in-exercises-1-4-the-matrix-a-is-followed-by-a-sequence-leftmathbfx_kright-produced-by-the-power-met

Question

In Exercises $1-4,$ the matrix $A$ is followed by a sequence $\left\{\mathbf{x}_{k}ight\}$ produced by the power method. Use these data to estimate the largest eigenvalue of $A,$ and give a corresponding eigenvector.
$$
\begin{array}{l}{A=\left[\begin{array}{ll}{4} & {3} \ {1} & {2}\end{array}ight]} \ {\left[\begin{array}{l}{1} \ {0}\end{array}ight],\left[\begin{array}{l}{1} \ {.25}\end{array}ight],\left[\begin{array}{c}{1} \ {.3158}\end{array}ight],\left[\begin{array}{c}{1} \ {.3298}\end{array}ight],\left[\begin{array}{c}{1} \ {.3326}\end{array}ight]}\end{array}
$$

James Chok · Accepted Answer

Okay, So in this question, you want to approximate our Ivan values and Ivan vectors off this a metrix right here. And by the way, So this is a This is really important. So if you want to use it, just use plastic. So this is a matrix right here on the day of that is given to us is zero is equal to 11 x zero x one is 10.736 and so on. So on, so on. And so what this step does here is we multiply a and zero x want eggs, Irving this, then you do X one x to x ray and sometimes awe. Then what this stuff does it is that it essentially returns the largest valley in here in absolute terms. Yeah, and so would print out the results. So first thing that would print out is the multiplication together, and we&#x27;ll print out in your second. So just be careful when I wanna print out that I printed out the transpose not be better. So you&#x27;ll get a one by two vector instead of two by two. Victor. The reason why because just by the sanitation is a lot easier to read, so what we&#x27;ll see is x zero is negative. 1.9 and a times by x zero is negative 1.9 and negative 1.4 large entreating next, 1.9 and so on. So on, so on and finally x for a times x four is negative. Zero point negative 0.0.4012 on negative 0.3008 with life&#x27;s entry being negative. 04012

James Chok · Answer

Okay, So in this question, we want to basically estimate the lives item value and Becca off this aim Agents. Right here. So what I have right here is a heightened Smoot where we basically do the calculations. So a is 1.818 minus 0.8 minus trip 12 and 412 and then these out out X dollars. Right here. So what we do in this step is we perform a and multiply it by X Y. So for x zero, we have 01 and x one. We have minus 0.5 and one l can t to do that. And then for the same, I will also print out the maximum value in the entry so the results can be seen here, where a eight times zero is split up. This vector and the large sentry being forward to sending for this as any for base. Yes, as a night. You had to be careful when interpreting this because I took the transfers here. So this is a one robot to calm. So one by two matrix. But you should actually be a two by one. But I just put the transfer there because it&#x27;s easy to be so. What you see is that mule zero is for one to new 16 Well, two is 5.1 and so on. So on, so on. And so I. Invicta is Expo, which is negative. 0.2. Factor zero and one, or a X four is equal to minus 1.2536 and 50064 with Reagan Value 5.64

James Chok · Answer

Okay, So in this question, we want to basically approximate that again vectors. And I&#x27;m also Ivan values questions era and life spied in about. So in this example right here, I&#x27;ve only given the I&#x27;ve been very close to zero closest to zero. If you want the Ivan value life value, you can just simply said athletic 20. So just change this line of code right here, and you will gets the lives talking about, which should be 19.18 So what I&#x27;ve done here is basically I&#x27;ve cut it up the invest power method to solve this to to find the ivy value. So in this case, just to go through it in case you don&#x27;t understand what I did or how it works. So we start off without initial X and we we set an Alfa and perform this step. So first of that we do is we have to solve a minus Alfa identity. Why is equal to X? So we want to sell for Why so moving a minus? I after invest, I got to decide you have Why Easy to a minus off a after I Inglis X. So what? The sepia is simply this stuff. You&#x27;re eight minus alpha times. Why? Taken in vessel exit? Inglis and we have multiplied by X. Then we want to find about the entreating. Why? Which absolute value. The absolute value is as light as possible. So then we compute new alcohol, plus one of them you&#x27;ll and then we compute x, which is why divided five new. So that&#x27;s what this step, that&#x27;s what this whole thing does here. So the printed out results is I&#x27;m here as a concerned then the wagon value close to zero is 0.122

Gennady Notowidigdo · Answer

Okay, so we are given This are four by four matrix, which I will hard light in green. So what, we&#x27;re gonna do it be going to use a computer package? So I used Mabel. If you want to use Matt, Matt or Mathematica, I feel free to do so or something else off that, like again, feel free to use whatever computer package you desire. So once I have this four by formation, so I want to find the Eiken values and the corresponding Aiken vectors, or sort of a basis for the Eiken space, so to speak. All right, so two. So, once I used the computer package, so let me just sort of figure out what this character is. It polynomial is. So if I obtained the characteristic polynomial, it&#x27;s going to be our to the four miles to are cute, minus 311 are square plus 312 aw, plus, 24,000 336. And I&#x27;m sure that this fact is nicely, um, so factors into, uh, plus 12 times our minds 13. A love that square good. And when we solved the zeros off the characteristic polynomial which is in this case minus 12 and 13 are then what we called the Aiken Valleys. So for this particular matrix, we have to Aiken values minus 12 and 13 case are so that our oppressed it yet. Okay, So for the Eiken battle, it&#x27;s for the Eiken value. Say Lambda equals two. I&#x27;m with these arms. That lab, just to make it easier, are equal to 13. Then if we compute, uh, our time, the identity matrix minus the matrix that we have, let&#x27;s call it a Okay, so let&#x27;s call it a Are I minus A. This ends up being for 4 to 4. 56 minus 19 28 minus 44. And then we just raised this just to make make it need us to be minus 19. Um, just dish it black. It&#x27;s minus 19. Then you got 14. 14 seven on 14. That should be 14 uh, minus 42. 33 Mona&#x27;s 21 and 58. Okay, now the Eiken space for our the Eigen value, our eagles 13 is pretty much just the null space off our times. I minus a or in other words, the Colonel. Okay, if you want you can. You can interject two things. Although I reserve the word colonel for lenient maps and no spaces for matrices. Okay, Now this matrix reduces too to to want. So I&#x27;ve done it in its simplest form. So it ends up reducing to something like this. Okay, so at this point, you can use back substitution and you&#x27;ll end up for the basis for the null space ends up being It will be minus 10 to 0 and one minus 40 and three. So obviously, if you did back substitution, you would set these two quantities to be one, and you work backwards, but I&#x27;ve just sort of normalized. It&#x27;s another normal, but I&#x27;ve made it fraction free just for our simplicity purposes. Okay, so we&#x27;ve got a basis for the eye Gance base for articles 13 and similarly, I will do in the next page, sir. I&#x27;ll keep it in black financing for the Eigen value. Let our used ob for I mean, it doesn&#x27;t really matter. I just like using are because we&#x27;re sticking to English letters rather than moving to Greek letters. But in any case, so for the Eiken value are equal to minus 12. Are I modest, eh? You know, when you write out what its people to sort of figure that out on your own. But it reduces to Mona&#x27;s 21 for 24 Sara Wan. Modest one of one. And in two bars, off zeros on a basis for the null space. So for the Eigen space for article 2 12 minus 12. Sorry is 2770 and zero. Modest one, sirrah. One. Okay, so I&#x27;ll leave it there. Uh, please verify older cat places that I&#x27;ve made Adlai Vilo on the computer or by hand, But in any case, thank you very much.

Chris Trentman · Answer

were given a matrix, a vector V and Eigen value of day and we were asked to show That&#x27;s five is an Italian value of A and V is an associate Eigen vector. And then we were asked to or fog Nelly Diagne eyes a check the see if five is an I in value of a we know that and Eigen value satisfies the characteristic equation of a that is you want to check to see what the determinant a minus five I is. This gives us the determines of the Matrix. Same is okay, except for now, the diagonals or negative 14 So negative one negative one negative one negative one negative one negative one negative one negative one negative one and you can see this determined. It is clearly going to be zero since we have duplicate rose and therefore it follows that lambda is an Eigen violet value of a check to see if V is an Eigen vector in this wagon space. What to see if these satisfies the equation? A minus five i times V equals zero or alternatively, we could compute eight times V to see if it&#x27;s equal to five. I times V yeah, that a Times V is equal to for a negative one. Negative one negative 14 Negative one. What we have. We have four times. One is four minus one minus ones of two negative one times one is negative, 14 times one is four negative times. One is negative, one again to and finally negative one times one is negative, one negative one times one is negative, 14 times when it&#x27;s four. So again, too, we have that. This is equal to two times I times. Well, still doing that. We calculate a minus five i. V. We&#x27;ll get something a little bit different since equation implies that a minus five I&#x27;d be is equal to this is matrix that negative ones times V. So we have negative one negative one negative one so negative three all the way down. So it&#x27;s clear that V is not an Eigen vector actually associate with this Eigen value. But we have shown that a V is equal to 222 which is equal to two times V, and so B is an Eigen vector of a associate ID with and I didn&#x27;t value Lambda equals two. Now I noticed that transpose of a is equal to the Agnos stays the same four for four in exchange negative with negative one negative one with negative one negative with negative one. We recognize this is the same as a So we see that A is a symmetric matrix Since a transpose equals a and therefore follows that a is orthe orginally the ag analyze herbal tour following Lee diagonal eyes a You to find a complete set of more than normal lighting victors. We have one Eigen vector. He is Thea connector associate with the item value landed too. We don&#x27;t know. However, the multiplicity of fighting value landed too, which we could calculate. You can also calculate the multiplicity of backing value Linda five and use that to find multiplicity of Viking value. Lambert to Let&#x27;s do that. We have it a minus saying Linda one is equal to five Eigen value must satisfy a minus five by V one equals zero. So we have that this coefficient matrix negative one negative One negative One negative. One negative. One negative One negative one negative with a negative one Reduces to the matrix 111 000000 So we have two zeros which tells us that despite in space has mentioned to and we obtain the system of equations V 11 plus V 12 plus view 13 equals zero. And so we have that view 13 is equal to negative. The 11 minus B 12 And if we take V 11 to be the perimeter s and V 12 to be the parameter t in the general form for the I connector V one is yes t negative s minus t which can be rewritten as the linear combination s times 10 negative one plus tee times 01 negative one. So we obtain the Eigen vector V one if we take has to be one in TV zero She&#x27;s 10 negative one and we obtained Eigen Vector B two if s zero and t is were one which is the vector 01 negative one. By construction we have that V one and V two or when nearly independent and Stan the Agon space. So it follows that you want to be too are a basis for this Agon space. The basis of Eigen vectors and we have the norm of V. One is equal to the square root of one plus one, which is to route to Norma. V. Two is the square root of one plus one, which is rare too. And so the unit vector you one is the vector V one, divided by the enormous V one, which is equal to one of a route to zero negative one word to and unit vector. You too is the vector V two, divided by three normal V two, which is the Vector zero. Whatever route to negative one over route to. And so because these are unit vectors and they&#x27;re still Eigen vectors we have that you wanted you to is in before I do this to check something. So I went ahead and I calculated thes unit vectors not checking something. I need to check if they were orthogonal who really wants You know what the normal set from this Eigen space. So we have that V one died with me too is equal to zero plus zero plus 1 90 quarter zero. So it follows that the one is not to put a dog in all to be to and so we&#x27;re going to have to use Gram Schmidt or thought generalization procedure to obtain a second orthogonal member of the basis. So we&#x27;ll take you want to be director V one? Let you too be the vector V two minus the projection of the two which is on the view one which is v two dot v one over v one dot you one sometimes the vector V one. This is equal to you Have defector V two is 01 negative one minus and then v two dot over the V one we determined was one and then v one diver to be one Mrs to times fy one is 10 negative one. This gives us zero minus 1/2 is negative on half one minus zero is one and negative one plus one have is negative 1/2 instead of you here She probably had made these w one and w two And now my Gram Schmidt we have that These vectors w one and w two not only are still Eigen vectors because that&#x27;s a linear combination of my conductors, but are and orthogonal said we have the norm of w one is going to be the normal V one, which is the square root of one plus one which is written to And we have that the norm of to and likewise we have that to w two is also and orthogonal specter to w one normal to w two. This is the norm of negative one to negative one which is the square root of one plus four plus one which is Route six. And so we have the unit vector you one which is w one over. The norm of W one is equal to 10 Negative one divine favor too. We&#x27;re one of her 20 negative. Whatever route to and the unit vector, you too is the doctor to w two divided by the norm of defector to W two. This is equal to you know you two is the vector negative 11 negative one or negative one to negative one. So negative one of a route six two of a route six negative 1/6 evidence and now we have that you want you to is a north a normal basis for this wagon space. Returning to previous results showed that v was in Eigen Vector. So she with the Eigen value, landed two, which could be the only other Eigen value. And it must have a multiplicity Get one since a is a three by three matrix. So we have the already The itself is a basis of Eigen vectors in the Eid in space. And so the normal V is equal to square root of one plus one plus one, which is three. And we have the unit vector you three, which is be over. The normal V is the Vector 1/3, 1/3, 1/3. And since a is symmetric, it follows that you three is orthogonal to both you wanting you to and therefore we have that the set you one you too. You three is a complete Ortho normal set of I Connectors of the Matrix A And so if we take P to be the Matrix with Colin vectors u one you too And you three this is the matrix one of around 20 negative one or two. You too is negative whenever Route six to a very six native wonder over route six and then you three is one of a route three 1/3, 1/3 and we&#x27;ll take matrix D to be the diagram Matrixes entries of the Eigen values. So this is going to be five five to the rest of the entries are zeros and using these to maitresse ease you can are thought Nelly diagonal eyes.

Problem

In Exercises $1-4,$ the matrix $A$ is followed by…

Problem 1 Easy Difficulty

Answer

Eigenvector $x_{4}=\left[\begin{array}{c}{1} \\ {3326}\end{array}\right]$ or $A x_{4}\left[\begin{array}{c}{4.9978} \\ {1.6652}\end{array}\right]$
Eigenvalue $\lambda \approx 4.9978$

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

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Eigenvector $x_{4}=\left[\begin{array}{c}{1} \\ {3326}\end{array}\right]$ or $A x_{4}\left[\begin{array}{c}{4.9978} \\ {1.6652}\end{array}\right]$Eigenvalue $\lambda \approx 4.9978$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Caleb E.

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.

Caleb E.

Michael J.

Eigenvector $x_{4}=\left[\begin{array}{c}{1} \\ {3326}\end{array}\right]$ or $A x_{4}\left[\begin{array}{c}{4.9978} \\ {1.6652}\end{array}\right]$
Eigenvalue $\lambda \approx 4.9978$

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

Linear Algebra and Its Applications