Were given real matrices that define a linear transformation of our two. And for each matrix were asked to find all the Eigen values and a maximum set s of linearly independent Eigen vectors were then asked which of these linear operators are diagonal Izabal, in other words, which can be represented by a diagonal matrix in part A. We are given the real matrix A with components 563 negative two. Well, first we'll find the characteristic polynomial. This matrix delta T apartment he's been riding. It's a two by two real matrix. This is simply T squared minus this trace of of a which is three times T plus the determinant today, which is negative 10 minus 18, which is negative 28. Oh, is that what happen? Which we can factor as t minus seven times T plus four. The roots lambda equals seven and lambda equals negative four. Are the Eigen values of this matrix A. You want to find the corresponding Eigen vectors is the Eigen values helping to find the corresponding Eigen vectors. Well, first we'll subtract seven down the diagonal of a. So the matrix and this is a minus seven I. And this gives us the two by two matrix negative 263 negative nine. Which corresponds to the homogeneous system negative two X plus six Y equals zero. And three X minus nine. Y equals zero, which corresponds to the system, X minus three Y equals zero. Now we see that Vector V one with components 3, 1 is a solution non zero solution. In fact, it's an Eigen vector. Now, on the other hand, will subtract lambda equals negative four down the diagonal of a. Find our matrix M. So M is a plus for I Which is the Matrix 9632. And this corresponds to the modernist system. Nine X plus six Y equals +03 X plus two Y equals zero. Which just corresponds to a single equation. Three X plus two Y equals zero. And so we obtained the solution the two with coordinates to -3. Yeah. And these are our Eigen vectors. And therefore the set S. Which is vectors V. One and V. Two for 31 and two -3. Is a maximal set of uh linearly independent Eigen vectors. Each is from a different item space since S. Is the basis of our two. Yeah. We should get him. Dude. We should threaten uh D. O. D. Officials. Dude. Yeah. We got the power of podcasting on our side. Yeah. Yeah. We got combination. Stop. Come on. Do someone's gonna get pinkeye one of these days? No, you shut up. Well, S is a basis for our two. And therefore it follows that. Our Matrix A. Sorry, one second. Because you're Ronald Mcnair the cloud. Yeah. Matrix A. Is diagonal. Izabal. Well be careful because she's made out of balloons. Yeah. And she might pop on the way over to your new place. That's good timing. Yeah. You got that in writing. Hey buddy. You want hey I got to make sure a white face paint. In fact using the basis S. A. Is represented by the diagonal matrix D. Yeah. Which is the diagonal matrix with entries 7 -4. Yeah. Uh Best friends, All of us. Okay. We're so close. Um When you guys you guys uniforms Then in part b. were given the Matrix 1, -1 2 -1. No that's it. Which we call B. But Jack in the box. That's not a clown. No, it's literally a jack. Once again, we'll find the characteristic polynomial. This is a two by two. Matrix. The characteristic point of the old delta of T is t squared minus the trace of B which is zero tense T plus the determinant of B which is mm Positive one. But notice that there are no real roots. Think of. Yeah, I guess so. But maybe technically besides you. Therefore it follows that B is a real matrix, no Eigen values and no I don vectors. It's funny, it's funny that that's the worst one and therefore it follows in particular that B is not diagonal. Izabal. Alright finally, in part C were given the matrix five negative 113 first or find the characteristic polynomial delta T. This is a two by two matrix. So this is T squared minus the trace of C, which is eight times T plus the determinant of C which is 15 minus negative one or plus 16, which we can factor as t minus four squared. Therefore it follows that the roots lambda equals four is the only Eigen value of our matrix C. Now to find Eigen vectors will subtract forward down the diagnose or C. To get our matrix M. So M is C minus four I. Which is the matrix one negative 11 negative one corresponding to the linear system. X minus Y equals zero. Homogeneous system. Yes, has only one independent solution. For example, we could take The equals 1 1 as a solution and therefore this is an Eigen vector. Especially stop, you should be looking right now but I'm gonna you're listening to Adams deal corner of C. There are no other Eigen values. Therefore it follows that the set S. Which is the Singleton set 1. 1 is a maximum set. Yeah, a linearly independent Eigen vectors of C. But notice that S. Is only one vector and therefore is not a basis of our two. Therefore it follows that our matrix C is not diagonal. Izabal luxuriously, now that I have the walls, you should, I think. Do you know what kind of mattress you fucking with? I have a full already have a full. You should clean. You're an adult, you know?