called Basal problem, named after a city in Switzerland, which was actually unsolved for about 100 years. And then the question was to calculate explicitly what this is. I didn't even have a guess. And then earlier found it in I think 17 34 and he reported it in 17 35. He used a different method. You use some real analysis, but this is really the fastest and the most direct way. I really wanted to do it just to show you how powerful these methods are that you can find first of all, first of all, you have an explicit expression of a particular function, express function t as a linear combination of harmonics of the functions. Maybe I should write this down. It follows from the expression is that t is equal to the sum and not equal to zero. One divided by minus N and ad here it to the IN t plus i. In this more precisely, they're not equal This is an equality for almost all almost all values of t. There is a subtlety here that actually all of these formulas work out. If you don't insist that the left hand side are defined everywhere. They defined outside of a set of measure zero. So it could be outside of some finitely many points. I think in this case, maybe outside of 0.0 and two Pi. That's just an example. But let's get back to our regular scheduled programming and talk about What do we want to talk about? Yes. Now we want to talk about operators. For now, the question what we discussed is how we have used now inner product in particular way. So we came up with the notion of tonmal set or list and the orthonormal basis. We saw that normal basis behave in a much nicer way generic basis, we can find easily the coefficients are the fully eqficient. But now, our next project is to understand what can we learn about operators acting on inner product vector spaces? So, in fact, we produce a very nice family next week. We will produce a very nice family of operators acting on It's a nice family. Of operators acting on an inner product tra space. Tra space was in the product. Which are called Sjoint So let me give you a kind of a quick and dirty definition of what it means. It means the following that for any orthonormal basis, Beta of V. If you take its matrix representation and call it A with respect to Beta, Then if you take the transpose of this matrix and apply cops conjugation. This will denote this as a star. You will get back A. The matrix obtained by taking the transpose and conjugation is called adjoint matrix. It is a matrix of what's called adjoint operator, self adjoint means that adjoint is equal to, which means that the adjoint matrix is equal to the original matrix. Adjoint operators are those operators, which with respect to t normal basis have a self adjoint matrix. A is called self adjoint. We're going to prove a beautiful result eigenvalues and eigenvectors of such an operator. It turns out that every operator like this has a eigen diagnozable, number one. It's base on basis can be made tonrmal. You see? We have a marriage of two chapters of the book. We had this chapter about diagonalization. But we know in general, an operator may not be diagnizable. Here, this property assures both over the real numbers and over complex numbers. It assures that this aparator is diagozable Moreover, it's eigen basis can be made tonmal can choose an tnrmal eigen basis. It's a basis, consists of eigenvectors and it's tnrmal. It's like the best of all worlds. And Moreover, the eigen values are real numbers. Even if it's a complex matrix, which is very nice because in fact, in quantum mechanics, such operator, self adjoint operators, represent absorbables and absorbable such as coordinate momentum, spin, angular momentum, and so on. The idea is that the result of the observation is an eigenvalue of such an operator. In operator itself can be complex, the matrix could have complex entries. But physically, we don't observe complex coordinate, Mmenta and so on. And this explains why there is no contradiction because for such an operator, eigen values are necessarily real numbers, even if it itself has complex entries. This is a very powerful result and probably get to it on Tuesday next week and then on Thursday, I hope to give you an outline of how these things realize in quantum mechanics. This is a preview of what we're going to do. Just to, we're going to prove some preliminary results, which will enable us next week to prove that A is diagno or T is diagnzable has orthonormal eigenbasis And Eigen values are real numbers. Very powerful result. Let's discuss some things about operators on meets discuss some preliminary results concerning operators acting on inner product spaces. Finite dimension in product space. So here I want to need some preliminary results. And to recall the result, first result, I want to explain the first result. I want to recall something we proved before. Which is the following. That suppose you have any vector space dimensio. Over F, same as real complex numbers. T is an operator from V to V. Then we know that the matrix representation of this operator, relative some basis Beta, where Beta is some basis is upper triangular. I and only if the minimum polynomal splits. Split into linear factors, minus Lambda where Lambda elements of F. I from one to M. That will be something that we will use now. To prove the following result. So here the existed basis. Sorry, I should have said the existed basis. For some, should have proven. For some basis. For some basis. Beta, let me. I did not state it properly. Is upper triangular for some basis. Okay. Now, of course, if it is a aragar for some basis, it doesn't mean that it's pangle for every basis because you know what happens when you change basis. You multiply by Q and the left Q inverse on the right and you completely mess it up. You have something up atangular unless it's a multiple of the identity of matrix, you're going to mess it up in general. We're not saying that it's operangle for every basis. We're saying that exists some basis Beta. For some basis, it's up Triangular. The property is equivalent For example, if you have a diagonalizable operator, then there is an gon basis. Relative to gon basis is diagonal. A diagonal means piangular. That's one example. Second example, if F is a field of complex numbers, we know that there is a Jordan basis, and with respect to which the matrix has a Jordan form. Jordan form is apangular. Jordan basis works, gives you an example of such a basis. The statement is that this happens if and only if the no polyoma splits into linear factors. Okay. Now we can improve on this and this improvement will be crucial for proving this result. We will prove now that in fact, this minimal polynomial splits, if not only there is some basis, but there is some tonomal basis in which the matrix is upangular. It is a much stronger statement. We are able to find a basis which both gives you a pangar form and is tonmal both properties can coexist. Suppose now with an inner product is an inner product space over F. Then T is an operator. A operator. Then of Beta, I will formulate like this. There exists an orthonormal basis. Beta of v such that the matrix of the aparat relative to this basis relative to this basis, is upper triangular. This is equivalent to the minimal polynomial that stands for men polynom Z is the product of linear factors. Okay. So now, So this is if and only if statement, we have to prove it both sides. But this way is obvious. Follows from the previous result. Because if you have an orthonormal basis, it's a basis. There is a basis in which the matrix is a triangular. According to a general result, it means that the menial polynomo splits. What we need to prove the only nontrival part here is going back. And we're going to use Gram Schmidt procedure for that because it's very well adapted to triangular things. You see. This follows from from the Gram Schmidt procedure. Because, according to our result, let's call it star. So suppose P of Z splits Then star implies that there exists a basis ga, some basis, not necessarily at normal, such that the matrix representation with respect to this basis is upper triangular. Let's say Gamma is equal to y one to y n. Now we're constructing following theorem one, which is a Graham Smith procedure. We are constructing an orthonormal basis out of it, which we denote one E N. I claim that this apgar form will remain with respect to this guy as well. For the simple reason that This matrix, having a pngular form is exactly equivalent to the spans of the first few vectors of the basis to be T invariant. Is equivalent. Gamma is y one to y n. That the span of y one to y k is T invariant. For all one. Because what does it mean that it has upper triangular form? It means there are zeros everywhere here, which means that this is t of y one. It means that it doesn't have any components in y23 and so on, so that span of y one is preserved. This is t of y two. It means that the linear combination of y one and y two will still be a linear combination of y one and y two. And so on. It's precisely saying that if you take the first K colons, the span is going to be consists of vectors which are have zero below position. And conversely. But we've just proved that the span of y one, y k is equal to the span of one K. That's also T invariant. And that is equivalent to m of t Beta to be a triangular. It's absolutely almost obvious. In other words, we're not spoiling triangularity of the matrix by switching from the basis y1yn to the basis one N because at each step, the span of the first k of y is the same. That's what we proved we proved it in our theorem one. These two spans are the same. Because this is t variant, and this is T invariant, but this t variance of those successive spins is precisely the statement that the matrix is a parangular. Okay, great. But now a simple consequence of this is that if we are over the complex numbers, every polynomial splits into linear factors. Therefore, always be an orthonormal basis for which the matrix will be upper triangular. That is actually a famous result, which is called the theorem. Now we got it with very little effort because of all the groundwork that we have laid before minimal polynomials. The crollry of this which is called Shres theorem. Not to be confused with Shor's lemma. S was very prolific. To call Shor's theorem. If F C then then given any eperator on V, there exists an orthonormal basis Beta of such that is apgula. We will use this result next week to prove those statements about SFAjoint operators, and more. There's one more thing, the connection to the dual spaces that we should discuss, I'll just leave it for you to read. Maybe I'll say a few words next time. Basically, the inner product allows you to identify V and V prime. There is a bijection between them, which is an isomorphism if the vector space is defined over the real numbers and is almost an isomorphism. Is with a twist that it's a semi linear map, meaning that the sum goes to the sum, but scalar multiple goes to complex conjugation of scalar multiple. In the case of complex numbers, but in any case, you get a bijection. It clarifies a little bit the meaning of the structure that we're talking about inner product. Inner product is something that enables you to canonically identified V and V prime. In general V and V prime cannot be cononically identified. You have to choose a basis for that. They have the same dimensions, obviously they are isomorphic, but there's no canonical isomorphism. But inner product is an additional piece of information, additional data which give a particular isomorphism and V and V prime. That's just the curiosity not really such a big deal, so I'll just leave it for you to read. One more thing, which is the artogonal complement. Again, it's something that is well was well understood already in the original discussion of the dot product, which is that if you have a vetter space within the product and you have a subspace, then you define what's called the rtognal complement. And if you identify if you identify your or space with a dual, this will correspond to to what to the annihilator. There is a counterpart of the anilator which we discussed in the context of those spaces, and that's called the Artogn complement. So I'll just this is