a-common-misconception-is-that-if-a-has-a-strictly-dominant-eigenvalue-then-for-any-sufficiently-lar

Question

A common misconception is that if $A$ has a strictly dominant eigenvalue, then, for any sufficiently large value of $k,$ the vector $A^{k} \mathbf{x}$ is approximately equal to an eigenvector of $A .$ For the three matrices below, study what happens to $A^{k} \mathbf{x}$ when $\mathbf{x}=(.5, .5),$ and try to draw general conclusions (for a $2 	imes 2$ matrix).
a. $\quad A=\left[\begin{array}{ll}{.8} & {0} \ {0} & {.2}\end{array}ight]$
b. $A=\left[\begin{array}{ll}{1} & {0} \ {0} & {.8}\end{array}ight]$
c. $A=\left[\begin{array}{ll}{8} & {0} \ {0} & {2}\end{array}ight]$

Chris Trentman · Accepted 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.

Nicholas Barvinok · Answer

we have the matrix three minus 1 to 0, and we want to see if the vector one, too is an icon vector. Well, to check that, we can just apply the majors to it. We get three minus two is one and two times one plus zero times two is two. So this isn&#x27;t Aiken Vector with Aiken Value one. Uh, the next one. We have the matrix minus three, minus 15 minus 212 minus two minus one for 211 Now this gets sent to minus six. Minus one is seven plus five minus two. The second entry is minus four plus one is minus. Three plus two is minus one and finally minus four. Minus one is minus five plus four minus one. So this isn&#x27;t Aiken Vector with Aiken value minus one. All right. Next one is one minus two minus 10 Applied to 10 This gets us one minus two. So this is not Nike in Vector, because if it waas the first entry Well, the first century goes toe one. So it was an icon. Vector would have Eigen value one, because that&#x27;s what&#x27;s gonna multiplied by. But zero times One is not negative juice. This is not a night in vector Next, One 12 minus 21 two to I This gets sent to two plus for I and this gets sent to minus for plus two I. So let&#x27;s see, in the complex plane, we start with the victor two too. I so right here. And this gets sent to So this is on. So the diagonal y equals X and this gets sent to also that doesn&#x27;t make sense, because this is ah right. So they the director too gets sent to two plus for I. So from here, thio here and then to I gets sent to minus for plus two I. So as we can see, they&#x27;re being rotated and stretched in the same way. So this is in fact, an icon vector. Let&#x27;s try to figure out its wagon value. So two we multiply by one plus two eye to get the two plus four I And then if we take too I and multiply by one plus to I, then we get to I minus four. So it&#x27;s an I convey vector. With this, I can value the next matrix is two plus three a minus two minus two A three plus three a minus three minus two. A. Applied to 23 Now this equals four plus six a minus six minus six A and six plus six. A minus nine minus six A, which simplifies to minus two minus three. So this isn&#x27;t Aiken vector again with iving value minus one. And the last one. You&#x27;re the Matrix minus nine for six minus 634 minus 946 applied. And we want check of the vector 203 is Anakin Vector. Well, this equals C minus 18 plus 18 is zero. Then we have minus 12 plus 12 0 and third rose the same as the first row. So this has an icon value. 00 And it is an icon vector.

Nicholas Barvinok · Answer

we want to find an Ivan Vector of the Matrix 9023 with wagon value. Nine. Well, let&#x27;s say our factories X y. On one hand, this equals nine x and two x plus three. Why, on the other hand, because the night in Vector with Ivan value nine. This equals nine x and nine. Why so nine X equals nine. That&#x27;s no matter what X is. And we have two x plus three. Why equals nine. Why? Which we can simplify to two. X equals six. Why and X equals three. Why or better yet, why equals X over three. So when I can vectors anything is any vector which has any value of X and 1/3 that value of Why So, for example, 31 would be such a nag factor. No. For the next Matrix, we do the same thing we have 1257 We wanted Ivan Vector that has an icon value of four plus square root of 19. So we have X. Why on one hand this equals X plus five. Why and two X plus seven. Why? And on the other hand, this equals work whispered of 19 x times for four plus squared of 19. Why so we have two equations. X plus five. Why equals four plus swear it of 19 x and X two x plus seven Why equals four plus squared of 19? Why? So what can we do? What would be the easiest way to solve this? Well, let&#x27;s say Let&#x27;s solve this first equation for Let&#x27;s actually solve the second equation for X. That seems to be the simplest thing. So we have X equals four plus squared of 19 minus seven times. Why divided by two, which simplifies to squared of 19 minus three. Why over two? And plug that back into here. So we have, I swear it of 19 minus three. Why all of that over to Plus five. Why equals four plus squared of 19 and then multiplied by squared of 19 minus three all over, too. Why so no, we can isolate why we can factor out why and actually get why multiplied by squared of 19 minus three over two minus five minus four plus squared of 19 squared of 19 minus three all of that over too. And in fact, this equal zero, which means that Why? Well, let&#x27;s see. Does this equal zero? If this factor is not zero, then why must be zero? So let&#x27;s take a look at this factor. It&#x27;s squared of 19 minus three and I want to write everything over to someone to make the minus five minus 10 and then minus for swear it of 19 plus 19 minus. I&#x27;m sorry. That should be a minus 19 and then plus three squared of 19 and finally plus 12. So all of this over to now this numerator simplifies to wave squared of 19 minus four, squared of 19 plus three squared of 19. So those actually canceled and we have minus three minus 10 is minus 13 plus 12 is minus one minus 19 is minus 20. Divided by two is actually minus 10. So we do. In fact, get that Why equal zero? And if y equals zero, then we have that to X equals zero. And, uh, from this equation, and if two X equals here, then X equals zero, and the vector 00 is not considered an Eigen vector. So we don&#x27;t have an I in vector for this value. Next, we have 103200 0 to 1, with bye in Vector, which has Aiken Value. Three. Let&#x27;s do X y zem. It gets us X plus three z two x and two X plus z should equal three x three Why three z We have X plus three z equals three x two x equals three Why and two x plus z equals three z So from this first equation, we can simplify thio three z equals two x or X equals three halves Z um, from the second equation, we have X equals three halves. Why? So we in fact have three halves? Why equals three have Z which means that why equals e And so now in this 30 equation, we can substitute in X three halves Z for X so I get three z plus Z equals four z So we get four z Abel&#x27;s for zine. So what we get is that Z can be any value. Why has to equal to Z that X has to equal to three have busy and any Eigen vector of this form. We&#x27;ll work the next one 0312 We want Aiken value of minus one. So again. We do the same thing I get. Why three X plus two Why? Which should equal minus X minus y. So you get why equals minus x three X plus two y equals minus y so I can substitute in minus X for y. In the second equation, I get three X minus two X equals X. Yeah, so X, as we see can be This gets X equals X so acts could be any value. And why just has to be the negative Any vector of this form will satisfy will be an I in vector with wagon value minus one. Next we have 0111 times x y with Well, let&#x27;s first write it out as why X plus y. And this should equal one plus square to five over too. X one plus square to five over two. Why? So I get why equals one plus square to five over to X and X plus y equals one plus square to five over too. Why so again I can plug in wantto plus square to five over to X in tow. Why here? And get X plus one plus square to five over too X equals one plus Where 25 over two Squared X And so what do I get? I can factor out the X and get X one minus one plus square to five over too. I&#x27;m sorry. That&#x27;s a plus. Minus one plus square to five. Over too. Squared. Equal zero. So again, we need to check whether this factor is zero. If it&#x27;s not zero than X, must be zero. So let&#x27;s see, we have I&#x27;m gonna write everything. Well, let&#x27;s write everything over four The end. So we have four plus two plus two square to five, minus one minus two square to five plus five. And this simplifies to four till two. Square to five, minus two square to five. Cancels four plus two minus one is five and then I&#x27;m sorry, this should be a minus five. So, in fact, this factor is zero, which means that ex convey anything and then why just has to be one close square to five over too X. And this is our wagon Vector or anything of this form is an I in vector and our last one. 123017 0 to 1 We wanted I can value of one plus square to 14 x y z equals X plus two y plus three z Why plus seven z to why plus z equals And I&#x27;m just gonna keep his land for now. Lambda X Landa y land Izzy, where land is one plus square to 14. So we have X plus to why plus three z equals one plus square to 14 x Why plus seven z equals one plus square to 14. Why? To why? Plus c equals one plus square to 14. Izzy. So what can we do? We can solve for why, in terms of Z in this last equation and we get that, why equals I&#x27;m gonna use Lambda instead? Could shorter land of minus one Z over too now. So we know what wise in terms of Z and so we can now substitute that into the first equation. Get Lambda minus one Z over too. Plus lambda minus one Z plus three Z equals Land X. And if we expand that where if we simplify that factor out the sea, we get Z land of minus one over too, plus lambda minus one plus three and then all of that divided by Lambda equals X. So again we can pick axe to be anything. And then I&#x27;m sorry. The opposite. We can pick we pig zine. Then why is Lambda minus one over two Z and finally X is Z times lambda minus one over two, plus Lambda minus one plus three. She&#x27;s really land. Ah, plus two and all of that divided by Lambda And that&#x27;s our final Agon vector.

James Chok · Answer

So in this question, given that we know this, we basically want to find on item value off a in terms of you and so physical. Next, multiply that invest on this line So it&#x27;s a left hand multiplication. So this is equal to a minus. Alpha identity, new X expanding brackets You have i a new X last. Why Alfa Identity has a new X. He was a constant second. Just threw that out and get new a X minus Phi Well, identity times about anything is just the same thing. So it&#x27;s just how far new X So now we move a extra itself. So we have Yes, you could see one of them. You&#x27;ll and ex Ellen plus Alfa New X factory out X. We have one ever knew what&#x27;s one plus by Alfa you Thanks. Which can be written as I hope, plus one over New X so excellent. So what we&#x27;ve just done. We&#x27;re just showing. Is that a X zika too Alfa plus buy one from you. Thanks. And so therefore, this is your wagon value

Runpeng Li · Answer

Yeah. Okay, so four problem, too. Work even three I can values with the Cory&#x27;s money, I get vectors and we need to find a solution off the of the equation. Escape plus one. He was a Xscape. So the first thing is to find accidental and by definition, in our textbook, we have X zero, these people to see one. She&#x27;s a scaler times. Uh, V one C, two times you too. See? Three times feet. Okay, so four problem too work even because we only have three inspectors in this problem way only write down three terms. Okay, so it turns out to be x not X, not a Steven, which is connected to negative five and three. And we have First I get vector which it iss 10 negative. Three and three. I can values with the Cory&#x27;s money, I get vectors and we need to find a solution off the of the equation. Escape plus one. He was a Xscape. So the first thing is to find accidental and by definition, in our textbook, we have x zero, these people to see one. She&#x27;s a scaler times uh, V one C two times. You too. buzz See? Three times feet because I can. Again Victor, which is 21 negative five. Excuse me. Now that last again, Victor, she&#x27;s an active 33 negative. Three of seven. So that is equivalent to find out a solution off this, then your assistance. We on Lee have three of inspectors in this Problems way only write down three terms. Okay, so it turns out to be x not X, not a Steven, which is connected to negative five and three. And we have first I get vector which it iss 10 connective three and second 12 collective three 01 Collective three Elective 3 to 5, seven times the column vector with C one C two C three, which is equal to yeah, effective too. Elected five. I mean it back and again. Victor, which is 21 negative five. Excuse me. Now that last are you Victor? She&#x27;s an active 33 negative. Three of seven. So that is equivalent to find out a solution off this, then your sister is solving this system. I was keeping the process. So the solution is see Juan si tu c three iss to want to so that we fund. We have courage on our wishes. Either of her ridge. Excuse me for you. Check on Victor. It turns out to be why so? One, 12 collective three 01 collective three elective 3 to 5, seven times the column vector with C one C to C three, which is equal to yeah, effective too. Next, it&#x27;s five. If either or negative three US. One times 21 Act if I and ah, see three. See the reason too. Times did you again, Victor Nectar three Negative. Three at seven. Me solving this system, I was keep the process. So the solution is see Juan Si tu c three iss 212 So that we fund we have heard on our agenda wishes either off Ridge. Excuse me for each item, Victor, it turns out to be. Why? Said one so that our x k we just need to add our uh huh. Give value to dead. So we have two times. First I can value wishes. Three. So I have a three to the power of K times again. Victor 10 Negative. B plus 23 us one times to one negative five and ah, see three. See the reason too. Times to you again, Victor. Nectar three Negative three at seven. Uh, the second I get back that way, it is four. Fifth to the power of K to one active fight. And it&#x27;s me or so I have two times 3 50 to the power. Okay, times again, Victor. So that our x k we just need to add our, uh, give value to dead. So we have two times. First I can value wishes. Three. So I have a three to the power of K times. Yeah. Give Victor 10 negative. B plus negative. Three nectar three and seven. Okay, so here&#x27;s our That&#x27;s the solution off of the equation. X capers won was because I have a ex king. And so when we take K to be infinity, then the first term will be infinity. But the last two turns to terms will be zero. So our when the second I get back that way is four. Fifth to the power of K to one negative five and it&#x27;s me or so have two times 3 50 to the power. Okay, times again. Victor, cake tends to be infinity. Are x k Will just be two times 32 power Okay. Times 10 Negative three

Problem 21 Easy Difficulty

Answer

Conclusion: If the eigervalues of A are all greater than 1 in magnitude, and if $x$ is not an eigenvector, then the distance from $A^{k} x$ to the nearest eigenvector will increase as.

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

Linear Algebra and Its Applications

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Caleb E.

Michael J.

Joseph L.

Vectors Intro

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Conclusion: If the eigervalues of A are all greater than 1 in magnitude, and if $x$ is not an eigenvector, then the distance from $A^{k} x$ to the nearest eigenvector will increase as.

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Caleb E.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications