classify-the-origin-as-an-attractor-repeller-or-saddle-point-of-the-dynamical-system-mathbfx_k1a-m-3

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.
$A=\left[\begin{array}{rr}{.4} & {.5} \ {-.4} & {1.3}\end{array}ight]$

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.

John Mcalister · Answer

So in this question were given a following system. We get X prime is equal to two X plus five y squared and that Why prime they too? Why? Times three minutes for X. Now, we&#x27;re interested in finding the equilibrium of these. And so, in order to do that, the equal re of the system is going to be wear both ex prime in white primer equal to zero. So we&#x27;re gonna set both of these equal to zero. Uh, and now this becomes, ah, system of equations, and we can solve this system on defined, uh, this points exit. Why were both ex prime in UAE? Prime are zero. Um, So when we sit, why private equals zero. Think about this bottom equation. Uh, that means that I&#x27;ll reread it. Why? Times three minutes for X equals zero, and we can tell immediately that you know why Must equal zero or X equals 3/4? Great. Um so these are the only two ways that why prime can be equal to zero. So now that we have this, but we can break down the rest of our investigation into cases. Eso will say we know that ex prime must be equal to zero. So you can say two X plus five y squared equals zero. If why equals zero? Then we get two X equals zero, right? We just play Why in and from this week we find X equals zero. So when y equals zero equilibrium point is 00 If why does not equal zero then X must equal 3/4 right X equals 3/4. So, uh, we&#x27;ll say else X equals 3/4. Then we played that in and what we&#x27;re left with is three halves plus five y squared equals zero, and we may already see that there&#x27;s gonna be no riel answers for Why here, If we sell for why we get that y squared is gonna equal negative 3/10. Um, we need Teoh. Think about Riel. Why we&#x27;re not interested in imaginary solutions for why complex solutions for why we&#x27;re only interested in Really? Why So, uh, in the case that X equals 3/4. There are no equilibrium. So, uh, that means we have discussed every case, you know, every case where, uh, f of X. Well, we&#x27;re ex private. And why prime both equal zero. So this is an exhausted proof. And we found only one, uh, equilibrium. So we say that 00 is our only equilibrium. Now that we&#x27;ve found this equilibrium, we can determine its stability by a sanity that Jakub Ian. And remember the Jacoby in, uh, label these functions first, I&#x27;m gonna call Ex Prime if, if x and y And why Prime is g of X and y Now, where the Jacoby in is, uh, in terms of exit. Why going to be, uh, the extra bit of f at the point x y, the UAE derivative of F at the point x y the extroverted of genius. The point next, why? And the UAE derivative of G at the point x y. So that&#x27;s the general form for the Jacoby in In this case, Uh, look at F the extra native of F is too. The wide derivative of F is 10. Why the extra motive of G? It&#x27;s going to be negative for ninth. Excuse me? Negative four times. Why? Negative for why? And then, uh, the Y derivative of G is just three minus four x cool. So this is our, um this is our Jacoby in and Now, what we&#x27;re interested in is determining. But the Jakub Ian is at 00 in particular. When we&#x27;re thinking about the stability of 00 you wanna find the Eigen values of our Jacoby in at that point. So, um, we&#x27;re going to investigate the determinant of the Jacoby in at 00 minus the identity matrix, times, land and land that will be our Ivan values when we set this equal to zero. So our first step is to find the determinant of this, uh, matrix. The Jacoby at 00 is going to be too zero zero three. And then we subtract, um, land along the along the day. I know here. So we get to minus Amanda three, massive. And now the determinant of this is is rather easy to compute. Um, for all to buy team. Interesting. That is rather easy, but in particular, this is nice, because we get just two minus claimed a times three minutes, Linda. And then we set that equal to zero on. When we do that, we see immediately that Lambda equals two and Lambda equals three. So we&#x27;ll call this one the one limb to 10 respectively. It&#x27;s a land of one equals to lend a two equals three. Uh, and from this, we can determine the stability of our, uh, six point right are equal. Aram 00 has to Eigen values which are positive. Right? Lambda one is greater than zero William to to rid and zero a civilised means, um, more our 0.0 when our system is gonna be approximated well, by just the limiter system, all solutions, they&#x27;re going toe move away from 00 So that means we&#x27;re gonna have a unstable note movement to know, because both of these are real numbers. Um, so we haven&#x27;t constable note at 00 and we know that because the idea values are both real and positive.

John Mcalister · Answer

in this problem were given the system. Ex prime is equal to four X minus. Why? Minus why sign of X and, uh, why prime is equal to X plus two y and that we are interested in finding the equilibrium and determining the stability of the equilibrium in this system. So, um, for us, just for a notation sick, I&#x27;m going to label these as f and G respectively. And now recall that an equilibrium of the system will have ex prom. And why prime equal to zero. So that that is to say that our equilibrium will be the set of all points X y in our two such that f of X y equals zero and g of x y equals zero, right? Eso to find these were just going to you have to solve this system that we get when we set both of these expressions equals zero. So our system is four x minus. Why, minus why sign of X equals zero and X plus two y equals zero. Um, now, there isn&#x27;t an easy way to do this by addition subtraction, right? We can&#x27;t run reduction because we have this, uh, sign of extra here, but I can easily solve for X in my second equation, right? Exceeded. Clearly X is equal to negative two. Why, uh, now I could just make a substitution in here. So when I make that substitution, when I get is that, um 84 times. Negative to why? Which is negative? Eight. Why? Minus y minus? Why? Times sign of negative two y is going to equal zero. And now I&#x27;m going. Teoh, simplify this. I said this was negative. Eight y minus wise. Negative. Nine y minus. Why? Sign of negative two y equals zero. And I can further factor this. I get pull. Why? Out Get negative nine minus sign of negative two y equals zero on This is his factors. That factor does I can get this eso and but it tells us all the information we need, right? The only way for this equals zero is if why equals zero, right? It would be if y equals zero or, uh, negative. Nine minus sign. Okay, negative. Two y equals zero. But if we think about this expression, let me write it out fully. Excuse me. This would have to equal zero, but if we think about that. Uh, that means that negative nine would have to equal sign of negative to what? And we know that the range of signs between negative one and one. So, uh, certainly there are no such Why that would make this expression true. So, uh, weaken, We know that that cannot happen. So that means the only way for this system to be true. The only solution satisfies the system is why equals zero. And now, when y equals zero, come back here. Remember both f of x f of x y energy mix. So I have to of the equals zero. So when Michael zero weaken, just a plug that into Seiji of X y will call that G of X zero that&#x27;s going to equal X plus two times zero, which is just two. And a rubber that has two people zero. So that that tells us that X equals zero. So we like. Well, zero x must equal zero. This is the only ordered pair. Uh, that satisfies, um, this expression. This is when we found that by setting up the system of equations, finding a solution for why we only have one solution in this case, and because of that, uh, we can&#x27;t immediately find you the the solution for X. So we know that our equilibrium is that 00 and we can now use the Jacoby in of our system to him determine its stability. And I recall that the Jacoby in is a matrix in terms of X and Y, And it is just the, uh, matrix of partial derivatives. So what I meant by that is it&#x27;s going to be the first equation with respect to the first variable. Taking the derivative with respect of the first variable that X and Y on the first equation derived with respect to the second variable at X Y and then in the second row will be the second expression second equation, uh, with a derivative with respect to the first variable. And the second equation with the derivative with respect to the second variable is in the bottom right here. And so this is our Jacoby in eso For this expression in particular, this is going to look like eso. If we take the derivative with respect to X of our first equation, we get four in our first term, right? Uh, zero in our second term, and we take the derivative with respect to X of this third term, we get Cho sent of X, and we keep that negative. Why? They&#x27;re so we get negative y co sign next four minus y co se necks. Now, when you take the derivative with respect, why our first term? Because the zeros this second term is negative one. In this third term, we treat sign of execute, uh, come efficient. So we have negative one, um, minus sign of X. Now, for the extroverted of Jeannie, this was a little bit simpler. It&#x27;s clearly just one. And the wider riveted is just gonna be too. So this is our Jacoby in, uh, for our system in terms of X. And y, uh, we&#x27;re interested in now is the Jacoby in at our our equilibrium point on and in particular, we want to know about the, uh, Eigen values of that Jacoby in. So how we&#x27;re gonna find that out is we&#x27;re gonna take the determinant of the Jacoby in at 00 So the determinant of Kobe in it 00 minus the identity matrix times Lambda Lambda is going to be our Eigen value when we set this equal to zero eso This is going equal. Think about the Jacoby in at 00 we plug in zero. For For all of these, we see Fuller minus zero times the coast in a zero. So just four. Ah, here we get negative one minus the sign of zero Santa zero recall is just zero. So this is negative One on the bottom row. We have one into, and I don&#x27;t forget that we&#x27;re subtracting lambda along the diagonal. So this is our, um our metrics that working that the determinative and with just a two by two matrix taking the determinant is quite easy. It&#x27;s just getting the difference of the products of the diagonal. So we take four minus linda times to my slammed up and then subtract the product of this tact. So one times negative one is negative one, um, so minus negative one is plus one. So this is our characteristic polynomial. And we can use this if we expand this out instead of equal to zero. When you do that, we get we have two squared, um, minus six Lander waas eight, my plus one and that&#x27;s gonna equals zero. So I can combine like terms and make this just plus nine. And now I see have, actually perfect square train. Oh, meal. Right. This is, uh, just wind minus three squared. That&#x27;s gonna equal zero. That means that Lambda minus three equals zero. So loomed, uh, equals three end with a multiplicity of two. So we can say that Lambda one equals three and Lambda two equals three. Right? We expect to Eigen values from a two by two matrix on, and we&#x27;ve shown that both of the Eigen values are equal to three. Um, So because we have multiplicity, our social is gonna look a little bit different than, uh, we would typically expect when we can still use argon. Becky, I didn&#x27;t values to determine what the determine the stability of the, um, fixed point. Right. We have the fixed 0.0.0 Um, both of our values are positive and real. Right? So, um, the real part of lambda is greater than zero. Um, so we know it&#x27;s unstable, and we know that neither these have are complex eso because none of them are complex. We expected this to be a node um, because the rial and because it&#x27;s positive, we know it&#x27;s gonna be unstable. So this is unstable node. So now that, uh, we&#x27;ve seen this thing is actually all that we can do without proceeding to sketch the face plane for, um, for this system. So we will visit that later on, but for right now, we need to do is find the find equilibrium. We did that by solving this system equations, just like normal. This is just algebra here. And then once we have that figured out, we can use the Jacoby in to determine the stability by looking at the idea of values for the Jacoby in at this fixed.

Problem

Classify the origin as an attractor, repeller, or…

Problem 11 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.
$A=\left[\begin{array}{rr}{.4} & {.5} \\ {-.4} & {1.3}\end{array}\right]$

Answer

There is no eigenvalue that is greater than 1 in magnitude, so no direction of greatest repulsion.

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Samuel H.

Michael J.

Vectors Intro

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There is no eigenvalue that is greater than 1 in magnitude, so no direction of greatest repulsion.

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Heather Z.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications