operatornamefor-aleftbeginarrayccc1-2-3-1-2-3-1-2-3endarrayright-find-one-eigenvalue-with-no-calcula

Question

$\operatorname{For} A=\left[\begin{array}{ccc}{1} & {2} & {3} \ {1} & {2} & {3} \ {1} & {2} & {3}\end{array}ight],$ find one eigenvalue, with no calculation. Justify your answer.

Elias Lozano · Accepted Answer

So in this question, we&#x27;re gonna find on I get value of a specific A matrix, this matrix being 111 222 That&#x27;s the one, by the way. 333 Now it&#x27;s a rule of linear algebra matrices that if a matrix is linearly dependent, I can value is equal to zero. And that&#x27;s the case here. A is linearly dependent linearly dependent means that the rows or columns are multiples of one another. So in this case, you can see Row one through. Three are exactly the same, meeting their multiples of each other. You can also see this in the columns, so call him three. It&#x27;s just three times column one calm to It&#x27;s just two Times column, too. Now it&#x27;s a rule that if a matrix is a nearly dependent, I can value always equal zero

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.

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.

Sarah Klein · Answer

this problem wants us to verify that this wagon valued Eigen Vector correspond to the given matrix. We can do this using the equation eight times, Lee equal to Lambda Time. See plugging in what we&#x27;re given. We get 13 to 2 times the vector which can be written critically like this 11 is equal to four times 11 So on the left side we get that this equals for for when he multiplied in the constant into the vector, then for vector multiplication on the left side we get long times one plus three times one on the top and two times one plus two times one on the bottom which becomes one plus three and two plus two. So then we get the Vector 44 which is equivalent to the right side meaning that this Eigen value and Eigen Vector does correspond to the Matrix

Sarah Klein · Answer

this problem asks ist friend Eigen. Values of the Matrix A with the given Eigen vectors could do this using the equation. A times V is equal to Lambda Times e So first for the first Eigen vector V one, you can plug in a 4213 then V one vertically. One negative too is equal to lambda Warn times one negative, too. Then doing matrix multiplication on the left side, we got the top row to be four times one, which is four plus one times negative two, which is negative two and two times one which is two and three times negative two, which is negative. Six is equal to to negative four and that is equal toe lambda one times one negative too. So we could see the constant Lambda 11 Dividing to divide by one and negative four by the bynegative two gives us land. One is equal to two. Then for the second Eigen vector view, too, we can plug in again a 4213 Then the vector V two wishes 11 is equivalent to lambda two times 11 again matrix multiplication of the left side, The top row. We get four times what it&#x27;s for. Last one times one, which is one 1212 plus three times one is three, which gives us 55 which is equivalent to land a to times 11 that if we divide five Vector 55 by the Vector 11 We find that landed too is equal to five, and those are two Eigen values.

R M · Answer

problem Number 30. Ah, question A for London One is E quickly, soon and am line us. I&#x27;m that one. The vector is a prince zeal. It&#x27;s here that we want those veto My enough bigotry is equal to deal B one is London negative 11 and zero plus s one, you know, And one where London and s your long who oh, are are the basis of flying in space you want Ah, let&#x27;s, uh he won better equipped to you one better. Which is one view on one And the two better is equal to re negative 11 And and you by applying grams that poses by applying Ram No there process. We can say that you two rector is equal toe be toe vector moments. Uh, be too. And you won uber They&#x27;re nowhere ordered The magnet you would off you won better it spreads not the blind Buy you one a letter. It is equal toe Have negative one to and one Ah So also we can hear that or so gonna bases off. The one is equal to one, you know on one on negative one Teoh on one being we sold from Lunda two is equal to one. It was going to get one as a minor&#x27;s on the time in the vector is equal to zero to give that we vector is negative. One negative one and one is the pace is off the agonist space here too. Ah, then ah, B one on B A factor is equal to zero, which is equal to you. Toe that, uh, on b have that. So he&#x27;s, uh, vectors in you want? Oh, are or a goner. So the vector in E to

Problem

Without calculation, find one eigenvalue and two …

Problem 19 Easy Difficulty

$\operatorname{For} A=\left[\begin{array}{ccc}{1} & {2} & {3} \\ {1} & {2} & {3} \\ {1} & {2} & {3}\end{array}\right],$ find one eigenvalue, with no calculation. Justify your answer.

Answer

$A$ is not invertible. Hence, $\lambda=0$ is an eigen value of the matrix $A$

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$A$ is not invertible. Hence, $\lambda=0$ is an eigen value of the matrix $A$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Kayleah T.

Kristen K.

Joseph L.

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

Linear Algebra and Its Applications