show-that-if-a-has-n-linearly-independent-eigenvectors-then-so-does-at-text-hint-use-the-diagonaliza

Question

Show that if $A$ has $n$ linearly independent eigenvectors, then so does $A^{T} .[	ext { Hint: Use the Diagonalization Theorem. }]$

Amy Jiang · Accepted Answer

Okay, So if the 80 very transpose that semen he DP interests transposed and then that&#x27;s it equal to p inverse transposed the transformers and be transposed, that&#x27;s able to pee transposed in verse, transpose and transpose. But this is a sequel to Q B G and cue in verse, where Q is equal to P. T and GT is a diagonal rejects. So based on this, we know that a T is diagonal Izabal.

Jimmy Yao · Answer

don&#x27;t just the problem here we want to argue are a Is that many independent? So So there The corner of a linear independents is equivalent to say a s incredible. So it is actually a is inevitable if, in the only if agents pose a it is also you, Bertel. So they said speakers If we see this terrorist in so the so so here for this demolition If a the colon is they&#x27;re Carlin&#x27;s of a it&#x27;s linear independence. Then we know a is you&#x27;re irritable. So if a is the convertible and so we also know a transpose is inevitable So this is because the rink of a it&#x27;s ricotta rink of transposed So yes, so they have the humors so safe. But he knows the U S of a transport be ancient spell beavers and the humorous of a will be a uber&#x27;s. So we say a transpose Huber&#x27;s and a universe we&#x27;ll be the humors of a transpose eight. This is because it&#x27;s their time. Together, it&#x27;s echo to a uber&#x27;s in a transpose universe, a transposed in a and become to a transpose i A. If you go to a inverse, a issue could you buy. So this means it is the inverse matrix of a chance was eight. So agents pose a is convertible, so for two this direction so we know a chance. Close A. It&#x27;s the Invincible due to their rink knowledge t serum. We know that the rink of agents pose a trust that not a team of aid you This chance was a It&#x27;s Nico to in the hair agents will agency in veritable So we know this rink. It&#x27;s because you in so the knowledge he should be sure that you come to narrow So we know if agents pose a eggs. If you co 20 you have to have X You cut your hero, right? They were going to consider the now space of eighth. So for the now space of a so it looks like it&#x27;s satisfies a X equal to zero. So we choose any experience now space okay thing than now of age. So this aches had a property. A ex coaches there. So also, if we transposed, there&#x27;s we&#x27;re gets shows X transferal agents bill should also be there. Oh, if we time that to them together we will get X transport a gentle a X if you cut through there. Mm, No, I don&#x27;t want to do that into this is enough. So if X it&#x27;s lifting this now space a X is equal to zero and we times a transpose to two this vector. So we know agents will A X is equal to zero since we already know if there now space for a transpose a is contents in the their oh, so we can derive that extra butte there. So if exit zero then we means this means this now space for any the X in the snow space We can deduce that X is zero. So this means there&#x27;s now space only contend content zero So according to their well so well, I don&#x27;t want to do that. So for any So we know there no space here, so only condemned zero. So the knowledge t for a it&#x27;s their so called into the rank nal ity. Mr Room. We know the rank of a it&#x27;s in. So here we know the the colon, the colon that&#x27;s a his denia independence

R M · Answer

58. The condition all right, is see one view, one last see to be to is equal to zero. It&#x27;s just one. So a see one you want better close a see toe be through Vector is equal to zero. So see one, not the one. The one better plus C two London toe be to vector is equal to zero and this is 11 you was see through veto Bester from one and substitute into 11 we get That&#x27;s not the one minus lander to time C one B one vector is equal to zero as we know that bandana is not equal under two. So c one b one better. It was zero as we know that you on vector it&#x27;s not with zero. So see, one could be with zero and also see toe would be the questions here. So you want and b two Victor, Are Lanier really independent

Runpeng Li · Answer

okay for a problem in 28 were given a matrix eight that is in my inn and symmetric and we want to show Oh, also the Matrix A is in vertical asymmetric and the inverse exists Inverse exists Now we want to show the quadratic form off. The inverse inverse matrix is also positive. Definite. If the key then we&#x27;re ready. Form off the matrix. A is positive. Definite definite. All right, um two. To prove this to prove this statement, we have two parts too. Take care of. First part is to show that the inverse of a is symmetric, right? Because we if the if the inverse matrix is an awesome it&#x27;s not symmetric, then we don&#x27;t have any. We don&#x27;t have any like way. Don&#x27;t we can&#x27;t. We cannot say that this court radical form off this off, Nancy Metric Matrix is positive. Definite right. We have to required this matrix to be symmetrical. Then we can talk about a positive definite or not. And the second, the second part we need to take care of is just a positive, definite thing To show X transpose times a times acts is positive. Definite definite. Okay, that&#x27;s first and catch the first The first part. So first recall if we have you weigh have a convertible matrix. We always have this equality at this a inverse transpose is equal to hey transpose and in verse. And also we&#x27;re given the Matrix A which he says mentally matrix And this is transpose is also equal to a so we have also a inverse. So these tells us that the inverse matrix he&#x27;s symmetric because the transports off this inverse matrix is equal to the matrix itself. So symmetric and we are down for for sparks. Now, the second part, the thing we&#x27;re looking to is the quadratic form off the inverse. Sorry, we missed negative. Why here? So we that the court uniform we need to consider is tthe e is for the members Inverse matrix. That&#x27;s first, um, as first lookat the matrix A. Since this matrix ihsaa positive definite data means every single I can value is us too, right? This by our dear, I&#x27;m giving the textbook. Now if we consider the inverse matrix recall that the Eiken battery is just won over Lum di since London, guys always part of so one of alumni&#x27;s definitely positive. So that means every single again battery off this inverse matrix is positive ends the inverse of a hiss. Positive. Definite, Definite. So we&#x27;re done.

Rakvi . · Answer

So suppose the A one feel 31 plus your toe Feel free to all the way a pool. Okay, Be all freaky Is zero That is some linear combination off these vectors. Steel free I zero Okay, inside W Then you want to show that all the eyes are in fact zero. That is all the weights are zero. That is with what you want to show. Okay, so let&#x27;s show this. So using the property of linear transformation, you know that a one off P off Leven is equal to p off even everyone. Okay, Similarly, second term is equality off. It will be too similarly all the way up to give it. Okay, this is 1/4 here. I use the property that he respects tradition 60 off. Even we won, plus de off. It&#x27;ll be it all. The way is just equal to tee off. Even we even close it. We do close, you carry. This is 1/4 0 now you know, used effect that de is 1 to 1. So 121 will tell you that if he off a vector is equal to zero, the vector itself has to be zero. So this implies that even we even plus a 22. Thus all the were toe a Cherokee is actually called a zero. Okay. No used effect that the set of even week A is a linearly independent certain side we so linearly independent. So if some combination off revised zero, that means all the weeds have to be equal to zero. So you have shown that the 80

Donald Albin · Answer

All right. So prove from definition 4.5 point four that if v one v two So this is a set containing V one v two dot dot dot v an is linearly independent. Um, if they are linearly independent, then and I didn&#x27;t finish reading it, but anyway, I&#x27;m gonna work on this linearly independent thing for now. If those air linearly independent, then um d one v one plus de two v two plus dot, dot dot plus de and the and does not equal zero for ah d one d two that ah de and that are rial numbers. Okay, Now, um, and if a is an in vertebral n by n matrix than the set a V one a V two that thought all the way up Teoh a V an is linearly independent. So we&#x27;re gonna prove this by contradiction. Let that be then nearly dependent. Okay, so we&#x27;re trying to prove that that is linear Lee independent that that set. So we&#x27;re going to start with the contradiction with the opposite statement that the contradiction with the opposite statement, which is that, um, that is the nearly dependent, If that is linearly dependent then some see one times a times V one plus some see two times a times V two all the way up to some. See an times a times V an must equal the zero factor and noticed that Ah, C one is just a constant so we can switch this around and just write a time. See one V one plus a time See two v two. Plus that that that plus a CE envy and must equal the zero vector noticed that a is in every term. So a see one V one plus c two v Teoh that dot c N v n must equal zero of actor. That&#x27;s not very neat. If we take the in verse of a times both sides than the left side is still going to be the zero vector. But the inverse of eight times a is the identity matrix and the identity matrix times Another matrix is just itself for some c one c two see, and that are rial numbers. But that contradicts something that we know to be true here and here are contradictory and therefore a V one a V two God, I don&#x27;t a v n are linear Lee, independent

Problem

$A$ factorization $A=P D P^{-1}$ is not unique. D…

Problem 28 Hard Difficulty

Show that if $A$ has $n$ linearly independent eigenvectors, then so does $A^{T} .[\text { Hint: Use the Diagonalization Theorem. }]$

Answer

Thus, $A^{T}$ is diagonalizable.

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

Linear Algebra and Its Applications

Catherine R.

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Kristen K.

Samuel H.

Vectors Intro

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Thus, $A^{T}$ is diagonalizable.

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Heather Z.

Kristen K.

Samuel H.

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

Linear Algebra and Its Applications