Okay guys, let's go back to what we discussed last time.  We have a finite dimensional, vector space. Mostly we consider the key, the field of real numbers, or the keys of the field of complex numbers.  The time I proved the following theorem, that there exists a unique monic polynomial, P of Z, sorry. In addition, we have to have an operator from V to V operator. There exists a unique monic polynomial p of z of degree less than equal to the dimension of such that if we substitute for Z into this polynomial, we'll get 00 operator.  This is a notation for the vector space of operators from V to V of t, the minimal polynomial. Of V, since is defined as a linear operator on V.  Maybe I'll give you another example. Last time we saw 22 examples.  One was if t is given by a matrix, let's say acting 2-2 the matrix of this form, then the polynomial p of z is z squared. Indeed we have squared zero.  And I referenced problem number six from the mid term exam which discusses operators of this type. That's the first example.  The second example is an example of rotation by 90 degrees counterclockwise, where matrix is like this. In this case, the polynomial is p of z z squared plus one. In fact, you can check the squared plus identity matrix is equal to zero. Okay?  This can be generalized to the case of rotation by an arbitrary angle theta, counterclockwise rotation by angle theta.  We discussed this example before, and we saw that the matrix of this operator has the following form, Cosine theta sine theta, minus sine theta, cosine theta.  In this case, you can check that the polynomial is ESP is squared minus two cosine theta z plus one.  Indeed, if you substitute into this polynomial, you can verify that you're going to get the zero matrix.  Now example, this is a special case when theta is by 90 degrees, theta is pi over two. In this case the is not equal to zero or in this polyoma does not have real roots.  That polynomial p of z has complex roots. What they are actually is e to the minus theta. It's only if theta is zero that we get real numbers, namely one. If theta is pi, we get negative one. But otherwise, these are complex numbers.  This can be written as a product of linear factors but only oversee, this is what I mentioned last time, is that if your vector space is defined with real numbers, this polynomial that you get is going to be polynomial, real coefficients.  But in general, we know that such a Paloma does not have real roots. This is a prime example of that.  Now the question is, first of all, what is the connection of this to eigenvalues? This is going to be addressed in the next theorem.  We will see that actually eigen values are the roots of the polynomial.  If the polynomial, the polynomial can be factored as a product of linear factors over your field, then it means it has all of its eigenvalues in your field. You get the list of all the eigen values as the roots which appear in from these linear factors.  But in general, you may not be able to factor. You don't have eigenvalues, as many eigenvalues as you can happen. Counting with multiplicity, which as we discussed earlier, would be the dimension of. Before I get to that theorem though, I want to give you A more general example where you can easily infer what the minimum polynomial is. You see that these are ad hoc cases. In each case you can figure it out just by playing with it. For example, if it's two by two matrix, then we know that this polynoma is either linear degree one point or degree two. Because the minimum polynomas going to have degree less than equal to the dimensions. If dimensions two, it can be either degree one or degree two. Degree one is easy to rule out because a degree one polynomial would occur only if your matrix is proportional to the identity matrix, because that's what a degree one polynomial will give you. Then clearly it has to be a degree two. You just play with the T squared. You take and you take the identity. There has to be a relation between these three matrices. This relation is unique. Yes, I is the identity matrix. This matrix is by definition, this matrix. So I'm not sure what the question is. T squared plus this matrix is zero. Take t squared, you will get a matrix with negative ones. How is it T squared is non zero. What is t squared? Just calculate this, T squared. What do you mean by t squared? Two t squared by definition is the matrix which is obtained by multiplying this matrix with itself. In this case, t squared is zero. This matrix is different from this matrix. For this matrix, t squared is zero. T squared by definition in this case is this always just take the matrix and multiply by itself. That's the definition of T squared. If you multiply these two matrices according to the rule of multiplication, you will get the zero matrix. However, if you do the same with this matrix, you will get negative 100, negative one. You see now it's not zero, but if you add to it the identity matrix, you will get zero, okay? So now in this case you have to take T squared in the same way product of the Mets with itself. Then you take t and you have identity. It turns out that there is a unique, the unique efficient you want amonicolyono. It means that coefficient in front of skirt is one. Then you have two free coefficients. One is in front of t and one in front of the constant cofficientr. This theorem applied in this case means that there are unique numbers which turned out to be negative two cosine theta and one for which this equation is satisfied. That's how you can find ad hoc example, you can find the minimally. Here is another case where you can find, instead of writing, I'll talk about the matrix will be the matrix of the operator acting from a fend. It has the following form. Where is it? Oh yes, one on one. So the zero. So you have one just below the diagonal. This is diagonal, okay? Almost all entries are zero. In particular, all entries on the diagonal except the last one is zero. And then there are entries like this. Okay? So you see this is the main diagonal. So for example, for three by three, that would be like this. There are zeros here everywhere. Also, this matrix only has entries just below the diagonal and there to one the last colon, which could be arbitrary numbers. But for convenience I put negative sign in front. I claim that in this case of z is going to be the n. In fact, all the coefficients I claim that if we substitute for in this pm, we will get zero matrix, I call it A. If I substitute a, z is zero matrix. Matrix. Okay, so how do see that? This is like a whole family of matrices. You can think in terms of matrices or you can think in terms of linear operators. A matrix is the same as a linear operator from Eff, right? Because vector space fan has a canonical basis and we can represent a linear operator from by matrix. And conversely, given a matrix, we consider the operator of multiplying on the left by this matrix right here, you see that every polynomial actually can be realized as a minimal polynomial of a particular linear operator, arbitrary monic polynomial of degree n, where n is a dimension of the space. This is an arbitrary polynomial because we can take arbitrary coefficients. It turns out that it indeed is a minimal polynomial with specific matrix. And this can be constructed from these coefficients in this way by putting these coefficients, or more precisely the negation, the negatives of those coefficients in arranging them in the last colon, and then putting one below the diagonal. Okay, so the question is why? Why is this so how to see that? Here's a very nice argument. Let's apply. Let's take a vector vector which has the first basis vector from the canonical basis. It has one and the first component and zero everywhere else. If you take V, this is now. Let's take apply to V, multiply this vector by this matrix. We know that when you multiply this, that vector by this matrix, we get the first colon, right? The first colon has one and the next component then apply one more time, that's the second colon. But guess what? Second colon is actually the third basis vector. You see where it's going. So if you apply N minus two times the colon, this is minus first colon. Minus two times will be, you'll have this one here. All right? It is going to be finally e to the minus one is the last basis vector. You see, all of them are linearly independent. All right? Because we get all the basis, all the basis vectors. So that means there is no relation applied to V. There is no relation between these vectors. If there was a general, in general, the minimal polynomial could have degree less than n. Yes. So this is the claim, I'm trying to prove. This claim. I haven't gotten through it yet. Okay. This is a statement and this is a proof. So I'm trying to explain what the statement is. So the statement or the statement is in particular that it is degree n, right? This degree N in general in Pina could have degree less than n. But I'm going to rule this out by the fact that this vectors are linearly independent. If the p of degree less than n, if I apply it to any vector, I'll get zero. But that would mean that, in particular, if I apply to this vector, I would get a linear dependence relation on them. Therefore, it cannot be degree less than n to be consistent. Let me put to the end here. Now what does I want to repeat one more time? But I explained before to say that this expression is a zero operator is equivalent to saying that if I apply to any vector, I get zero vector V, which in this case is a fan for any vector F. You see, but here is a vector, the first basis vector that I know that it's impossible to have a polynomial, so there is no polynomial. A polynomial or any polynomial for that man non zero polynomial of degree n minus one to n minus one. Z to the n minus one plus one. Let's call it tilde, I substitute for and apply it to any vector. I get zero. Because if that were the case, that would mean, this would mean that the vectors are linearly independent. If we choose to B one. This is an equation of linear dependence. The above vectors are linearly dependent. You see therefore. But if there is no polter of degree and minus one, there is no monic polynomial of degree and minus one, or minus two, and so on with that property. That implies that the minimal polynomial, it necessarily has degree n. The maximum possible degree, minimal polynomial has degree N. Right? But also we know that there is a unique monic poly polynomial which satisfies this equation. If we find which polynomial actually applied to this vector kills it or gives zero, then we know for sure that that's the minimal polynom. Which polynomial, which polynomial is going to annihilate this vector? For that, we need to find out what is T to the n. Up to now we calculated TV and up to the n minus we got all the basis vectors. But the polynomial of degree n will also include the term to the n. What's t to the V? We just read the last colon. You read the last colon and you get minus zero, minus one, minus a minus one. You see that? If I take to the N plus, well N minus one to N minus one plus et ce plus one plus a zero identity and apply to, then I will get zero. Do you see that switch? I did not switch. It's exactly the same numbers as here. The last line, where is a and times da middle board here. Okay. Oh, I'm sorry. Yeah, this is typo. Sorry about that. Yeah. Right. Okay. Yeah. Our contention is that in front of the you have I. Right. So this was a twins mistake. Thank you for catching. Okay. So the upshot of this is that there is a unique polynomial of degree N, mononomial of degree N, which annihilates this vector. You can see clearly just by calculating what is, what TV is and so on. Just because of the uniqueness of minimal polynomial, we can derive from this. This is a minimal polynom. There can be no other minimal polynomial. There could be no other polynomials having the same property because we know that it's unique. This is true in general. In general, if you have a situation where you have a specific vector in your N dimensional vector space for which you know for sure that there is a unique polynomial of degree n which annihilates it. Right? And there is no polynomial of smaller degree which annihilates it. Then you know that that's a minimal polynoma because there is no other choice. You see from this we conclude that this is a minimal polyoma in this se, now you see the question becomes, why are we so interested in these polynomials in the first place? Because it seems like the ad hoc game. So yes, for this family of matrices, we do see right away what the minimal polynomial is. For those other examples, it's a little bit more difficult, it's more ad hoc looking. But the reason is that we really want to have a good, good understanding of a good conceptual approach to the problem of diagonalization of describing eigenvalues. What are the eigen values? The main point of this discussion is that the eigenvalues, Are precisely the roots of this Minimal Paloma. Let me explain this. And then based on that, I will explain what, what this Minal Paloma could look like. In general, if you have any questions you welcome to ask me. There's a theorem. Let's call them two because I just erased the theorem again. So you have this vector space and you have this operator, then the O, so that you have this minimum point. No, Emil, the zeros of P of Z are exactly the eigen of a. In other words, lambda is a zero. Lambda is a zero of z, meaning that of lambda is zero. This statement is equivalent. So the statement is lambda is an eigenvalue of, that is to say that exists vector v such that V is non zero. Remember, the definition of eigenvalue applied to V is lambda. These two statements are equivalent. Therefore, it's actually, we have to prove it in both directions if we go this way. Suppose lambda is a zero of lambda zero. Now we know that this is equivalent actually, to p of z being equal to z minus d times some q of z where this has degree and maybe degree m, where m is less than equal to the dimension of V. According to what we know about minimal, this would have the degree minus one. Right? Because we have chipped off one linear factor. Obviously, if you have a polynomial which is a product of minus lamb that times Q, then if you substitute Zq lambda, you get zero because the first factor will be zero. But actually we have seen when we discuss polynomials just before, the term this is, the converse is also true that if lambda is a zero of P, then must be divisible by Zm. Lamb. Okay. So this statement, that means that P of Z is equal to this product. Now, when we substitute t into this polynomial, what are we going to get? We get P of t applied to some vector, is equal to minus lambda times d, right? Times Q of t applied to V. Now you see we have found this is true for any V. And moreover, it is equal to zero because by the property of minimal pal, the property of minimal polynomial is precisely that you take PFT and apply to any vector, you get zero vector, right? You see we have found, if we call this vector, then we see that t minus lambda I times applied to W is zero. Which means if we open the brackets and take this to their side, that equals lambda. You attempted to say, great, so we have found an eigenvector but not so fast because we have to also check the W is non zero because we discussed this. That's why I always emphasize when we talk about eigenvalues or eigenvectors, you have to check two conditions. Not just this condition, the applied to the vector is lambda times this vector, but also that it is non zero because otherwise it's not very useful. A zero vector satisfies this equation. Trivially, for every lambda we need to show, because here we're lucky we can take an arbitrary V in this equation, because mino Pomal works. For every vector, there exists at least some for which this is non zero. You see this is the crucial point of the proof. How can we show that? Obviously this is mo, this is also mon, that is to say, highest cohicient is one, right? So I wonder if you see that, I've already see this argument. How can you show that this is non zero for some V? You can argue by contradiction. You say, okay, let's suppose it is equal to zero for every V. Where is this? This is double that. Then Q of t applied to V is equal to zero for all V. But that means that Q of t can be chosen as a minimal polynomial and it has a degree less than. That's a contradiction when P of Z has a property that no monic polynomial, smaller degree can have this property. Therefore, this cannot be true for every V. This cannot be true for every V, but that's our W. For some V, this is non zero. Therefore, this equation will be a tg vector and eigenvalue equation. You see, you'll see what is the yes, but we know that has the smallest degree for this property. There are no polynomials of degree less than N M for which this is true because it's a minimal polynomial. Mini polyoma is the one which has the smallest possible degree amongst all polynomis which have this property. This one, This is impossible. P of Z here is the one which is called the minimal pallynomial. And it is called the minimal Pal for a reason because it has the smallest possible degree amongst all paladomialQ having this property. You see, therefore this cannot be satisfied for all V. But if it is not satisfied for all V, then there exists a v such that Q of T applied to v is non zero. Call this then from the previous blackboard is non zero, is lambda. That means lambda is an genvalue, right? So it's very nice, it's a very nice argument this way. We prove it in this direction, that if lambda is zero of this polynomial, then lambda is eigen value. Let's prove the opposite statement, the converse. Suppose lambda is an eigenvalue of so then you have, you have applied to Mdv is non zero and you have V is equal to lambda, right? Yes. Then next, what about t squared V squared? V is of TV, but that's of lambda V. And that lambaV that's lambda squared V. Continuing in this way, we find that all for all K, the KV given this property is lambda to the V, right, and so on. But that means that if I take P of t and apply to V, it's the same as substituting lambda instead of t into the polynomial, because this is a combination of linear combination of powers of t. Namely, you have the t squared and so on. But each time you apply to the k v, you simply multiplying it by lambda to deka, that means to the k has to be replaced by lambda to the k. That means substituting into this polynomial. But lambda, because that's what happens for each monomial, that means that, right, but we have the condition. Our condition on minimal polynomial then becomes P of lambda applied to V is equal to zero. Now, this is an operator, this is an operator on V. But because we have a vector on which this operator acts merely by multiplying by sum, by number, the action of this operator on this vector can be written as a multiplication by this number. Since V is non zero of lambda times vehicle zero means that P of lambda is zero. And we're done because that's what we want to show. We're going from here to here. We're showing that if you have a lambdas and eigen value, then it is necessarily a zero or a root of the minimum pom. At this point, we need to discuss the framework. What are we trying to do? The, the crucial idea here is the idea of eigenvectors and eigenvalues. First, I want to explain why it's something that we need to pay attention to be a general operator, a generator for general operator on vector space. If you take a random vector, it's not going to be rescaled by the operator. It's going to move, It could be rotated right. It's really, if you think about, it's rare that you have, you have a vector on which a given operator, just by rescaling it, right? Why is this priced? Because for the following reason, general remark, you have a linear operator from V to V. Suppose that there is a basis of eigenvectors, let's call it v1v of eigen vectors. That means that I is equal to lambda I I for some lambda I, right? These are the eigen values. Now suppose, let's see what the matrix looks like in this basis. Call it beta. This is a crucial point. Remember, what is this matrix? Its first colon is the result of the action of On V one, written in terms of this basis, right? Its second colon is two, written in terms of this basis. Its case colon is VK, written in terms of this basis and so on. Right? These are the colons. But what is V one? According to our assumption, it is just lamb 11. But what is lamb 11 written in this basis is just lamb one and then zero, Right? Well, anyway. Clear, this is lambda 22, Therefore it's represented by this colon lambda two. This is lamb VK. According to our assumption, that's going to be again the diagonal entry. You see the only non zero entries here are diagonal entries. We conclude that this matrix is diagonal. That's why this situation is called the situation where we have diagonalized our operator. We have diagonalized it because we have found a basis in which the matrix of this operator is a diagonal matrix. A diagonal matrix is one for which non zero entries are also constraed on a diagonal. Next, you ask, why do we care? Why does it help us? It helps us because if you have a diagonal matrix, it's much easier to deal with it. For instance, it's very easy to find its ends. Power diagno matrices are great, We can easily find what N's power is. For instance, let's say to the matrix of t. To the M is going to be, let's call this A. It is going to be just the power of this. If you have a general by matrix, you want to take it power. Good luck with that. You know that each time to take a product, you have to take each entry, you take a sum of n terms. Now you have to do it n squared times. And then you have to iterate it M times. But now for the Enomtrix, we know how simple it is. It's just raising each value to this power. For instance, you can use it to find out what the power of any matrix is. Suppose you have a matrix B. Suppose you have a matrix B, you can write you think about it. A multiplication by this matrix is a linear operator. Suppose that you can diagonalyze it. Then there's a standard basis in say or N, and then there's your Regan basis. But we know that you can relate the two matrix, you can write it times A times Q. Inverse is a change of, well, it depends how you, if you write Q, Q is a change of basis matrix which consists of the agenvectors. Okay? So now if you want to take B to the M, then you write a QverseQverse, then a miracle happens that this middle canceled. You see, you end up with just times A to the M times Q inverse, for instance. You can now raise any matrix which has this property. That is to say that it has an eigenbasis, a basis of eigen vectors. Right? This equation means that the apirator from FN to FN sending vector V to B times v, has an eigenbasis. Eigenbasis means that there is a basis of eigen vectors. Let's say those eigen values are lambda I. Then the fact that this matrix B, or rather the operator of left multiplication by B has an eigenbasis can be interpreted in matrix algebra by simply saying that this matrix and by matrix is equal to QA Q inverse for some diagonal matrix, namely this one, The original one, right? Where actually we know what Q is also is a matrix whose colons are the eigenvectors. Look how much things simplify. You don't have to evaluate the power of B of your original matrix by multiply, multiply how many times t. Each time you multiply by, you have to evaluate n square entries. Each evaluation consists of taking the sum of things. It's more or less be operations. And Q times m operations. But here you only have to find the eigen basis and that's just multiply 2 meters. If m is equal to 10,000 there's a great improvement. Or if it's 1 billion and so on. This has tremendous applications, for instance, in the study of systems of linear equations, which I think you have learned in 54. I'm just recalling this stuff has been presented already in 54. But I'm try to remind you why it is essential, why it's important, why it's useful to have eigenvectors and eigenvalues. The best case scenario is when you don't have just a single eigenvector and eigenvalue. But when you actually have a basis of eigenvectors, in this case, things simplify tremendously, For instance, in this way, that's how this starts. And then the question is, how do we approach this problem? The traditional approach is to define what's called a characteristic polynomial. Okay, the traditional approach, we find eigen values from the retispalinomial. What is the Krityspainomial? First, we define the determinant of ByN matrix. Given a matrix with entries IJ, where I and j from one to n, there is a formula alternating sum of products of the I's. There is a recursive formula for by N, where you use the composition along the first row, reverse colon, and so on. This function determinant has several important properties. One of them is that the determinant of two matrices, of the product of two matrix is the product of the determinants. It's a way to assign the determinant. A determinant gives you a way to assign a number to a matrix and by the matrix. First of all, this function has this property that the determinant of the product is the product of the determinants. And the second property is that the determinant is non zero. If, if A is invertible, okay, you get zero value for the determinant precisely for those matrices which are not invertible. For instance, if an equal to two, then you have a two by two matrix A 11a1. Two to 122 determinant is A 11 times A 22 minus A 1221. Let see, it turns out that this, this by two matrix is invertible if and only if this value is non zero. Now that's the first step. The second step, we apply it to the problem of finding eigenvalues. The idea is to calculate the determinant, not of the original matrix, but first of all, how do we apply it? Suppose you have a linear operator on some data space of you mentioned N choose any basis beta of V. And let A be the matrix of relative to this basis, with respect to the basis. Then we take the determinant of A. Then we find out that if you take another basis, we will get another matrix. You beta, or you can take bet to prime. Prime means you take to prime. But the consequence of the property of the determinant is that the result will be the same. Why? Because a prime is going to be Q times A times Q inverse. The determinant of this is equal to determinant of Q times determinant of a times determinant of Q inverse. But this is the inverse of the determinant. First of all, this is invertible, obviously otherwise. You see, these two things cancel out. And you see that the determinant of a prime, the matrix relative to the basis beta prime, is equal to the determinant of A relative to the original basis beta. Therefore, to each appiator you can assign a number unambiguously. You calculated by using a specific basis, Converting your operator into a matrix, and taking the determinant which is a particular formula. But then you show that the result is independent of the choice. I should mention that here you can take determinant of q times q inverse, and according to this formula, it will be determinant of Q times the determinant of Q inverse. But this is identity matrix and the determinant of the identity matrix is one. That's how you know that this is the inverse of the determinant of Q. Because the product of the determinative Q and the term of Q inverse has to be equal to one. This is why in this formula, these two terms cancel out. The result is again determinant of a first of all. This way you can assign the determinant not to just a matrix, but to any operator which is independent of the choice of basis. Next, you consider the following. Parate minus lambda minus z. You consider T minus z times dt and you consider the determinant of this. This is going to be a polynomial in z because now there's a parameter, it's going to be pomal in z of degree. In fact, the leading term is going to be not one, but minus one to the n to the n. Let's call it p of z. Okay? Now, the claim is that lambda is an eigenvalue of P of Z, a root, a zero root. That is to say, P of lambda is equal to zero if and only if lambda is an eigen of. Why? Because you see, let me just show it in one direction. Suppose that lambda is an eigenvalue. That means that times there exists some non zero vector v such that TV is lambda V, but then minus lambda I is zero. That means that this is not invertible because this matrix now has a non zero element in its no space. If you have an operator which has a non zero non space, it is not invertible, it's not injective, and then it's non subjective either. This implies that minus lambda I is not invertible. Then, according to the property of the determinant, the second property of the determinant, it means that the determinant of the matrix is zero. This implies that the determinant of minus lambda I is zero. But that's precisely saying that if you substitute lambda, this number lambda, this is number into this polynomial P of Z, you get zero. This is a, I'm not telling you anything new because this is definitely bread and butter of math 54. It's the core chapter of 54 diagonizationI'm recalling what I'm going to explain to compare our present formalism, our present framework, to what is traditionally done. Then the theory proceeds in this way. He says, okay, first of all, from this we find out that tigon values are precisely the roots of this is called characteristic plyomalat. The ion values are just the roots of the critics plyinomial, okay? Given an operator, for example an operator on a fan, given left modification by, by matrix, you follow the following algorithm. You calculate the critic plyinomial, you find zero, right? Then if you are of the real numbers you may not get. It may not have enough zeros, but let's say you are the complex numbers. So then we know that a polynomial of degree n will have zeros counted with multiplicities. Then for each of them, you try to find a coresponing eigenvector by trying to solve this equation. If you're lucky, you can solve them all, and then you get an eigenbasis and then you diagonalyze the matrix. That's how we usually proceed. Then we start discussing obstacles to that. It turns out that you cannot always do that. Then the next best scenario is called Jordan form. Now, how is this approach different from what we have been doing this week? This goes to the heart of the textbook that we're using. The author of the textbook, Axler said that I want to do it without using the determinants. That was his idea. Why? Because his experience and I concur is that usually the determinant is introduced out of the blue without explaining where it comes from. It's magic. Okay? It's a god given function, polynomial in matrix efficiency, which has these properties. Then the question, how do we know this where, who came up with this and so on? This is left unanswered. We're given this magic wand after that with this magic want. Indeed we get a tool for describing eigenvectors and eigenvalues. He says, I don't like this. I want to introduce the concepts of eigenvalues and have a conceptual framework without using the determinants. This is what we are doing. Instead of Curtis polynomial, we have the minimal polynomial. You see we have not used the determinants in showing that exists minimal polynomial. Then we have shown that the agen values are precisely zero. In the traditional approach, you construct a polynomial given the matrix, or given one of many matrix representations of your operator, construct the decreased polynomial by using the determinant. You prove that all the eigenvalues are the zero of the polynomial, which is what I wrote here. Now we do it differently. We are proving that there exists a polynomial of smallest possible degree whose zeros are all the eigen values. You see the advantage of this approach is that we don't have to use the determinant. In other words, we're not using something that we don't understand. This advantage is that it is much less explicit because we have proven the existence of the operator, but we're not given the tools to find it. But what I'd like to explain to you what he's not explaining it in the book. In this chapter, obviously because he wants to avoid the determinants. However, he introduces them in the last chapter. What I'm planning to do is to come back at the end of the semester and spend a couple of lectures on given a conceptual derivation of this determinant by using what's called multilinear maps. This way we have it both ways. We can have both a conceptual understanding of eigen values and eigen vectors without the usage of the determinant. But then also we will be able to give the foundations conceptual foundation to the notion of the determinant and criticise polynomial. Okay, But for now I want to give you some pointers. Since you have been exposed to Crits polynomials already in 54, now we're doing something different in a sense. I feel like if I was looking at the first time, I'll be confused. What is the connection between mineral Poomas and Curtis Pos, right? We need to see it because here is an explicit tool. Here is something else we have come up with and we know in both cases the eigenvalues are the zeros of those polyoma equal are equal. This's what I want to explain the equal in a generic vanilla case, but in general they are not equal. And I will show you what the difference is. Okay, So what is the difference between the characteristic polynomial, polynomial of tea, and minimal polynomial? Both of them. The zeros of both. Let's call this is P of Z, let's call this characteristic of. Although I think transitional, maybe the Greek letters to separate them. More characteristic point, the zeros of and at the same are the same. The set of zero is the same, Set is the set of all eigen set of all eigen values of t. Okay? So how interesting, Right? So you have your vector space of dimension n, right? You have an operator on the sector space. You attach to it two polynomials. One of them is the minimal palnomial, which we have been talking about this week. The other one is hidden from us at the moment, but you have been exposed to it before. And I have just given a summary of this theory is the arsicnoow, the difference is that the degree degree of Klay cars Paloma always has the same degree as a dimension. It starts with minus one to the end, z to the n. Whereas the degree of P is some number which is less than qual to, it could be less than n. P of z starts with the first cofion one. This difference is a trivial difference. We can always multiply by minus one. They differ by sin if n is odd. But it's not a big deal, of course we can. The problem is that it actually could have smaller degree what's going on, quality or difference between them. What phenomena are responsible for this? This is what I want to address so that you can this material better navigate this material also being informed by what you learned before. Okay, so let me explain. When we talk about these things, there is one aspect of it which has nothing to do with linear algebra, but has everything to do with the existence or non existence of zeros, of polynomials over the real numbers. The fact that the field of real numbers is what's called algebraically closed, that there exists polynomials over the real, such as quadratic polynomials we have discussed earlier which do not have or real roots. On the flip side of it, if you field is a field of complex numbers, the polynomial can be factored into product of linear factors. I will explain the connection between the two, Curtis Pom and minimal polynomial in the case of complex numbers. If you have real numbers, it's more difficult to discuss if you want to stay within the confines of the field of real numbers. But you can always treat a real polynomial as a complex polynomial, will complex zeros. We're not losing any generality in this discussion. Let's assume and that the field is a field of complex numbers. That means that means that both polynomials can be factored into linear factors of z is going to be minus one to the n. Then here you have a product of Z minus lambda I. There should be factors, but we have to be careful because there could be multiplicities. The devil is in the multiplicities. In this case, I'll explain in a moment. This is called a multiplicity. Okay? So that's ki. Because the total degree of this polynomial is n, it means that n is equal to the sum of MIs. A generic situation is when all MI is article to one. Most polynomials over complex numbers factor into a product of different factors. This is a special case when one of those numbers is greater than one, but it's an important case. Nonetheless, this is where the discrepancy shows up. Each MI is greater than aricle to one, obviously. Okay, that's the critistic polynomial. Now P of Z is going to be the product also. But here you have different numbers, which I'll call I. Linear terms appear in both, which is exactly what I just said, that they share the set of zero. The set of zero for both of Z and of Z is precisely the set of all eigen values of our operator. In decomposition into a product of linear factors, the factors themselves are the same. What could be different is the powers. These are the same, but this could be different. Okay? So the first case. The simplest case. The simplest case, all I article to one. That means that comia, the product of Zm and sum just to the power one. Sum is now one from one to N. You have distinct tag values. This case is also distinct values of t in this case. The theorum in this case is that there is an aging basis. Let's call it beta then the matrix. We are in the best case scenario which I described on those blackboards, where you can diagonalize the matrix. But there is a basis in which the matrix of this apparatus literally diagonal and then with all the perks that come with it. That's state second statement. In this case, guess Mino polynoma is equal to the critically in the generic case, simplest. But actually what do I mean? Generic generic means that u. Let's think in terms of matrices. You can think of all the operators from acting on V. We know that the isomorphic to the space of linear operators from V to V, isomorphic to the space of bio matrices over right over. In this case, the generic means that out of all bio matrices, the complement set, the cases which do not fall into this case, they have smaller dimension than n squared. They're given by some equations. Like for instance, it could be some equation on the coefficients. But each time you have an equation, the dimension of the space drops in the space of by and matrices. If you drop a pin randomly, you will hit this generic key, right? So you can also think in terms of probability theory that if you bound the values of the matrix, let's say from -100 to 100, or complex numbers, absolute value from -100 to 100, there will be, this case will be realized with probability 99.99% So that's what this generic means, it's really, really the most common. Okay, so in this case, it's a best case scenario in two ways. First of all, there is an aging basis. The Mets can be diagonalized. Second, there's no difference between cretius polyomas and minimum polynomas, so we actually do equivalent things. I'm Tim out of time. So maybe I'll keep the proof for now. I'll explain the beginning of next lecture. But I want to explain to you what happens in a non generic situation when the multiplicities are greater than one. I is greater than one for two things can happen. To illustrate this, I will just consider the example of two by two. N is two of two by two matrix. Now suppose it's two by two matrix, then Pomal has to have degree two. I'm sorry, I made a mistake to sine, nobody corrected me. It cannot be equal because one starts with z to the n, and the other one starts with minus one to the equal, up to that trivial change of the sine. Okay? But for equal two, this sine is equal to one. Anyway, it disappears. Then we have z minus lambda squared. That's in the case equal two. You either have a product of two distinct factors, z minus one, minus two, or you have a square of a linear factor. There is no other possibilities. This is non generic case. Okay. So what are the possibilities? The possibility number one is that operator is such that it can be diagonalized. I will drop, I will not distinguish now between operators and I will just take the case of two. I will treat apparatus as mats. There's no difference between the two notions. So I will write for the, and not write M of T, but just for simplicity. So this is one case when that is a trial No, right? Because if you calculate the determinant of minus lambda minus lambda zero, the critics pomal in this case is Z minus Lumber squared minus Z is lambda minus lambda minus this determinant is z minus Mb squared. This is a case when that is diagonalizable, but that's not the only case. Wait, I have to say what happens in this case is that in this case, P of Z is. What do you think? What are the options for P of Z? P of Z has to have the same zeros, right? It could be either Zminslamda or Zemin's M squared. In this case it's actually different Ziminsmda. Why? Because if I take and lambda times identity, what do I get? I get zero. Do you see that? Take this matrix and subtract lambda times the identity matrix. What are you going to get? You'll get the zero matrix. Okay, in this case there is a difference and so you see why it's called minimal polynomial. In this case it raises the degree, it's not necessary for the principal polyoma to have it for the minimal polynomial. But trisa polynomial is something that always has degree n. Therefore the multiplicities will always show up if something is off. In other words, if there are not n distinct tigon values, it would have to be a term where you have some power over a linear factor. Okay. So that's the first case. The second. Any questions, by the way? Yes. Actually, you first and then you next. Yeah. Oh, it's because in the definition of, for instance, if it's a one by one matrix, Poma is defined as a determinant of t minus z I. If it's a one by one case, then the matrix is just one by one, is just one number. Let's call it a. Then minus the determinant will be a. Minus a is just a number. You see It's efficient. Is minus. It's because in the formal is minus I on the diagonal, we'll have minus z times. When you calculate determinant, you'll get minus z to the n if you will. If you want, I can write it more like this. Minus minus d. It follows from the definition that it starts out with minus d. This is a cretisic polyoma, which is what we usually study, the determinant of T minus minus I. Okay, you have a question. Yeah. So are the powers of the factors of the, That's what I'm going to display. The first gas from this is that oh, it's probably just going to be the product of linear factors in degree one, right? The natural gas. I'm going to show you that that's not the case in general. This already can be seen in for two by two matrices, namely fake. Which is lambda. Lambda one. It seems like not a big deal. We put one above the diagonal, but it changes everything. First of all, in this case is not diagonalizable. The Ctsiclynomal is still z minus lumber squared. This one has a zero as a counterpart. It does not affect the determinant because the only eigen vectors, in this case the eigen vectors were 10.01 But in fact, every vector is an eigenvector. But here the eigen vectors are all multiple alpha times 10. Because if you take 10, you get lamb times 10. It's an eigenvector, but if you take 01, then you get lamb times 01. But plus an extra turn, it kicks it to multiple, the first component. This is not an eigenvector, there's nothing we can do about it in this case. The matrix is not diagonalizable. That's number one here is diagonalizable. These polyoma has a smaller degree, minimal polynomial has a smaller degree. Now is not diagnozable. Minimal polynomial is equal to the carts P. That's how we can distinguish these cases you see in the remaining minute or so. Let me explain why the principle minimal polynom minus lambda squared. If I take minus lambda, what do I get? This I get lambda 10 lambda minus lambda 00 lambda. I get the operator with which we already familiar. We know that for this operator the min paranormal is not but t squared. Which means that this operator, in other words minus lambda, this is non zero by minus lambda squared is a square of this matrix which we have discussed at the very beginning of this lecture. We know that this is zero. That means that this is a minimal polynomial, not z minus lambda. That's how it always is. In general, you can always find a basis in which the matrix has a block form. With blocks either like this which is the diagonal or they have this once above the diagonal That affects from the point of view of Tis polynomials. Polyoma is always going to be the same. You count how many times lambda appears on the diagonal or lambda for each, but the minimal polyoma will depend a more finer structure, namely whether or not operator is diagonalizable. Okay, I'm out of time, so I should stop here. But I think this already gives you a good idea. This week we'll have just.