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Problem 11Letwhere A is to be considered a8 & matrix over Ca) Determine the minimal and characteristic polynomials of A and the Jordan form of A. (b) Determine ...

Question

Problem 11Letwhere A is to be considered a8 & matrix over Ca) Determine the minimal and characteristic polynomials of A and the Jordan form of A. (b) Determine all generalized eigenvectors of Aand a basis B of C' with respect to which the operator TA Ar has Jordan form_

Problem 11 Let where A is to be considered a8 & matrix over C a) Determine the minimal and characteristic polynomials of A and the Jordan form of A. (b) Determine all generalized eigenvectors of Aand a basis B of C' with respect to which the operator TA Ar has Jordan form_



Answers

Repeat Problem 9.14 for the matrix $B=\left[\begin{array}{ccc}3 & -1 & 1 \\ 7 & -5 & 1 \\ 6 & -6 & 2\end{array}\right]$ (a) First find the characteristic polynomial $\Delta(t)$ of $B$. We have (b) Find a basis for the eigenspace of each eigenvalue of $B$.

Do you like? We're told to suppose that the matrix B in problem 9 11 represents a linear operator on complex space C. Two. Whereas to show that in this case B is in fact diagonal. Izabal by finding a basis S of C two consisting of Eigen vectors of B. Well from problem 9 11. The matrix B Has Elements 1, -1 2 -1. Air walks. Just as we found in the previous exercise. The characteristic polynomial of B is still felt of T equals T squared plus one. Yours as a polynomial over the complex number. It does factor as T -1 times T plus I using difference of squares. And therefore it follows that the roots landed equals I and landed equals negative. I are the Eigen values of our matrix B. Report like an accurate accurate tigers rsx. That was like the next. Okay, now for our first Eigen value, I'll subtract I down the main diagonal of B. So we get the matrix M which is d minus I times I, which is a matrix ah one minus I negative 12 negative one minus I. This corresponds to the homogeneous system, 1 -1 times x -Y equals zero and two X uh plus negative one minus I times Y equals zero. And this reduces to the single equation one minus I times x minus Y equals zero. Now the system has only one independent solution. Take, for example X to be one and why to be one minus I. And so the vector V one with coordinates 11 minus I is an icon vector. It's exactly uh that belongs to in fact spans the Eigen space of land. It equals I. Now for the other Eigen value negative, I we subtract negative by down the diagonal of be. So we get the matrix M which is B plus I times big guy, which is uh What are you doing a bit? All right, well there we go. It's back to normal. So this gives us the matrix one plus I negative 12 -1 Plus I. Which corresponds to the homogeneous system. I just live one plus i times x minus Y equals zero and two times X plus negative one plus I times Y equals zero. This reduces to the single equation one plus I times X minus Y equals zero. And we see that this system has only one independent solution. For example, take X equal to one and y equal to one plus I. Then it follows that the doctor the two is an Eigen vector belonging to and in fact also spanning the Eigen space of Eigen value. Lambda equals negative. I you said leave her now as a complex matrix, we see that D. Is now diagonal. Izabal. Typically we have a set S with vectors V. One V. Two or 11 minus I. And 11 plus I can't have it. This is a basis of c. two. Yeah. We're going to do a house and I am the mummy birthday hit in my ass carrying. And it went all the way up to the rubber bend consisting of Eigen vectors of matrix B. And therefore it follows that B is in fact diagonal. Izabal don't do cream cheese please as a complex matrix. And using this basis we can actually represent B. I love it's a fucking addiction. I can't help myself. Were traveling all this is also fat. Yeah, yeah. By the diagonal matrix D. Which is the diagonal matrix which entries I and negative I.

Yeah. Yeah I submit that. We're asked to find necessary and sufficient conditions on the elements of A two by two real matrix. So the matrix is diagonal. Izabal in other words so that it has to real well nearly independent hiding vectors. So we'll take a to be the matrix with real entries A B c D mix as well. Oh just slow cure like like well first let's find the characteristic polynomial of a. Take the artisans. This is since it's a two by two matrix. T squared minus the trace of a times T. Plus the determinant today. No work. Still pissed. Mm And in order to have two real well nearly independent Eigen vectors and the whole it's just going to shit. And then I smoked my CVD and I'll go time to go back to sleep and that's where you get all the it follows that. This needs to be Yeah. Jack. Well asian replicants that pussy's yeah. Sex. Mhm. This quadratic has had to real roots which is true if and only if the discriminate it of delta which is negative trace of a squared minus four times the determinant today. Yeah is Greater than equal to zero I guess in this case they are allowing repeat real roots. Yes plugging in what this means. This is trace of a squared which is a plus D squared minus four times the determinant which is a D minus Bc Must be greater than or equal to zero. We're getting eight in the case. And if you square things out this is a squared plus two. A D plus D squared minus for a D. Plus for D. C. Is greater than equal to zero. Which we can rearrange to write as a minus D squared plus four times be Tennessee is greater than or equal to zero. Therefore A is diagonal. Izabal If and only if the entry a minus D squared plus four times B times C is greater than or equal to zero. Yeah.


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