a-widely-used-method-for-estimating-eigenvalues-of-a-general-matrix-a-is-the-q-r-algorithm-under-sui

Question

A widely used method for estimating eigenvalues of a general matrix $A$ is the $Q R$ algorithm. Under suitable conditions, this algorithm produces a sequence of matrices, all similar to $A$ , that become almost upper triangular, with diagonal entries that approach the eigenvalues of $A$ . The main idea is to factor $A$ (or another matrix similar to $A$ ) in the form $A=Q_{1} R_{1},$ where $Q_{1}^{T}=Q_{1}^{-1}$
and $R_{1}$ is upper triangular. The factors are interchanged to form $A_{1}=R_{1} Q_{1},$ which is again factored as $A_{1}=Q_{2} R_{2} ;$ then to form $A_{2}=R_{2} Q_{2}$ and so on. The similarity of $A, A_{1}, \ldots$ follows from the more general result in Exercise $23 .$
Show that if $A=Q R$ with $Q$ invertible, then $A$ is similar to $A_{1}=R Q .$

Sam Sohn · Accepted Answer

all right, so we&#x27;re given that A is equal Thio Q R and notes that we can say this is also equal to Q R. I, since multiplied by the identity matrix doesn&#x27;t really do anything. And we can also express the I as que que in verse because we&#x27;re given that cute is in vertebral and by the associative property we can groups. Demetris is like this with our Q in the middle. So we&#x27;re giving que our Q que members, and we are also given that our Q is equal to a one. So we&#x27;re gonna do that. So basically, we get that A is equal to Q A one Q in verse. And that&#x27;s basically the whole proof, because this is exactly what the definition of similarity is. So, since a is equal, Thio Q A. One cute in verse by the definition of similarity, A is similar to a one, and that&#x27;s it.

R M · Answer

problem. Number 44 we get London is equal to zero and zero Multiplicity is equal to two. Uh, London two is equal to see on London. Three is equal to negative two. From here, we can get the basis is equal to Ah dear. One zero. And they get one and 10 negative one and zero. And from here the pieces for crying in a space using London toe It is equal to 111 and one. And from here the faces off e three where I&#x27;m gonna spray three is equal to one negative one one and negative ending it.

R M · Answer

problem Number 32. Ah, we say that Let that I know. And the characteristic my normal binomial Ah be Lunda Ah is equal to the determinant off a minus London time The unity, um using fear him Ah 4.7 point five and four foreign 7.6 we have be Lunda is equal to along the one minus tanda but the blind by London to minus nanda until we reach Ah Lunda and buying us number which is equal to Lunda, who&#x27;s quit miners Lunda one plus London to times Lunda plus Lunda one none. The tool. Lots of lying by down the three miners Lunda, there were each Lund the end minus London uh which is equal to negative Lunda three plus London, where Mother Blowing by London one last round The room bluff on the three minus ah under one learned a tune plus along the one along the three plus London toe under three. Not a blowing boy Lunda plus Lunda. One London toe learned the three a bunch of lying by until we reach Lunda and minus number, which is equal to I think it to one brother and along the border and plus negative one O r and minus one. Lunda in minus one. Number one bus London through plus through me. Rich from the end. Plus already Lunda one London through run. The three still run the end. Yeah, boy it waiting The right hand sign of the equation. Ah, and the right side of the equation. Still the right hand side. Um, the equation. All right, 0.7 point five. We would have negative one for And London or an plus. Do you want longa for? In minus one plus p and is equal to ruin. Negative one for Lunda on the poor. And plus, you get the one or in minus numbers. This is the right hand side. London one bus, Lunda. And for Lunda in minus one plus bus run the one number two. Nothing will change. Lunda. Monday the same the same wry pants high that this equipment from equating the two equation we gets Ah, by equating Onda coefficients in port side with the same power of Lunda. Um, full size. Ah, terms with the same. Our, uh, Lunda we get that be one is equal to Lunda. One blast done That too. Plus run the end Poor negative one and minus one Be to is equal to on the one London to bust under one on the three plus Lunda n minus one on the end. Negative one and minus to until we&#x27;re age be. And it is equal to mom. The one done the tour Lamba. And on the one with the flying bile of the room of blind Bible on the three. And in Breach London, we can see that these ah base coefficients You can see the This is, uh yet the Syria Do you You&#x27;re being this Ah, aggression. We can say that because the London is equal to the determinants off a minus Love that one for London time is the unity I We say that the or zero is equal to that determine and off a but from region 5.7 point five be no is equal toe be And so we CEO Urbino will be equal toe the end will be equipped or the German and off a ah using the simple Ah see vita So ah, using you&#x27;re the your sample uh, new u n c v and let in me I i is a quinto a J a J north of London. Oh, I we get that determinant or a minus number. I is equal to see 11 that see in one C one and see and and which is equal to with some mission. Oh, incident. Oh, I one I to then I and I&#x27;ll see one. I won See? Oh, so to this is 11 11 See, I to to Well, we reach Ah, See I one. And for all eyes, all you want who I and you know, 21 Teoh. And here we know that we have the maximum power off London the maximum Well, we&#x27;re, uh Lunda. Uh, You see, I won One is equal through C 11 see I and and an end, which is equal to see. And then, uh, for example, I am is equal to and and, um, is equal to and ring. Uh, that&#x27;s accident. One don&#x27;t know three until and is equal to one. There were Bickell fishing&#x27;s oh, that in our, uh, Lunda is be Lunda is the little Oh, the pro decked. Um, anyone? One minus Lunda 2 a.m. and minus rondo, which is equal to a 11 a two to a 33 I&#x27;m sorry. A 118 or two minus a 11 lost a 22 rundown minus lander squared but the blind by a two to minus number. You know, we reach a and and minus London, which is equal through Ah, negative one Good and Lunda More. And plus I think that one or in minus one seeking man from I went to one so sorry is equal to end well, a I I along the and lightness one plus sigma Well, I is not equal. J is equal to one to an a i a J Jane of Lambda and minus two negative one and minus two plus until we reach a 11 a two to to a And so all other factors have orbiter lower than in minus one. So we can say that be one is equal to negative one n minus one A 11 plus a +22 los A and which is equal to ruining it. One for end minus one. You are or a the result in 12 and three and poor say that that determinants of a is equal to the some off North publication from I is equal to one toe, always equal to 10 for London I I which is a product. Oh, old Eigen values where that you are a is equal to the submission from I d Quit one. Try deeper to 10 for all the again or buying and values.

Nicholas Barvinok · Answer

we want to verify very arbitrary two by two matrix A B C D that let&#x27;s call it A. That it&#x27;s trace is the sum of the Eigen values, and it&#x27;s determinant is the product of the Eiken values. So in this case, the trace, which is the sum of the diagonal elements, is a plus D. So let&#x27;s find the Eigen values well. And I guess let&#x27;s actually write up this. How first, in the determinant of a is a D minus. BC So what are the values that this matrix we first take the determinant off this matrix? A. But subtracting the identity times lambda, this gets us a minus lambda D minus Lambda minus be seen which we can expand out, too. Landis squared minus D lambda minus a lambda plus minus a minus. D lambda plus a D minus. BC No, the Agon values are the roots of this equation. To find the roots, we can use the quadratic formula so we get a plus D plus or minus square root of a plus D squared minus four times a D minus BC and all of this over too, right? So let&#x27;s first ad up thes two Eigen values. So when we add these up, we&#x27;re going to be adding a plus D plus the square root added to all that over to added to a plus D minus the square root. So we get exactly a Prince de plus a plus D so twice a plus d and all of that. Oops, this is not equal. Well, let&#x27;s say that the say this is equal to ah Lambda one plus Lambda too, which is a plus D, which is the trace, as we wanted. Now let&#x27;s multiply. Used to Eigen values together So again, Lambda one times Lambda, too, is a plus D plus square root of a plus D squared minus four a. D minus. BC all over too. So this multiplied Bye. A plus D minus the square root of a plus the squared minus 4 80 minus BC All of that over too. So when we multiply this out, if we group this a plus D as a single term, we get a plus de squared, plus a plus de times the square root term minus a plus T times the square in terms of that cancels out and you get minus the square root terms squared. So minus a plus D squared plus for a D minus BC and all of that over four. And so the April&#x27;s D squared cancels with the minus April&#x27;s D squared and we&#x27;re left with 4 80 minus BC over four, which is a d minus BC, which is precisely the determinant of a So we&#x27;re done or the determined of our original matrix that I call it a Yes, I did.

R M · Answer

problem. Number 41. Ah, we say that Lambda want is equal to zero and zero on London, who is able to three. So these are the faces we&#x27;re from here. We can get the faces of negative one one and zero. I&#x27;m not get the 10 on one. And from here we can get the faces, uh, 11 on one.

Jack Chen · Answer

Okay, so UNICEF program We&#x27;ve already known that the metrics eight has the following decomposition You Sigma Wien transports. So, um, to show that, um to show that to the columns off the argument actors off agents post Wednesday, we can compute h is post A So it just because they you caused to you Sigma the transpose Onda We put analogy Suppose outside times us Sigmund the transports Because to the sick much is suppose you transpose you seem RV transports and that we know that&#x27;s both you and, uh me. Uh, I thought, you know. So the definition off talking a matrix will be you just oppose equals to you, you numerous on the re transpose he was to the interest eso by this condition h is posed a equals two v tons. Six matches Post sigma because you transpose times U equals two And I didn&#x27;t Leah magics. So we can you know this part of the universe Now we move this term the universe to the left that gives us agents pose a V equals toe three times seeking matches pose sigma and we already know that Ah sigma are diagonal matrix So sigma because to something like, Let&#x27;s check in my one have to see my, um And there are some zero terms here. So sick a ma just post time sick. My close to, um something like sick of my wife. Square Sigma to a square. And the sigma is square. Okay, So with this result, we know that if we write out the close to a lot of column vectors, the one I&#x27;ll be too the three up to the end. Then we can decompose this equation as ancients post eight times the I equals two sigma High Square times v i. And this is exactly the ag of the definition of the Eigen vector and again values. So Sigma Chi Square, our, um, again values off a chance post 10 say, and the re eyes. Ah, corresponding Eigen vectors. So this proof on the so if Sigma Square Aiken values off, patron supposed time, say, um, there&#x27;s sticking. My eyes are, ah, singular values off a This proof, the first end of the last statement and full a second statement. We just do it exactly the same thing for a times a transpose. But in this case eight times, they transpose equals Teoh, you sigma The transpose times, the times, stigma chest post temps you And again we used effect that you transport of the transpose equals to the verse. So the protect off this two metrics will be an identity magics. So we can you know this then we have you Sigma Sigma matches post types. You Sorry Decisions. Suppose here. But we can I&#x27;ll rewrite it as your humorous. Then we have a HS post tempts you. You close to a your time signal my time sticking my transposed. So the column vectors off you have you eyes? Are I get better again? Victor&#x27;s off a agent. Suppose so. This proof, the second statement and then we finish on proof here.

Problem

Show that if $A$ and $B$ are similar, then $\oper…

Problem 23 Medium Difficulty

Answer

Since $Q$ is invertible we can write
$A=Q R=Q R Q Q^{-1}=Q(R Q) Q^{-1}=Q A_{1} Q^{-1}$
So, by the definition of similarity $A$ is similar to $A_{1}$

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Kristen K.

Michael J.

Vectors Intro

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Since $Q$ is invertible we can write$A=Q R=Q R Q Q^{-1}=Q(R Q) Q^{-1}=Q A_{1} Q^{-1}$So, by the definition of similarity $A$ is similar to $A_{1}$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Kristen K.

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.

Kristen K.

Michael J.

Since $Q$ is invertible we can write
$A=Q R=Q R Q Q^{-1}=Q(R Q) Q^{-1}=Q A_{1} Q^{-1}$
So, by the definition of similarity $A$ is similar to $A_{1}$

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

Linear Algebra and Its Applications