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[M] Produce the general solution of the dynamical system $\mathbf{x}_{k+1}=A \mathbf{x}_{k}$ when $A$ is the stochastic matrix for the Hertz Rent $\mathrm{A}$ Car model in Exercise 16 of $\mathrm{Section} 4.9 .$

The general solution of the dynamical system is $\mathbf{x}_{k}=c_{1} \mathbf{v}_{1}+c_{2}(.89)^{k} \mathbf{v}_{2}+c_{3}(.81)^{k} \mathbf{v}_{3}$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 6

Discrete Dynamical Systems

Vectors

Missouri State University

Harvey Mudd College

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Idaho State University

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Hey, everyone. So Ah, the question we're looking at today gives us ah, matrix A. And, uh, it wants us to solve the ah system of differential equations given by the ex prime equals a X, which will have three components. So three different possibly mixed differential equations wants us to use during canonical form. So how we do that is the whole basis of the method is that we set X equal to some in vertebral s. That's gonna be a three by three matrix times a, uh, three column vector y plan. Just change of variables there. And the reason we do that is because it allows us to puts the expand equation into this form. Right? And if you recognize this, it's because, ah, for a certain value of s that we can uniquely determine given a and we will we can have this equal J, which is the Jordan canonical form of the, uh, a matrix and say we will have That's an X. We will have a new equation. Why prime equals j times. Why, and we'll see that this is solvable. It's the analogous to the diagonal ization method, except for non Dagnall non diagonal Izabal Major sees it works as well. And so these air two equations that we need. And if we solve this second one here, then we can, uh, solve for y. And if we solve for y and we know s that we can solve this first one here. So what do we need to find? Well, let's get to work Finding J and s. That will be all we need. So how do we find the Jordan canonical form for something? Well, we take the Eigen values first, so that's gonna be your agon value equation that a minus slammed I the characteristic polynomial of a set that to zero. So what's a minus? Slammed? I diagonals take determinant of that, set it to zero. Well, the determinant of a three by three matrix, if you remember, is going to be It's gonna be a sum of three different parts. It's gonna be a some of this term times the determinant of this matrix. Plus this term times the determinant of this matrix, uh, starting here. And so you're sort of going around the back of the matrix if you want to think of it like that and Plus this term will that zero times the determinant of this matrix. But that's gonna cancel out. So we don't have to worry about that. Your let me cleanups factory in a little bit. Cool. Okay, so to start, that's ah, and then minus one. Perfect right, That's just negative. Landed times. Thea, do you buy two determinant? That's this Times this minus this times. This should be all familiar. And now the green one. Well, that's one we don't need to worry about it. Uh, times this times this minus this times this Well, that's negative. One time negative one minus Lambda Times. Here is just Syria. So it's one times one times one that's just one. We're gonna some these up plus equals Well, expanding out. It's, um, negative Lambda Times. That's going to be okay. You want to set that equal to zero Now, if you just look at this, you can see that ah, one is going to be an Eigen value. And also negative one is going Teoh he and Eigen value. And as it turns out, I'll save you the math of Destry writing this it's terribly hard. It's just some algebra, but we're going to get, uh, lambda minus one times one minus lambda squared equals zero. Let me just check my notes. Sorry. Yep, that's it. So, um, just make sure to keep these which to Which one is minus in which one is plus in ah, in your head. So we have the two agon value, slammed a one equals one, and that has a multiplicity. Ah, one. It's only repeated one time. It's not squared and land to It's gonna equal negative one. And that has a multiplicity of to. Is this squared right here? Okay, so we have the Eigen values here. Come back to that. And, um so that gives us the diagonal for Jordan Matrix Jordan canonical form of our matrix. A. But we need to find out how many Jordan blocks it's going to have. So we know that there will be one linear linearly independent Eigen vector for Lambda One. And that's the Mexican have, because only as multiplicity of one. But for Lambda two, we have multiplicity of two. So we could either have one or two linearly. Independent Eigen values dragon vectors. We know that the number of Jordan blocks is going to equal the number of linearly independent Eigen vectors. So we need to determine Ah, the Agon vectors for lamb to Okay, So how do we do that? Well, the Agon vector equation a minus slammed it to which we can sub in as negative one. So a plus one i a plus r times it's Eigen vector and we want that equal to zero. So what's a plus? I Well, that's just going to be one one shoe 0110 110 011 Is that what Crites Yes, it does. Okay. And who didn't mean to draw that? And we're going to solve the system of equations for components of e gives us This is zero. So that's going to go to well, we can quickly see that the top cancels with the bottom and we can get that if we subtract this from that, then we can get and that's our role reduced form. There's there's nothing else we can do there. So, uh, that gives a solution that gives us a V. So we see there is one, um, one completely zero. So that means there will be one free variable. And if there's only one free variable. There's only one, uh, Eigen vector that we can yet that satisfies this, right, Because, uh, if you had to, then you could have, uh, the linear some. So let's write it like this. V equals Alfa Times some the one. And if it had to linearly independent Eigen vectors than we could, right? This right where v one is not in the span of v two, but, uh, it's only one free variable. So we can't have a beta which only have an Alfa, and we have that The dimension of the wagons face for Lambda two is one. So that means what we have. One Eigen vector and one Eigen vector. So we have to Eigen vectors to Jordan blocks so we can write out Jay now, Sir, J equals negative one negative one one and zeros down here as usual. Well, this is already a Jordan block, this one, so we can't do anything there, so that's a zero. And we need one more Jordan block. So we have to combine these two. Otherwise they'd be three, and that's just gonna be a zero. And there, that's our Jordan canonical form. Okay, So that's half the battle we have now. Found this, Jay. Now we need This s okay. So what is the s? Well, it's the Matrix. That, uh, sort of gives you j from a right. We have found it before in the chapter satisfies that equation. Um, So it will be basically the equivalent of, ah, diagonal ization matrix with the Eigen vectors for a fully diagonal A, um, but where a is defective, where it doesn't fill the full possibility. So the dim of some Agon space does not equal the multiplicity of its agon value, which we have for Ah, Lambda too. So our section for Lambda Two is going to be instead of the Eigen vectors of Lambda two. It's gonna be a cycle of generalized Eigen vectors. So what does that mean? Well, a cycle is of the form. Um, we need a cycle of length to so it's a minus. Slammed it to ah, I times v prime, which is the generalized Eigen vector, which is different from the item vector, right, Because for an Eigen vector, this V zero v prime and those are gonna be our first two columns and then we're going to have the Eigen vector for Ah, the, uh, land equals one because that's just it's it's non defective in that I'd in space So it's right that as w so these are gonna be your columns in column one column to going three for s So let's start by finding the cycles. So basically, what this means is that this has to be non zero. So view problem can't be an Eigen value. So Eigen vector Sorry. And, ah, the term that would come here, which is a minus landed to you. I squared times of the prime That s t equals zero. So that's where that's our constraint. That's what we need to solve for right now. Coming there. It's not multiplied. So let's go ahead and calculate a minus. Lambda two I squared. So that's going to be well, we already did a minus lambda Too high. That was 110 Perfect, right? That's what we were reduced earlier. We need that squared so we just multiply it by itself equals one to on. Okay, so we need to find some V that brings that to zero. So we're just gonna do the same augmented matrix, and we're going to reduce. You'll see. This is a pretty easy, rare reduction. We know the it's already. And that's just that we, uh all the roads are the same. So we can just nickel the zeros one big zero down there. And so this is our equation. We have to free variables when we have that The prime, the first element plus two times a second element. Plus one times the third element is going to zero. So we can write. Um Well, it's just 211 negative to write. That works. Sorry. That's not quite what we need to dio. All right, so this is this is what we get and we they receive that their to free variables. So we can separate this to, um, two different vectors, right in Alfa one in a beta. Once we get Alfa Times. Negative 210 Uh, plus data times, negative. 10 one. So, uh, the second requirement is that neither of these air Eigen vectors So we need this to be non zero, so we can check these. So, uh, well, let's just do it out. So 110011 110 The times. Well, second one is gonna be a little easier. So to start with that medical on 01 it's getting just be some matrix multiplication equals. Ah, negative one one, negative one. So you see, that's non zero. So that's gonna work for us. Okay, so let's just choose that one. Doesn't matter. We just need one. And we need to put it in a cycle when we already calculated a minus lambda to be prime. You've already done that. We know v prime. And so we just need w. So we just need to find the I the actual Eigen vector for, um land war so that we do that well, the same way we found Eigen vectors for lead to. So I get vector equation. And since Linda one is one, it's just a minus. Sorry times Let's just say if you want equals zero. So a minus slammed I it's going to be perfect. Now we can ah go about reducing this. I'll save you guys Thea the tedium of that. Okay, so if we reduced it, we see that there's one Eigen vector as expected, this is one of three variable and V one equals well, some free variable fines. Well, negative. One of the first plus one of the last needs to equal negative one of the second plus one of the last. And that needs to be zero. So we can see that That hits that requirement pretty nicely. Um, the simple vector there. That's nice. So we have our w. Oh, I said that was w earlier. Why did I write If you want Sorry for the confusing notation. Everyone, That was very strange. I knew I had something written. Okay, so we have our columns of s You're ready to Seoul for us. Not even solves. Put. Put us together. Okay, so it's going to be to go back up here. We need it to be a minus. Landed to I, Tom's V prime. Then be prime then w used to be in that order. Okay. And that's just to do with, uh, because we put the one at the bottom. Thea actual Eigen vector for lambda. One needs to come last. And then for any Jordan block, you need to go the, uh, the order of this cycle. And that's how you determine that the order needs to be this. We had put the one first. If you put the one up here and then the Jordan block second, then the W would go first. But we did it. It's tradition to do it this way with the largest toward more. So it's gonna be this its first. This is second, and this is going to be third. OK, so step there for a second. S o s equals. Well, that's negative. 11 negative one, then Negative on 01 Finally. 111 That's that's golden. That's all we need. Okay, so that's that's there s lovely Circle that and green. And now we go way, way back up to the top to what we were originally solving for, um Seems like a long time ago. And we have this equation now, so we just need to plug everything in and we will solve the equations. Okay, So by Prime Nicholls, J times Why? Well, what does that give us that gives us a system of ah, differential equations running out of room here? I'm gonna raise some stuff. Sorry, guys. Only give us so much room in these, okay? And we're back here up at the top of this lean ice. Taking this girl appear real quick. Oof! All right, well, I messed something up there with overlay, but good thing is, we don't need that started a pet again. Uh, so this is our right prime in RJ. That's gonna be times y in resolving for the components of why, based on this equation, So remember, why is going to just be three components that were selling for Why one y two y three? My prime is gonna be the derivatives of these, so we can get that three equations. White three. Sorry. Why one prime equals negative y one us. Why, too, right? That's just this topper multiplied by why? And then why? To prime equals negative wide to and why three prime equals y three. Okay, so you'll see that unlike a diagonal ized form that we would use to solve this equation if they were diagonal Izabal. I'm like that We have this mixed term up here, but it's OK because we have these two that are unmixed, that we can just solve right off the bat and look into this and you'll see that it's pretty easy to solve actually. So let's do. Ah, this one first. So, uh, this one first? Why three Prime Michel's wife three. Well, what equals its own derivative? So we get that. Why three equals some constant of integration. C three for why, three times e to the t Perfect. That's good. Now this one. Well, it's just negative of that. So saving breath. Ah, you can just check. But this is just si two e to the negative t. All right. And now So we plugged it in up here for this one, and we get why one prime Plus why one equals C two e to the negative t. Okay. And now you might not realize that right now, but we have a chick to solve this. It's We're gonna have an integrating factor called eye of tea. You've learned this earlier in the book. Not sure how long ago, but you can go back and read this sections if you want. It's not too complicated. You'll see it works out and let that equal e to the T. So we cancel out this and we multiply through on both sides. So why one prime you do the teeth, Plus why won t to the t equals C two. Now, this is where the trick comes in. Well, what is that? That's just why one either The tea, that's the product. You love it. So just that whole thing derivative versus time. It's gonna equal C to right. And that's something we can solve. Just integrate both sides. Why? I want you to the t equals C two t for a C one. So why one is gonna equal perfect, And that is going to be our wise right. But we want to solve for X. So we have to plug them back into this equation where R. S, uh, matrix that we found is gonna come into play. All right, So, uh, we have this equation for X based on why X equals s times y remember, that's a column vector. That's a components of why so X equals what's R. S salt for it down here? It will check that. That's correct. And then plug in. Otherwise, why one that's right there. All right, Perfect. Now, this is just some easy matrix multiplication to get our final X and we see that that is going to be bit of a messy some, but you can simplify it out. Perfect. That's our X Feel free to send fight out. But for all intents and purposes, this is the solution. And every problem's gonna take this form. Um, we'll do all the other words. Don't make a video, but, uh, yeah, that's that's the process. It's just these guiding equations right up here. This is the most important part. Um, okay, I hope that is helpful. And everyone have a good day.

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