classify-the-origin-as-an-attractor-repeller-or-saddle-point-of-the-dynamical-system-mathbfx_k1a-m-6

Question

Classify the origin as an attractor, repeller, or saddle point of the dynamical system $\mathbf{x}_{k+1}=A \mathbf{x}_{k} .$ Find the directions of greatest attraction and/or repulsion.
$A=\left[\begin{array}{cc}{1.7} & {.6} \ {-.4} & {.7}\end{array}ight]$

James Chok · Accepted Answer

Okay. We want to solve this. Only e using this initial condition with a physical to this. So the first thing that we do is to find a iving values so determined. Oh, a month Land our identity. So one minus dander. Minus four. Mice land, uh, minus two and three. So then this is equal to minus one. Lambda is foremost land law plus six to this equal to them squared. Plus Amanda minus a plus. Plus. Pull, Amber. Wanna stand though? Minus four plus six. This is Lando. Swear plus three, huh? 03 llama plus two. Take it. Land a minus plus one. And the prostitute? The service is equal to zero. You have Lando exactitude minus one. Old mines, too. So what this means is that we have on a tractor now to find the island values. I mean, Victor&#x27;s, we need to Seoul a minus. Land our identity. That&#x27;s why Geeta Physics. So let&#x27;s start with you on this case. So now let&#x27;s do lamb. The easier to minus one. We have Well, a is equal to 13 minus two and minus four. So we have one minus minus one. So let&#x27;s give you two months to throwing into four minus minus one. So 84 +123 kinds of pita. You know, 12 uses, right? Taking the first right of the first caller. Yes to you. No one wants to eat a too busy to throw. So you don&#x27;t want is even to eat it too. Now, let&#x27;s eat out. One is one. So you have either is equal to one and one such a first island mountain. Now, for your second ardent victor, I&#x27;m not gonna do it here, and you can do it yourself. But you are in Victor&#x27;s. Our land, not one is equal to minus one they want is equal to 11 land. That, too, is equal to negative to be too. You think you two and three? Yeah. Now your general solution to your early E is then X is a function of tea. Is it good to see someone? One on one eats the most one or medium iced tea. Plus. See some too. 23 eats the minus two tea. See? So now we use our initial condition. Zero is equal to 32 So 32 he&#x27;s equal to you. Said 111 plus season to season two Cheers through. So then you have three Easy to cease of one plus two cities up to and to use it to see someone plus three c sub. Sure. Yeah. Now, subtracting these equations you have one is able to true seize up to minus three seats. Up to is negative, Caesar two. So seize up to his ex connected one. And that would imply that Season one, this one is equal to three minus, minus two. Look, now you see someone, you see three executives see someone minus two to see someone is X file. So get general solution. It is. Then it&#x27;s a function too. These people to you. Five climbs 11 each of the minus T loss minus one less minus. One times 23 each, The minus Sure to

James Chok · Answer

Okay, so the question us us to solve this already with a easy go to this. Ah, nde on this initial condition. So to find the Aryan values we need to solve determinants Oh, a month&#x27;s Lambda our identity. And we need to solve this sequences. So this is equal to negative to minus Lambda one negative. Why? Well minus Lambda terms. So this is negative to minus lambda all my slander and plus five. So this is Lambda Squared minus for lander plus two Landa minus eight plus five which is Lambda Squared Minds to land out my story. So Landa minus three Landa plus one which is even easier because of this fact. So we have no idea values land but one is even 23 on land too is equipped with a negative one. So we have a is equal to native to one Newtie five and four Landa one Is it 23 land too. So first off, we know that it is a saddle point. Now to find up again Victor&#x27;s we need to Seoul a minus y land our identity. It&#x27;s about some victor eater people, sister. So then gets so let&#x27;s do lander is good. 23 So called number one. So negative. Do you want a street using maybe to buy? That&#x27;s negative. By 14 months, Street gives you one. That&#x27;s why he won eater too. You take this job. So if you take this one and this one, you get either one minus either to zero. So you are one last idea to predict zero. So yes, eat. Ah, one is It was negative. Eat up to so then Geeta physically to you. No one and negative Beetle one. So you don&#x27;t want easy? Go to one. You have one and mice one. So that&#x27;s your first. I didn&#x27;t, Victor. Now I&#x27;m not gonna do it for the other one. But I will just say that I didn&#x27;t. Victors are 11 but one is even 23 And be one is equal to Mars 11 And Lando. She is a modest one. And be too music too minus five once. So to solve your body, you have X function of tea. Easy to see someone on Mars. 11 each of the three T plus C sub, too negative. 51 each of the minus t. Now our initial condition is excellent, by the way. About zero Musical 232 So we think we have 32 is eager to see you. So one minus 11 Let&#x27;s see some too minus five on one. So we get three easy to negative seas of one Mars five C&#x27;s two. And here is ecstasies of one plus C sub. Too often adding these two equations together you get five is equal to native four seats up to and see some too easy, too. But in 85 4 I didn&#x27;t see so, too would be equal to two plus 54 which is eight plus five is 13 over four. So our solution is then this is season one. A solution. He&#x27;s. Then you see X function of tea is equal to 13 over four. Be one smartest 11 e to the three t mice five over or huh minus 51 Either maintain t

Bobby Barnes · Answer

so they want us to solve this initial value problem here. So first, let&#x27;s go ahead and try to get our system set up. So remember what our goal is is to rewrite this to be where it&#x27;s the Vector X is equal to. So it be some constant C One times our first Eigen vector times e to the Landau ones are first. I could value t and then plus same thing. But we just used the two&#x27;s for sub scripts. So this is like our in goal for this. So we first need to hear what are architectures are and what our Eigen values are going to be. So let&#x27;s go ahead and find the Eigen values first. So over here, we&#x27;re going to do a minus slammed I, which is going to give, uh, tu minus lambda three minus one minus two minus lambda. And then remember, we want to take the determinant of this because this should be equal to zero. And so that would give us two minus lambda negative two minus lambda. And then plus three is equal to zero. And if we go ahead and foiled this out, so actually I&#x27;m just gonna factor this negative out to make this a little bit easier for me. So that&#x27;s the difference of squares. So it be four minus. I&#x27;m just squared. Plus three is equal to zero. Um, so we could distribute that negatives allowing two squared minus four. Uh, Waas three is equal to zero. So that&#x27;s Lambda Squared. Minus one is you go to zero. And if we factor this, we get Lambda minus one. I am the plus one. So that tells us Lambda is just gonna be plus or minus one. So we have those. Um, Now we can try to figure out what our Eigen vectors we&#x27;re going to be. So when Lambda is equal to one, we get a minus one eye and that&#x27;s going to give s o b 13 negative one negative three. And then we can go ahead and reproduce this to just 1300 So that would tell us X one plus three x two is equal to zero or X one is equal to negative X to. So then the one is going to be Well, it&#x27;s x one next to got ahead of myself x one x two So we replace X one, which is going to be negative. The reacts to next one. Yeah, still, x two. I&#x27;m getting all the numbers slipped around. Background the x two so we&#x27;d have negative 31 So this is going to be our first. Uh, I am Vector here. This negative 31 It&#x27;s actually let&#x27;s come up here and just replace this. So we have the one here being negative 31 and Lambda associated with that is supposed to be one. So now we repeat the same thing. But with Lambda to actually just go ahead and erase this, I&#x27;ll put this one in blue. So this should be negative one here and then we just need to figure out what I can. Vector is associated with that. So let&#x27;s come down and do. Lambda is able to negative one. So that means a um so it won&#x27;t be minus. Or actually, I guess it still is minus minus one I, And over here that gives us it looks like 33 negative one negative one, which would reduce down to just 1100 So that tells us x one plus x two is zero or X one is equal to negative x two. So then our B two This is gonna be x one x two. So this would be negative, X two, because explains even x two x two. So then we get next to negative 11 and so that we could come up here and plug that in for V two. Now what was it again? Negative. 11 Right. And at this point to follow what c one c two are. We can use the fact that we have this initial value here three to. So let me pick this up, scoot it down a little bit, and then we&#x27;ll use the fact that we know when t is equal to zero. So t equals zero. Then on the left side, we&#x27;re going to have 32 back to three, 32 and then this is equal to eso Eat the zero. Indeed, zero. Both of those were just gonna be one. So this is just see one negative 31 plus C two 11 And if we were to go ahead and actually distribute this, we get negative. See? Are negative three C one plus C two and then C one plus C two, but notice that this is essentially the same thing as negative. 3111 And then we have C one C two on the outside here. So we actually have a matrix that we could go ahead and revenues. Um, So I went ahead and did this earlier. So if we were to reproduce this whole thing, it would end up giving us negative 2.5, um, 4.5. And this is going to be equal to 1001 And then we have C one C two on the outside here. So this tells us that C one is able to negative 2.5 C two is going to be equal to 4.5, right, So we can come up here. I&#x27;ll just scoop this down again and we can replace. So see one, see one with negative 2.5 and C two with 4.5. Let me make this look more like negative 2.5, right. And so this is going to be our solution to start for the initial value problem. And then the next thing they want us to do, um, is determined. What is the origin? Well, at the origin. What we need to do is look at our Eigen values. So we had the Eigen values of plus or minus one. So since both of these are going to be different ones, positive ones negative. This implies we have a saddle point at the origin. Because remember, if both of them were positive, then it would be a repeller. If both of them are negative, it would be a tractor. And when we have, like this mixed going on, then it&#x27;s a saddle. Okay, so we have that. And then the next thing they want us to do is to determine the greatest attraction or repulsion. So in this case, the greatest attraction. Well, this is going to occur when we have a negative lambda. So this is Lambda Lust in zero. And in this case, we are most negative. One is just going to be negative one. So we look to say Okay, well, that matches up with the Eigen vector 11 So what we&#x27;re going to say is the greatest attraction is through the line with point. So with the Eigen vector there, so 11 and the origin. Okay. And then to get the greatest, uh, repulsion, the greatest propulsion, this will be when Lambda is strictly greater than their or largest Lambda, which in this case is going to be through the line with point. Um, so we look at our Eigen vector for our largest positive one. So this is just gonna be the Lambda one. So this is negative 31 So it&#x27;s negative 31 And the origin? Uh huh. And the last thing they tell us to do. So since we have a saddle point, they want us to, uh, graph or sketch the typical trajectories. So we&#x27;ll just use, like, negative three and 111 So let&#x27;s go ahead and use those. So let me write that over here. So we have negative 31 and this is repulsion. So this is going away from the origin, and then we have 11 which is going to be attraction. So maybe I&#x27;ll do this one in green and actually let me keep it consistent with what I had up there may be so I&#x27;ll do repulsion and red and attraction in green. Yeah, eso Over here, we&#x27;ll do 11 Let&#x27;s just go ahead and mark reefer everywhere. Okay, So or repulsion? We&#x27;re going to go negative three and one up. So it will be going out like this to start. So going away from here. And then let&#x27;s do it in the other direction as well. So it be three negative one with this, right? So we have that. And then for the attractor is going to be at 11 So then just look, something kind of like so, actually, not like that eso, since it&#x27;s a tractor is going to be going towards it. So it looks something like this. And then we also have, like, negative one negative one over here. And then again, this one is going towards it like that. And now to get the rest of the trajectories, what we can do, excuse me, is go ahead and just kind of draw some lines coming off of these, and they&#x27;ll kind of just run through everything else so you can see like Okay, well, this line would kind of be following here until we kind of get close to the origin. But then it&#x27;s gonna go like that. So it will be going towards it. And then once it hits serial, start going away. Maybe I should do these in blue toe. Make it stand out from the well black that we have for the cordon access. So it looks something kind of like this, this one just kind of like that. And then over here, same kind of idea gets really close to origin, then kind of sweeps out, and then it will just follow, Like, this path here and then Same thing over here. Kind of comes in that goes out, and you could be a little bit more accurate with this. But since this is just a sketch, I don&#x27;t think it really matters all that much. Yeah, so then something like this. Yeah. So this would be like our graph or a sketch of the trajectories.

Sriram Soundarrajan · Answer

Hello. Young workers are probably number 16 to from the surface areas off. Fear. You&#x27;re the given point, Teoh. No. Come on. One before. And okay. Minus three comma minus. Are it by foot. Here. He needs to find their distance between these two points. For example, if the door points, uh, eggs from my mom. And thanks to my bill, there stands between them is x two miners explored the whole square like the A minus climate or squat. This is everything We are not. Player. This 10 foot it Jake is extra is minus three and explore never. Why police Mind as that before and by burn this minus one before the horse quit. So it comes to me. My enough for where? Plus minus two quit. Which is because 15 plus foot are the group ready units. That&#x27;s the distance between the group Point did Kit. That&#x27;s a tough question. Thank you.

John Mcalister · Answer

So in this problem, we&#x27;re investigating this system. Ex prime equals X times one minus y And why Prime equals Well, high times X plus one on. We&#x27;re interested in finding the equilibrium of that sound in your system and determining the stability. So I&#x27;m gonna call this first equation f just for a notation sake. In the second equation, G and remember and defined the equilibrium were interested in all of the points X y, which are in the rial plane. Such that f of X y equals zero and g of X y equals zero. Right, So this is what we&#x27;re looking for and we do that is just by setting each of these equals zero. Um, So if we said it f equals zero f x y equals zero, that means that X times one minus y equals zero. Right? So that means that either X equals zero or why equals one, uh, eso Now we can do the same thing with G g of X y equals zero. That means that why Times X plus one equals zero. Which, of course, means that why equals zero or X equals negative one. Um, so these are the only ways where that we can have, um, an equilibrium point where both of these are true. So let&#x27;s say in the case that X equals zero right? In the case of X equals zero, we need genie of X y t equals zero x can equal negative one cause xrt equals zero. So why has to equal zero s? So that means that when X equals zero, then y equals zero and the 00.0 is a fixed point on. And that&#x27;s the only option when X equals zero when x is not equal zero that moans, Why must equal one. But in one wife was one. Uh, Jean of X Y now has two still equals zero And why I can&#x27;t be zero so access to equal negative one. So when y equals one, X must equal negative one and we get the point negative one one as our other equilibrium. Right. And this eyes are universities covers all of your options for finding these equilibrium. So, um, we know we&#x27;ve exhausted over options. These are only two equilibrium. We just have the 0.0 the origin and negative 11 So we&#x27;re gonna explore the stability of these by examining the Jacoby in now. And remember that the Jakub Ian is it metrics in terms of X and y, it&#x27;s just the matrix of partial derivatives. So who took the, uh, partial derivative with respect to X of our first equation? Uh, x y the person derivative with respect to why of our second question, right. The, uh, partial derivative with respect to X of our second equation exit what, in the partial derivative with respect to why of our second equation had exit line. And, uh, we do eso that&#x27;s in general with the Jacobean. Looks like with this example, we take a look at F The extra motive is one minus y The UAE derivative is negative X uh, for our equation G. The extra motive is why and for equation. What g The y derivative is X plus one. So this is our General Jacoby in for our system. And now we&#x27;re gonna look at our Jacoby in at each are equilibrium. Eso It&#x27;s examine our origin first, the 0.0 And, uh, how we&#x27;re gonna find stability is by looking at the Eigen values of the Jacoby in so to find the item values. We&#x27;re gonna take the determinant of the Jacoby in at 00 in, subtract the identity matrix times Lambda on Lambda will be our Eigen value. We&#x27;re going to set the sea below zero and sell for land. So, uh, this is going to equal the determinant of the matrix. Uh, you come in at 00 We put that in, we get one zero zero one on, then we&#x27;re gonna subtract lambda from the diagonals. We get the determinant of one minus. Landau, 001 minute. Slander on this is a two by two matrix. So taking the determinant is rather easy. We&#x27;re just gonna take the difference of the products of the that Wagnalls. And in this case that just comes out to be one minute slammed at times, one minutes land or one minus y squared. And remember, we&#x27;re sitting that equal to zero eso this because really quite easy toe to Saul. You get that? We have one minus. Landau, I wish is equal to zero. And, um, from this week, we know that Lambda must be equal toe one. In fact, both Landau one and land to to this this is a road to a multiplicity, too. So we have Linda one equal one and land to equal to one. So this, uh, in a number, since this, this has some behavior which is hard to capture just by looking at it and solving it is not a zoo. Nice as we would think of like a, uh a saddle point or ah, stable spiral or something like this. The behavior is a little hard to describe, but we see that both of the the Eigen values are really and they&#x27;re both positive, which means that we&#x27;re going to have an unstable note of sorts. It&#x27;s gonna look a little bit different because both of our values are the same. But we will happen. Unstable nude eso At 00 we have an unstable node. Cool. So now we can get a something with our next point, which was negative 11 So we&#x27;re gonna take the determinant of the Jacoby in at negative 11 minus the identity matrix times Lambda, and that is going to be equal to plug in negative 11 to the Jacoby in well, one must. Why? Which is 10 right? Negative X is negative. Negative one. So positive one and then on the bottom. Here. Why? Which is one and X plus one negative one plus one is zero. So we have this. And remember, we are subtracting lambda from the diagonal. So we&#x27;re gonna have that on. And to take the determinant of that right, it&#x27;s going to be the difference between the products of, uh, the diagonal. So we take my slammed attention when it slammed it to get learned, the square minus Wynton&#x27;s run is one. Then we&#x27;re gonna set that equal to zero. So once we have that, we see that Lambda squared is equal to one, which means the Lambda equals plus or minus one. Um, and this this tells us that we need to know about our, um are missed ability of this fixed point. Right, Because we have to hide in values which are both real, right? Number one and Linda to are both riel, but we see that Lambiel one. Well, I&#x27;ll say that when the one equals positive. One landed, two equals negative one muskets. Linda. One is greater than zero, but lame to choose less than zero. So we have an unstable and a stable direction. That means we&#x27;re gonna be talking about a saddle point. So the point negative 11 is a sad oh, equilibrium. And that completes the problem. We first identified both of the, um, fix phones by solving the system where both ex prime in White Chronicle does party go to zero. And then we explored the stability of each other&#x27;s by finding the Agon values, uh, each of the Jacoby ins at those points, unless we found that 00 is an unstable node and negative 11 is a saddle.

Problem

Let $A=\left[\begin{array}{ccc}{.4} & {0} & {.2} …

Problem 14 Easy Difficulty

Classify the origin as an attractor, repeller, or saddle point of the dynamical system $\mathbf{x}_{k+1}=A \mathbf{x}_{k} .$ Find the directions of greatest attraction and/or repulsion.
$A=\left[\begin{array}{cc}{1.7} & {.6} \\ {-.4} & {.7}\end{array}\right]$

Answer

No direction of greatest attraction because there is no eigenvalue that is less than 1 in magnitude.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Caleb E.

Kristen K.

Joseph L.

Vectors Intro

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No direction of greatest attraction because there is no eigenvalue that is less than 1 in magnitude.

Linear Algebra and Its Applications

Lily A.

Caleb E.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Caleb E.

Kristen K.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications