Okay, So there is a famous theorem called the Cayley-Hamilton theorem, which is associated with the characteristic polynomial. So here I'm going to state it and then the proof is actually a fairly elementary. But then what does its implications? And the reason it, the proof becomes elementaries because we have the theorem on the Jordan canonical forms. That is why the proof is elementary. All the work was in the Jordan canonical form theorem. Theorem. It's called the Cayley Hamilton here. Okay? So let me just first say something and then I will write it down. Okay? So suppose a is a square matrix with complex entries. All right? Now, you may recall that when you were doing your undergraduate work, you know that if you wanted to find the eigenvalues, you compute a determinant of a minus Lambda I, which is same thing as computing the determinant of Lambda I minus a. It just changes the sign. And you set it equal to 0 and you solve it. Okay? So that polynomial you get when you do determinant lambda I minus a or x minus a, That's called the characteristic polynomial. That's what you learned and its roots whether eigenvalues, right? Because that Lambda I minus a is invertible if and only if determinant is non-zero. You're using that, say if lambda is an eigenvalue, lambda I minus a not vertical. So that is one definition of characteristic polynomial. Okay? Then the other definition, which is a is a matrix. So a is also generates a map from C N to C N. So it is a linear operator on C and complex vector space. We've defined the characteristic polynomial for such things, which it seems to have nothing to do with the determinant. Okay? So the Cayley-Hamilton theorem, one way of thinking of it as saying that those two definitions are equivalent. You say, Well, what's the big deal there? Okay? The crucial thing about the Cayley-Hamilton theorem is not that the two polynomials are the same. The important thing is that if p of x or the characteristic polynomial, then we know that p of t is 0, right? In our definition of the characteristic polynomial, it was actually almost trivial that p of t 0, right? Or just from the definition because empty what zeros of p t becomes 0. But now you think of it another way, not P of x. I'm going to write all of this. So now here you have another way of getting p of x, which is just taking the determinant of that matrix. And from this, it is not clear that if you plug in P of a, that you'll get within this definition really not clear at all. Okay? So data's the strength of the Cayley-Hamilton theorem that if you construct this polynomial in this bizarre way with the detail and then some of the other, that polynomial p of 800 data is really the strength of the Cayley-Hamilton theorem. Not the fact that these two ways of defining characteristic polynomial are the same. All right, so I'm going to write down the mouse. I think when the memory in my laptop stopped declining with some file gets big because I've written so much. I think it starts going down. Because we're the Cayley-Hamilton theorem. Suppose a is a matrix with complex entries. Number 1, the characteristic polynomial of a linear operator. Number two. This is really the, right. Every square matrix generates a linear operator a on C and X going to X. So it has the characteristic polynomial. We have defined characteristic polynomials for things. What this is, this is another way to compute the characters same characteristic polynomial and this Q of u. And we know that, you know, when you have a characteristic polynomial, p of t is 0. So Q of a is here with automatic. Okay, So how is it, How are the two definition, the characteristic polynomial connected? So let us talk of the proof. So two is a consequence of one, because q of x is the characteristic polynomial. Then we proved above. Right? So this is direct consequences. If we just pull part one. Let Q of x equals the characteristic polynomial of a linear operator. Correct? So how do I know that this strange, complicated way of defining Q x with the determinant, which seems to have nothing to do with web. All the stuff we've been doing with generalised eigenspaces in anything. How is the characteristic polynomial? You defined the other way? Right? And the proof is going to be relatively easy because we have the Jordan canonical form theorem. Otherwise it's a very difficult proof. It's not trivial proof of one JBB Jordan canonical form for the operator. Okay? So now we proved in this theorem, you know that if you have the job, well that was the matrix version. They've kept the Jordan canonical form matrix, then the two are similar. So similar. So there is an invertible matrix. Q is similar to with Jordan canonical form. This is the crucial thing. Okay? So now I'm going to compute what is XI minus a? What I'm trying to show is that the characteristic polynomial of a and the j are the same. That's what I'm trying to head to the right. So I'm going to do a little trick. I'm just going to pull out the, okay, let's write one more step. X times u inverse. And then you pull out the U inverse and UN both sides. Okay, this is a very common trick. Okay? So let me come back here. So I want to prove that, you know, when you do this determinant thing, q of x and you get the characteristic polynomial which is defined in terms of the dimension of the generalised eigenspaces and the eigenvalues. Indeed, yeah, and the eigenvalues. So if J is the Jordan canonical form of a, then they are similar. And this J has actually, if you think of the Jordan canonical form, it has the eigenvalues on the diagonal. And the dimensions of the eigenspaces are the sizes of the blocks of the Jordan canonical form. Right? So the knee and the Lambda information is sitting in j. Okay? Let me remind you. Okay. Well anyway we'll come back to that. So the information which are there in the characteristic polynomial is sitting in j. It's the eigenvalues are on the diagonal. And the dimension of the Jordan block for each eigenvalue is the dimension of the eigenspace. So that knee and the Lambda information is right there in j. So I'm the connection between I and j. And so XI minus a is that you write the identity of this and then you pull out that we can, We're almost there. So determinant of XI minus a is the determinant of that. Right? Now, what properties of determinants we apply here? Make your life easy. Can we do some simplification? We do the product property on the right. So we get the determinant view inverse times the determinant of XI minus j times the determinant. You write determinant of a times B is determinant a times b. That's a property of determinants. Then once you have that, you're dealing with real numbers, right? So we can can you correct. And then these are real numbers. Yeah. And what else? Is that U inverse on the right? Yeah, right? Yes. So then you can, can you and me the first third determinants next to each other and then undo what we just did, make a determinant of U inverse U, right? Or in another way, determinant of U in versus how much, how is it related to you? It's just one divided by the determinant here, actually. But based on what you said just now, it's exactly a consequence of that. Yeah. Okay. Yeah. Right. Exactly what you said. And so it's really a consequence of that. Okay? So let's compute this determinant for this determinant. So what are the Jordan canonical form look like? There's a block for Lambda 1. Lambda k. Then what are the blocks look like? So if I did, all right. So I'm going to try to write this. Okay, so where did the blood lambda1. So there'll be some block here, lambda 1, lambda 1, and there'll be some ones here. The bigger lambda once they're there may be something here. You know, they may be someone. They may be something here. But they're all lambdas on the diagonal. And there'll be 0, 0 here, 0 here. And then this will have a block like this, lambda k and lambda k. But the important thing, and they're all zeros here. There may be some ones here, but the important point is, what is the nature of this matrix? There are everything, all the lambdas on the diagonal. Okay? What are the size of this block? Size of this block is the dimension of G lambda one is the T restricted to G lambda one that gave you that block. So that dimension is N1. And this time, this size is n k. We okay with that. All right. Now what is the nature of each of these matrices I've drawn it? Can you tell me, do you notice any structure other than the lambdas and the ones lambdas on the diagonal. What's below the diagonal? What's below the diagonal? Okay. Like if I drew any block like this, you know, So you had like J3 Lambda, J2 lambda and J2 Lambda. It would be lambda lambda, lambda one to lambda two lambda. So what's below the diagonal? Go ahead. Anyone? Is it too obvious or what's, what's below the diagonal zeros. So it's a lower triangular so you can just multiply the further you all agree, right? It's all zeros below the diagonal. That is all I'm asking for. Right? That's really, so what is X i minus lambda j going to look like? It's going to look like x minus Lambda 1, minus Lambda 1, 0. X minus lambda k, lambda k and zeros. All right? And this is the size N1, and this is the size n k. So this is what is called an upper triangular matrix. And when you take the determinant of an upper triangular matrix, meaning you know, which has, what are the determinant equal to 0, right? Let's say it was, make it for me. What's the determinant? What do you think? It's the product of the n? Why is that? Because if you, if you were just doing the rule of the zeros from the lower side will just cancel out. Okay? Do you have in mind like taking the A1 times the matrix? But you have to do more things, right? And then you'd have A1 times A2 times eight, which means you did an expansion along the column. That's what you mean? Yes. Yes. Okay. So it's L1 times determinant of a2 star 083 minus 0 times something plus 0 times, you'd expand along the first column. And then you repeat that. Expand along the first column. Just expand them and keep doing it. So whenever you have an upper triangular matrix, lower triangular matrix product. So what happens in this case is because we're just going to be 1 minus Lambda 1, N1. Exactly what the characteristic polynomial, that's totally okay with this. Lets us go back to this proof of the Cayley-Hamilton proof. Spent a lot of time on. Okay, Cayley-Hamilton says that if you do this determinant XI minus a, you get the same thing of the characteristic polynomial we defined earlier. So that's a canonical on the operator a. So then the homies with an a and G are similar. So you compute x I minus a and with this little algebraic trick, you get that's related to x I minus j. And then if you apply the property of the determinant, you get determinant x is the same thing as x I minus j. And then you Jordan canonical form has all the eigenvalue and be an inflammation. So when you compute determinant x I minus j, you get the exact right. So two very different ways of computing the characteristic polynomial. But they are the same. But that's not really the strength of the Cayley-Hamilton theorem. Cayley-hamilton term, ready? The important point is this, that you compute this polynomial and this determinant way. And then q of a is 0. That is really the Cayley-Hamilton theorem. Okay? So let us look at this for example. I just made up this example. The three by three matrix. I just made up something. Okay? So you just compute the characteristic polynomial. So basically you take the x i and you have to subtract it so you get x minus one. Okay? That's the characteristic polynomial. So to compute this determinant, so it takes some work. You can do various ways, but I'm just going to expand along the first column. Sometimes you do row deduction, but there's already a 0. So I'm going to use that 0 is there, you can expand along column or the row. So you get what? I'm just going to do. Part of the calculation. X minus 1 times 2. Cannot do it. You can just comes out to be. Okay, we can just check it. So remember a matrix here. So what's the significance of the Cayley-Hamilton theorem? If you take this matrix and you plug it in, that's equal to 0. Okay? Now, I don't really know any good application of this important fact that there is one very trivial sort of thing. This can be used to compute the inverse of a. Okay, So can you see how to compute that? This is the kind of question is sometimes ask on the prelims. Do you know how to compute from this? You can write it purely as powers of a. Can you see how? So? Not the actual value of the investment in terms of powers of a. You can write it like two lines from here. You're looking for a matrix B such that a times B is identity. That's what you're looking for, right? So can you find a b such that a times B is identity? You see now, I'm not sure it's very effective. So you can do that. So two is minus a cubed plus a t square minus DNA. So you can pull the a out. All right? I mean, it's just one thing. It's, you don't think of it. That n can be written as powers of a. But you can, that's 11. What other things you can see from the Cayley-Hamilton, if a is invertible can written as powers. Which if you don't know the Cayley-Hamilton theorem, it seems surprising that that is true. Question. Though. It seems like this wouldn't work if we didn't have like, okay, so the key thing is you can factor out an a, right? So if you had like 13 instead of 13 a, I guess you won't get that right. That's exactly it. So when will not hold its fatal down, fail. When you don't have this height, right? But when you don't have this item, what is the characteristic polynomial going to be? Which means x is a factor. So 0 is an eigenvalue. And if 0 is an eigenvalue, and it's not invertible, right? So it also tells you whether it's invertible or not. And are you with me on that? What I just said, if this constant term is not there, then this becomes x times something. So 0 is an eigenvalue. So which means there is a non-trivial null space and a is not invertible. Alright? Any question on this? Okay, actually, it's already 320 and they are only 10 minutes more. So I'm not going to start a new topic. I'll, I'll do the dual spaces tomorrow. And but if you have any question or anything, I know we covered a lot of ground in three days, lot of notation and lot of things. I'm used to it because I wrote the notes. I've been thinking about it for a month. That's all the notations my head and it's like any new abstract topic, you know, first thing you've got to get the structure of the thing in your head. What are these definitions and stuff? And then you focus on the mathematics. It's like anything new. It just takes time to get comfortable with the notation, what each thing stands for. Alright? So I've spent time here carefully doing things and sometimes it seems very long. Read the notes. I've just got the statements of the theorems in the beginning with explanations and examples in the middle. And I'm going to modify that this example I just did with the characteristic polynomial does now this matrix, I'll add it. I haven't done it. And I'll add one more question to the homework problem. There's one thing which I didn't talk about. So my definition of the mic, the minimal polynomial and the characteristic polynomial seem to be reliant on having that decomposition and the nullspace is and all that stuff. Which means that only worked for complex vector spaces which are complex. But actually you can define the minimal polynomials and stuff for any field. At least the minimal polynomial for vector spaces over any field in any tea. And I'm going to give a homework problem on that. That's nothing very complicated. Ideas are like this, actually simpler than what I've just done. So you can have any vector space over any field and a T and LV. And then one can define independently a notion of a minimal polynomial without knowing anything about the eigenvalues and stuff. Okay, and which is equivalent to this definition. So I've given a homework problem on that, and I'm going to add it. I haven't added yet. All right. So the notes, I'm going to change. One thing I'm going to add this example I've given does now just did. And I'll add one question to the homework. Alright? Okay, so see you tomorrow. Tomorrow we'll do dual spaces. And then the last thing I'm going to do is, you know, all these theorems, especially these norm turns for normal operators, they have an interpretation in terms of matrices. What is that interpretation? So there are no proofs. Just what does it mean? You need to know how to think in terms of operator then how to think in terms of matrices port. Alright? In applications typically you know, it's really the matrix version which has used and the exam sometime they give questions using the matrix version, so I'll just talk about that at the end. Okay? All right, so good luck. And I'm just going to stop the recording.
Lecture 5 Jordan canonical form
From Rakesh Rakesh January 05, 2022
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