in-exercises-7-12-use-example-6-to-list-the-eigenvalues-of-a-in-each-case-the-transformation-mathb-4

Question

In Exercises $7-12$ , use Example 6 to list the eigenvalues of $A$ . In each case, the transformation $\mathbf{x} \mapsto A \mathbf{x}$ is the composition of a rotation and a scaling. Give the angle $\varphi$ of the rotation, where $-\pi<\varphi \leq \pi,$ and give the scale factor $r .$
$$
\left[\begin{array}{rr}{-5} & {-5} \ {5} & {-5}\end{array}ight]
$$

Runpeng Li · Accepted Answer

so for probably 10. Well, given the Matrix that ISS next 55 life and make define, so can we need to apply our result in exactly six. So here we take Hey, I hate to be dignified, maybe five now so number will be making five minus five I and to scale The interpretation is fine hour, which is well, I was queer Time Fast five squared Take a square root so this will be by Times Square two. So see, we&#x27;ll be five times going to times the matrix well active, one over. Scared, too. So it&#x27;s just out right down as a screw to over two negative square to two. It&#x27;s 3 to 2, and negatives were to to, so we can take to be negative. So just posted Ri Pi over four so that our matrix will be five squared over square two times. No sign, it&#x27;s reply or by this sign, reply or sighing, reply over four and co sign reply or four

Michael Jacobsen · Answer

for this video, we&#x27;re going to be analyzing the the&#x27;re, um we see displayed here and to help us walk through this serum. Let&#x27;s use the example where a matrix say is given to be square 231 in Column one and then negative one square to three and call him, too Now for this matrix A. It could take some work to find its Eigen values. But if we have a special situation where the entries air the same here and then opposite on the main diagonal, then we can state the Eigen values out right. And it&#x27;s in fact, the situation we have that May diagonal were equal in the off diagonal. We have opposite values, and then this implies that the Eigen value lambda one for this matrix A. I&#x27;ll stay in column one and say it&#x27;s going to be first Pick your A from column one, which is squared of three. Then go. Plus the beat will be this entry, which will go in those positions So R. P is a one times I I&#x27;ll just write and I then that&#x27;s our first Eigen value. And the second Eigen value is always the Contra kit for two by two. Major sees with really entries, so we&#x27;ll have Route three minus I. So that takes us to the first part of the serum, where for matrices of this form, we can obtain the Eigen values immediately. Next, let&#x27;s determine what the value of our should be and the value of fee so that C can be written in this particular matrix factories ation that we see displayed here. Our goal here is to just simply produce RN fee. So now our is The module is of either Eigen value, since having a plus or a minus does not change the module ISS. So let&#x27;s take the modules of Lambda one and by definition, for absolute value or modules of a complex number. We first square this entry and then take the coefficient for I and square as well. In our case, that coefficient is one. So we have Route three squared for a three plus one, which is the square to four, resulting in two. Next, let&#x27;s find what the argument is for Lambda, which refer to as fee in this situation to find fee, we first look at the equation that tangent of fee, Whatever it is must be equal to be divide by a but recall RB value is one, and the A value is route three. So we have one over route three and we recognize that a solution is FY equals pi divide by six. So to summarize for this given matrix, the Eigen values are these two quantities. And if we were to write this matrix factory ization are would be equal to two, and fee would be equal to pi over six.

Michael Jacobsen · Answer

in this example. We have a long theorem here that&#x27;s provided, and we&#x27;re going to walk through it by illustrating its use with the Given Matrix. So, to that end, let&#x27;s start off by letting A Matrix A, which is of size to buy two, to be defined by 0.1 and negative 0.1 In Column one, then 10.1 and 2.1 in Column two. So now with our Matrix say, Let&#x27;s compare it with the Matrix C provided in the serum. The key feature with this matrix is that the main diagonal is equal, and that&#x27;s what we have going on here with this matrix A and likewise, the off diagonal must be opposites of each other, and that&#x27;s also what&#x27;s provided here. So since the theorem applies to this matrix, say, let&#x27;s list what the A and B should be. In this case will have that a is 0.1 coming from this entry and column one, and likewise be is negative 0.1 coming from the century here. Then, with that information, we&#x27;re ready to state the Eigen values. The first portion of the theorem we have that the first Eigen value Lambda one is a plus B I. In our case, that&#x27;s 10.1 minus 0.1 I, then lambda to is It&#x27;s conjure git, which is 0.1 plus 0.1 I. So this is the first portion off this theorem for the second portion. It provides us with the tools of using this matrix factory ization for our particular matrix. A. But here are mingle is to just determine what the modules are should be as well as the argument fee. So we&#x27;re solving for two things. Next, let&#x27;s start by solving for our so in this case, are is the module ist of either Eigen value. The minus and plus in the middle don&#x27;t give us a different value, and so this is going to be by definition of module ISS, the square root of 0.1 squared plus negative 0.1 squared, so this becomes 0.1 plus 0.1 which is the square root of 0.2 and it&#x27;s always difficult to determine how much we want to reduce here. But let&#x27;s reduce giving an exact value. So we have 0.2 or, in other words, that is to one hundreds, and we can say that that squirt of two in the numerator divide by 10. So this is our module. ISS are next. We&#x27;re going to determine what fee is, but we have to use the full equation to start out. They&#x27;re right. That tension tiffy is equal to be divide by a But we get that information here, so take negative 0.1 divided by 0.1. Now we get a negative one quantity altogether. Now, when we&#x27;re looking for fee, we&#x27;re just looking at what is the rotation for either Aiken Value? We&#x27;re focusing on the first case, Lambda One. And for this first Eigen value noticed that ends up in the fourth quadrant. That that implies that he must be equal to negative pira for since tension. A native PIRA four does indeed give us negative one. And that also sets us in the right quadrant. So to summarize for this matrix say that&#x27;s been provided thes are it&#x27;s Eigen values. And if we&#x27;re going to use that matrix Agon factory ization indicated above, this would be the module iss and this would be the argument

Michael Jacobsen · Answer

in this video, we&#x27;re going to be analyzing the theorem we see displayed here. It comes in two parts, and for the first part, let&#x27;s illustrate with the Matrix, which will determine to be a with columns Negative Route three over to negative 1/2 for column one and positive 1/2 negative, Route 3/2 for column, too. Now to use this, the&#x27;re, um, we must first start with the Matrix C that&#x27;s of the form. May diagonal is equal and looks like that&#x27;s what we have here. So we&#x27;re off to a good start and the off diagonal has to be opposites. And we have negative 1/2 here, positive 1/2. So we do, in fact, have opposites. It&#x27;s easy to be a bit hung up with this negative sign here, So if that bothers you, you can always say this is your be equals to negative 1/2 and the missing negative here can be inserted in this way with a double negation. So that&#x27;s extraneous. But it could help out with using this theorem. Also, lets indicate what are a value is here. We also have that a is negative square root three. Divide by two. With this information, we&#x27;re ready to analyze what the Eigen values will be immediately, which is a bit surprising. Since we&#x27;ve done no row reductions, we&#x27;ve computed no determinants and so on. But regardless we have that Lambda One is a plus B I. So it&#x27;s negative. Route 3/2, minus 1/2 times I and the second again value is just the congregate of the first, so be equal to negative Route 3/2 plus 1/2 times I. Next, let&#x27;s go to the second part of the theorem that tells us we can factor are matrix in this way where there&#x27;s a particular value of are found from this equation and the value of fee found from this equation here. So it&#x27;s determined our values and the fee values. Next we can say our which is defined to be the module ISS of either Aiken. Value is by definition take the first value a, which is negative, Route 3/2 square it and add to it to the B value, which we indicate as negative 1/2 and square. That what we obtained then is 3/4 from the first portion squared, plus 1/4. This results in 4/4 which means we&#x27;re taking the square root of one. And that says the our value is one. Next, let&#x27;s determine what the argument which is fee must be. To do that, we take the extra step of calculating 10 jeffy by formula. It&#x27;s equal to be over a where once again, this is the B value and this is the A value. So we have negative 1/2 divide by negative squared of 3/2, which is one over square root of three. Now I know FY equals pi. Over six would work for a solution here, but it turns out to be the incorrect value and the reason to see that is let&#x27;s go back to the congregate plane or the complex plane rather and plot Lambda one. If I call this a negative one going left, negative one going down then This value here is about at negative 0.8 and we have a negative 1/2 here. So we get a point about here, which is Lambda one in the complex plane. But if we put a ray through the origin in that point and determined this single, we see that pi over six does not fit the bill. It turns out that fee equals negative. Five. Pipe of I ever. Six gives us tension. FY equals positive 1/3 1 over square to three, but also lands us in the correct quadrant. Quadrant three. So do summarize for this particular matrix A that we&#x27;ve indicated The Eigen values are listed here. The our value is one and the argument fee is negative. Five. PI divide by six

Runpeng Li · Answer

kids. So four problem 12 Given the Matrix A, which is zero and 0.3 negative 0.30 So this case, we take a to B zero bey to be connected 00.0 point three So the lambda will be minus or worse, 0.3 i. Meanwhile, Thea, there are leads off the London so that would be 13 No scaling. And this is the Skilling and ordered a rotation takeout 0.3. Then we have duel one negative one and zero so we can take feed to be, um, negative high over to so that this becomes 2.3 times. Call sign Active high over too and connective sign collective pi over too. Sign negative by over two and co sign neck too high over too and

Runpeng Li · Answer

Okay, So for problem eight were given the Matrix Square with three three and negative three&#x27;s good three. So, by our result, in the example six. Now he tom along more textbook. Um, we know the Lando. Here we take. So first here we take a to B square three B to B. Negative three. So our landa will be, but with me minus or a plus, um three. Okay, so for the scaling and the rotation, So first we need to find out are which is the land of Lunda. So that&#x27;s three plus nine. Take the square root. Just a square root. Tough 12 Two times square three. Okay. And so See, we take out. Squared off three. We will have one have square with three over two. Negative square was three over two and one have now the and go. Just take negative three by over three. So no sign off. Connective high over three will be 1/2 and minus. Sign active by over three. Wilkie Positive square root. Three over two. Sign of collective fly over three. But just to be inactive were three over two and again Course I am negative. Piled on three will be 1/2. So here we take you to the connective. High over three

Problem

In Exercises $7-12$ , use Example 6 to list the e…

Problem 10 Easy Difficulty

Answer

$A = \left[ \begin{array} { c c } { - 5 } & { - 5 } \\ { 5 } & { - 5 } \end{array} \right]$

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$A = \left[ \begin{array} { c c } { - 5 } & { - 5 } \\ { 5 } & { - 5 } \end{array} \right]$

Linear Algebra and Its Applications

Lily A.

Heather Z.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Heather Z.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications