show-that-lambda-is-an-eigenvalue-of-a-if-and-only-if-lambda-is-an-eigenvalue-of-at-lefttext-hint-fi

Question

Show that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^{T} .\left[	ext { Hint: Find out how } A-\lambda I 	ext { and } A^{T}-\lambda Iight.$ are related.

Dharmendra Jain · Accepted Answer

Hello. One strand is given question. We have to show that Lambda is an Eigen value off on Lee. And only if, if and only if lambda is an icon. Value off, I suppose. Sophie. Right. So it is enough to show that the equation that they question a X is closed toe lamb decks has a non driver solution if and only if the equation a transpose X equals two Lambda X has a non travel solution. Right? So we can live. Right? This equation in the form off identity metrics that is 80 minus slammed. I yes, excessive cost, Ojeda. Okay, since that the dominant off this is request toe the determinant off transport C minus Lambda I is request to be dominate off a transports minus lambda making. Right here it I identity medics transposed right. So if the sequester lambda So the a minus slammed I don&#x27;t suppose right as you know, it is a questo Determinate off a minor. Slammed I Yeah, so the bad. This is in both the equations a simultaneously in vegetable or simultaneously not invaluable. It follows that the both equations or simultaneously have known travel solution or simultaneously have on Lee the driver solution, which means that Lambda is an ideal Eigen value off a if and only if it is an Eigen value off transpose off, correct as it is.

Nicholas Barvinok · Answer

we have that lambda isn&#x27;t Eigen value of the Matrix A. And we want to show that Lambda Squared is an icon value of the matrix. A squared? Well, what does it mean for eight to have? Ah, in vector of Lambda. It means that there&#x27;s some vector V such that when I apply a to it, I get Lambda B. Well, what happens if I apply a squared to be a squared is a time&#x27;s a applied to V, which is a applied Tau lambda v matrices our linear, so I can move the land outside and again I have a applied to be a apply to V is Lambda V. So I get Lambda Squared V. So we see that a squared has an wagon value of Lambda squared with Eigen Value V, where V was the Eiken value for a with dagon value Landa

Nicholas Barvinok · Answer

Yeah, we have a non singular matrix A with some wagon value Landa. And we want to show that the matrix a inverse which does exist because A is non singular has an Eigen value of one overland. Well, what does it mean? That a has an icon value of Lambda? That means that there is some Eigen vector. Who is he? When? When I apply a tow it, I get Lambda B. Well, when If I apply a inverse to Lambda V bye, that&#x27;s the same thing. I&#x27;m applying this a inverse to the right side of this equation. I can apply it to the left side of the equation as well, because they&#x27;re equal. So a inverse of a of the which equals V because the A inverse cancels with the A. So I get V. So looking from here to here, I see that a inverse of Lambda V equals V toe and verse multiplied the vector Lambda V by a factor of one over lambda and thus Lambda V is an icon vector for a inverse within Eigen value of one

Runpeng Li · Answer

All right, so, four problem 12. Um, we want to justify a statement meeting these sections. That that a little case. I am a smaller equal to Lunda. You go to a person, he&#x27;s all right. So far of em where I just write down the definition of famine. So legal case that our family is defined by the minimum off to quadratic form such that in the warm or fax is one. And, of course, fund me correspondingly. The uppercase am is defined by a maximum. Sorry, huh? All right, so four problem 12. Um, we want to testify. Statement made in these sections that that, uh, little case have a smaller equal to Lunda. You go to a perp is a similar affects transpose times x quadratic form with the normal fax is one. Okay, So here is what we want to show. The London is bigger, bigger, even two. AM since morning to m. So is this what we want to show now? The first thing to notice that Lum Di is equal to em where I just write down the definition. Mm. So legal case that our family is defined by the minimum off to quadratic form such that in the war more facts is one and a corresponding me correspondingly the uppercase am It&#x27;s defined by Mexico. I&#x27;m sorry. The I that is one the the Eiken value is equal to the corresponding. I can vector transpose times the matrix, a times the Eiken vector. Uh, the reason why this is true is because, um, you know, by our our definition of pagan values and I get vectors, we have we we always have this Now look similar text transpose times x The poetic form with the normal packs is one. Okay, so here is what we want to show The London is big, bigger, even two AM since morning to m So is this what we want to show now? The first thing to notice that loved I is equal to you three times three times the b b I transposed on both sides Then we will have transposed time Say I&#x27;m speedy I which is equal to B I transposed times Lum di times b I now love dies a skater So we can Who the scaler in front of toe vector multiplication. So we have V I transpose the I. That is one, the the Eiken value is equal to the corresponding. I can vector transpose times, the matrix, a times the Eiken vector. Uh, the reason why this is true is because, um, you know, by our our definition of pagan values and I get vectors, we have we we always have this now. Times Yeah, but B I&#x27;s normalized. So this should be one. So we have loved I That&#x27;s how we get this equality. Okay, So this equality implies that we can rewrite our definition of m to be the minimum off because this is a quadratic form and we can write as minimal off London eye. If we times three times the b b I transposed on both sides, then we will have transposed times. Hey, times be I which is equal to the I transposed times Lund I times b I now love dies a skater so we can through the skater in front upto vector multiplication. So we have b i transpose that is this smallest Eigen value in our amount of all the agg InBev&#x27;s correspondingly the up case. Sam can be reading that maximum me off them die. So it&#x27;s clear that Lunda were always bigger or able to minimum off there off the Londoners of the hall of all the again values a smaller you go to the maximum off all the argument times. Yeah, but B I&#x27;s normalized, So this should be one. So we have loved I That&#x27;s how we get this equality. Okay, So this equality implies that we can rewrite our definition of m to be the minimum off because this is a quadratic form and we can write as minimal off loved. I yes.

Gennady Notowidigdo · Answer

Okay, so this is a kind of a proof question. So we&#x27;re going to show that is that if a score so some matrix A If it&#x27;s square is the zero matrix, then the urn, the Eiken value of a is zero. Okay, So to do this, we&#x27;ll start with a sort of a more very fundamental statement. So if ah, so if Lander and X is on aiken value, I convict the pairing for a matrix A. Then up a X is equal to Lambda X. Okay, so we start with that statement, then if we multiply both sides on the left, So remember of matrix R multiplication? It&#x27;s not community. So it&#x27;s important to state which side you&#x27;re multiplying on this case. We&#x27;re going to multiply both sides on the left, so we&#x27;re going to put on a so probably wrong you sprung with color. I&#x27;m gonna put a matrix a on the left hand side. Yes, sir. Back to blue. They&#x27;re more buy bread side on the left by a So what we get So you&#x27;ve got a squared X is equal to lander times a times X. So what I&#x27;ve done is sort of you have probably take one step back for a Times Land X, which then changes to sew because landers a scaler, you can always take the scale of outs. So we landed times a X, so it&#x27;s sort of using this sort of compatibility slash communicative ity slash associative ity type operation on major season victims. Okay, now a squared. He is 20 matrix. Okay, but this become zero times X, which obviously becomes the zero vector that has to become so remember a to land. The X is an icon value I convict apparent. So this ends up just being lambda squared. Times x good are this leads to two possibilities. So either land a squared equals zero, which implies Lambda equals zero. Oh, sir, Or be implied that X equal to the sirah. Victor. Okay, now, because Latin X and I again Victor by death, by sort of specifications or definition, this cannot be true. There are probably used a different college. Just cross this out, that this cannot be true, since I convict er&#x27;s are non zero. So all that your left with is possibility that Lambda has to beezer on it. It&#x27;s the case that land equals zero, then has to be the only Aiken value for a so this is obviously a relatively detail proof older. I guess I could fleshed it out a little more, but it&#x27;s definitely better than a sort of a simple show that without many steps. So ultimately, we sure now that if a squared is zero, so maybe up always say this. If a sweat 00 matrix, then this becomes that, and then that reduces the possibilities. So I&#x27;ll leave it there. Thank you very much.

Runpeng Li · Answer

directs a four problem 10 6 First thing, we need to show that eight times a real part off vector V and is equal to eight times re part off vector Be a speed times eventually in victory be no. So the first thing we need to notice this debt um, our again, Victor v can be reading this really hard off e. Yes, I times. Uh, sorry. Yes. I times the imaginary part off Now the net that, um, condition the let&#x27;s just do the substitution things. We know he&#x27;s the connector. So and Linda is d I get value. So we have these really leadership. Now we substitute V by this expression. So we have eight times real part of the that&#x27;s I times imaginary part of and which is expended. We have eight times really hard off the best I times. Hey, times he imaginary part, huh? So this is for our step inside both this relation. Now we have the right hand side will be London a minus b I times really hard off me class I times imaginary partner. So we expand it. We have a times real part. Oh, yes, I Excuse me, I attempts a times he measuring hard off e minus you, sir. High times view that&#x27;s re apart. I&#x27;ll be minus Well, just pass be Tom Cement Remain powerful. Now we combine the memory those terms that contained I So we have a view he s off imaginary be. And also we have a i imaginary Do you minus e your heart off e times I So if we compare this part two our eight times part off this is that all the rial? I mean, the real part. So I compared this to terms you compute, I&#x27;ll start a new page. A l r e riel of b is equal to eight times really be, Yes, he times measuring off. And also we can compare this part. And here we are so contained the eye What? That is eight times imaginary e is equal to eight times imaginary e Why this and be times really so That&#x27;s our proof in the first part. Okay. And the second part we want to show that I want to verify that if PNC are given, has a theory room by the a. P is equal to P. C. So it&#x27;s just a from our theorem lie directly because we know, um, we check our room. Night can find that a is given by p times. See, time&#x27;s p you guys. So every time times, uh, if we multiply, pee on on the right hand side up. Both equation. So we have eight times p. Most pete I&#x27;m C then p inverse times p for the identity. So we have a p equal, which is equal to P. C.

Problem

Use Exercise 27 to complete the proof of Theorem …

Problem 27 Easy Difficulty

Show that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^{T} .\left[\text { Hint: Find out how } A-\lambda I \text { and } A^{T}-\lambda I\right.$ are related.

Answer

see the proof

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

Linear Algebra and Its Applications

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Heather Z.

Joseph L.

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see the proof

Linear Algebra and Its Applications

Lily A.

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Vectors Intro

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

Lily A.

Catherine R.

Heather Z.

Joseph L.

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

Linear Algebra and Its Applications