8,.2.1. Find the eigenvaulues Ad eigenvectors o the following Iatrices: (4 -) (b) 6- (~3 H), ( (~; 3), 614 (39o ( ! !...

For each of the following matrices, find all cigcnvalues and corresponding linearly independent eigenvectors: (a) $A=\left[\begin{array}{ll}2 & -3 \\ 2 & -5\end{array}\right]$ (b) $\quad B=\left[\begin{array}{rr}2 & 4 \\ -1 & 6\end{array}\right],(\mathrm{c}) \quad C=\left[\begin{array}{rr}1 & -4 \\ 3 & -7\end{array}\right]$

This problem Give Gives us a matrix and asked us to find its Eigen values and Eigen vectors. We do this by first taking the characteristic polynomial, which is found by the determinant. Ah, a minus land I, which equals zero. So that gives us the determine of matrix seven minus lambda a zero negative eight negative nine minus lambda zero and 66 Native one minus Linda Um, and that themselves out to be will be 17. Slander The Times Negative nine months Lambda Times native one minus lander plus 64 times negative one month's lambda which solves out to be Lambda plus one cute meaning. And since the should equal zero, that means our Eigen values equal to negative one with an ultra break Multiplicity three. In order to sell for the Eigen vector, we need to sell for a vector X such that a minor slammed I times that Eigen Vector X equals zero doctor for playing. So we're gonna plug in land equals negative one. Doing that gives us the matrix 880 negative eight negative 80 660 times X which is going to be equal to if you subtract this the first surfing the second row and then divide by two. We get the Matrix for negative for three and then zeros all on the balm to rose times a B C, which is equal 000 And from this we see that for a minus four B plus three C should equal to zero. And there's only going to be to, uh, matrices that will that are literally, literally independent. That will satisfy this. And those major cities are 110 and negative 30 four. Giving us our Eigen vectors for the Ivan Value. Negative one, and that's our final answer.

Were destroyed. So. Yeah. Whereas your people previous exercise 9 14 for a new matrix T. So we use the three x 3 matrix with entries 3 -1 1 7 -5 1. I mean I'm sure And 6 -6 2. No shit in part A wants the first time. The characteristic polynomial of B. To do this. We need a few things trace of our three by three. Matrix B is three minus five plus two or zero. Be determined of our matrix B 78. Well this is going to be wow negative 30 -6 -42. Yeah the 70 73. We talked plus 30. The 73-7 plus 18 plus 14. So yeah visit old caprices. We got headlights. This is a -16. We can see where you have a bubble bubble caprices like it's it's kind And we want to find the co factors be 1 1 which is the determinant of negative 51 negative 62 which is negative four 73- two. This is the co factor of 3162. Uh huh. Those are the cars those cars like which is zero worth. And the co factor B 33 is the determinant. Uh 3. -1 7 -5. But like which is not great. Get into the car. Therefore it follows that the sum of our co factors B II is negative 12. No, We know quickly by three. Nature sees our characteristic polynomial. Delta is given by t cubed minus the trace 12 times T plus the sum of the co factors negative 12. I'm sorry CQ minus the trace zero times T squared Plus some of the co factors negative 12 times two minus the determinant to be. Now you're 16 plus 16 and he ran equal. This is a cubic polynomial factor. This well notice that if delta has rational roofs, then it follows by the rational root theorem that has to be of the form plus or minus one. Plus or minus to plus or minus four plus or minus eight Or across your -16. If you're lucky you'll test two. You can sometimes division. So are coefficients are 10 negative 12 16, two times 1 is two, two times two is four negative eight seems negative, eight is negative 16 and zero. We have a remainder of 02 is a root of our characteristic polynomial. And therefore we can write a characteristic polynomial as delta t equals t minus two times the remainder T squared Quotient I should say plus two T -8. Which we can factor the quadratic and we get t minus two times t minus two to minus two squared times t plus four. See Yeah, now the zeros Lender one equals 2 And landed two equals -4. These are the Aydin values of our matrix B. I'd like to see, I'd like to see it. Dude chuck. Now in part B. Just as in the previous exercise. Mm hmm. We want find a maximum set s of linearly independent Eigen vectors of the Okay, this is in the previous exercise, we're trying to find the basis for the Eigen space of each item value of the It's working in the first Eigen space lend the one equals two. Mhm. Like, well, we're gonna subtract two down the diagonal of be. So our matrix M is b minus two. I I don't know. And this gives us the three by three. Matrix one negative 11 seven negative seven suck one and 6 -6 zero. That's a that's right. And this corresponds to the homogeneous system, X minus Y plus Z equals zero, seven, X -7. Wine Prince Plus z equals zero And six X -6 by Equal zero. And this simple system simplifies to X minus Y plus Z. Our first equation equals zero. And then the second equation is simply Z equals zero. Only here it's a bit. So we already see that any solution has to have Z equals zero. So we only have one independent solution. For example, take X equal one in life. Then it follows that the doctor you With coordinates 1 10. This is a basis. Siebert For the idea in space of λ one equals 2. Right? Yeah, comments are funnier than me and most Yeah. Now consider the other icon space for lambda two equals negative four. So we subtract negative four down the diagnosed B. To obtain M. So M. Is the matrix B plus for I. This gives us the three x 3. Matrix 7 -1 1 7 -1 1 she's And 6 90 of 6.6. She kept coming around after he found open. Mm This corresponds to the homogeneous system. Seven X minus Y plus Z equals zero. Seven, X minus Y plus Z equals zero and six X minus six. Y plus six Z equals zero. This system reduces to the two equations X porn site. All right minus Y plus Z equals zero. You divide the third equation by six and You only like the fucker for like 30 seconds. six Y -6. Z equals zero. So you do this. I just most pain. Yes. This one looks like eliminated the 2nd and 3rd equation right through my body. Yeah. Yeah right possible. And we see that the system actually has only one independent solution. So take Z to be one. You see that? Why must be one And the X must be zero. So we get the solution Z. Which is 011. Send somebody down here. And since the only independent solution this forms a basis for the Eigen space of land of two, which is negative. four garden hose in his ass. Therefore it follows the process the set S with vectors U. And V. You just Or 110 And 0 1. 1 elegant. This this is a maximum set like shows both P. T. Barnum of linearly independent audience. I've been vectors for B. one of them was where the what is it? Finally in part C. Just as in the previous exercise we're asking he is diagonal, Izabal. And if so to find a matrix P. Such that the diagonal make the matrix D, which is p inverse ap is diagonal. Okay, these are the current PM for cps diagonal on me. Sorry, some time. Well, we see that B has at most two linearly independent Eigen vectors. It's you and the. Yeah. Right. And so like you think like man people needs to be fucked up that they thought that was like a hand. Therefore it follows that our matrix B is not similar to a diagonal matrix, which is the same as a definition that B is not diagonal. Izabal War. Yeah. It's like, oh yeah, nobody fell for them. They all knew like this is just a retarded.

This problem asked us to find Ivan values and Aiken vectors of given matrix. You can do this by funding the characteristic polynomial, which this inviting the determinant of a minus land of times. I That would be the determinant of 10 minus lambda zero negative. Eight. Native 12 to minus lambda 12 80 96 minus Landau. Okay, solving this out. We get 10 minus lander times two months. Slammed other times a night of six months. Lambda minus eight times, two months. Lander times negative eight. And then the other two for, um, the subtraction are zero spoke due to the zeros in the Matrix Southern. This out a bit more we can get. Um well, we find that we get negative 60 months for Lambda Plus Lambda Square plus 64 0 sorry. We need to include the two months slammed up minus it becomes a plus 64 to minus slammed up, which simplifies to because you can add the 64 in yet for minus four lambda plus lander squared times two minus lambda, which is equivalent to tu minus lander. Cute. Then with that, that means that our Eigen value is two with an algebraic multiplicity of, um three. So this means that the Matrix could have or this AG value could have three Eigen vectors associated to it, but a maximum of three. So solving for the Eigen vectors, we confined this by finding anyone slammed I and the final director X that gives the zero vector so plugging in to, yeah, 10 minutes to wanna leave blanks for zeros for save time. Negative. Eight. Um native 12 to minus two 12 and then eight. Negative. Six mines, too. Times the Eigen Vector X that is equivalent to eight negative eight and 12 0 12 Negative eight Over here. Eight. When we add the zeros to this room, then with gash in elimination, if you add the top row to the bottom row, that gives us eight negative 12 8 Then zeros on both these bottom row's times X, which is equivalent to if you divide by four. That gives us two negative 32 all zeros against here times the vector A B C shaped equal 000 and without weekend fronts literally Independent solutions, ABC to the system of Earth to the Single equation to AG minus three b plus two C is equal to zero and she saw this. You'll find that the two linearly independent ah solutions you'll find from this are going to be like 101 and three 20 meaning that this adding value has geometric multiplicity of two. So this is our final solution for Agon vectors and our Ivan value is what it to