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[3 ~1 1 1. (5 points) Let A = 1 1 1 1 ~1 3 Find the characteristic polynomial of A. b Find the eigenvalues of A: C. Find a diagonal matrix D and an invertible matri...

Question

[3 ~1 1 1. (5 points) Let A = 1 1 1 1 ~1 3 Find the characteristic polynomial of A. b Find the eigenvalues of A: C. Find a diagonal matrix D and an invertible matrix P such P-1 AP = D.

[3 ~1 1 1. (5 points) Let A = 1 1 1 1 ~1 3 Find the characteristic polynomial of A. b Find the eigenvalues of A: C. Find a diagonal matrix D and an invertible matrix P such P-1 AP = D.



Answers

Let $A=\left[\begin{array}{rr}2 & -1 \\ -2 & 3\end{array}\right]$ (a) Find eigenvalues and corresponding eigenvectors. (b) Find a nonsingular matrix $P$ such that $D=P^{-1} A P$ is diagonal. (c) $\quad$ Find $A^{8}$ and $f(A)$ where $f(t)=t^{4}-5 t^{3}+7 t^{2}-2 t+5$ (d) Find a matrix $B$ such that $B^{2}=A$

Were given a matrix A A has elements to 213 Yeah. Well no that's in part they were asked to find all the item values and correspondent Eigen vectors of this matrix. Well, first we'll find the characteristic polynomial of A. So delta T. This is because this is a two by two matrix T squared minus the trace today. Oh okay. Times T plus the determinant today which is T squared minus five T plus four Which you can factor as T -1 times T -4. Now the roots lambda equals one and landed equals four of delta T. These are in fact the Eigen values of our matrix egg. Now let's find the correspondent Eigen vectors. So first we'll find the ones corresponding to lambda equals one. We'll subtract one down the diagonal of A. So we get the matrix M which is a minus, I beat a black man with it. And the corresponding homogeneous system m times X equals zero. Will yield the Eigen vectors belonging to land equals one. We have the matrix M is 1, 2, 1, 2. And to the corresponding homogeneous system is X plus two, Y equals zero, X plus two, Y equals zero. Which of course is just the same as X plus two, Y equals zero. So we have one degree of freedom and one of the non zero solutions is if we take X to be too and Y to be negative one. So for example, V one, the matrix two negative one is a non zero solution and therefore it's also an Eigen vector belonging to Eigen value lambda equals one. Now consider the Eigen value land equals four or subtract this down the diagonal of a. To get the matrix M. This is M is a minus four. I which gives us the matrix negative 2 to 1, negative one. This corresponds to the homogeneous system negative two, X plus two, Y equals zero and x minus Y equals zero, which corresponds to the system X minus Y equals zero, which has only one independent solution. So for example, you can take X and Y to both be one and therefore the two, which is the vector 11 is an Eigen vector. And then they crossed that belonging to the iron value. Lambda equals four. All right, then, in part B, we were asked to find a non singular matrix P such that the matrix D, which is P inverse times ap is diagonal, and also find the matrix p inverse, which we know exists because P will be non singular. Well, but PB the matrix whose columns are are Eigen vectors V one and V two. Then P is the matrix two negative 111 and the matrix D, which is P inverse. Ap Well, this is simply the matrix whose diagonal entries will be the Eigen values corresponding Megan values. One and four. Now mm key inverse. We can find simply using our formula for the inverse of a two by two matrix. So the determinant of P is two minus negative one, which is three. So we have one third times one -1, 1, 2, which is one third negative one third, one third two thirds. I love that. The answer. Just the right thing. Karma police in part C. Whereas to find A to the sixth power and F of a where F is given by the certain polynomial out of F. F T equals T to the 4th minus 32 huge minus 60 squared plus 70 plus three. That's nice. That's the type of rodeo. We'll use a diagonal factory Ization for a so we have that 8-6. Well, rearranging This is the same as uh p. d. p inverse to the 6th. Which we know is the same as P times d. to the 6th times p inverse. And plugging in. This is uh to one negative 11 times 100 So one of the fourth is 100 And then uh sorry, four to the sixth is 4096th. Real rodeo. Queer. Yeah, um mm Times the matrix. Uh One third negative. One third. One third, two thirds. And if you carry out these matrix multiplication, you'll get 1366 2230 1365 And 2731. Right? So this is the value of the age of the cysts. Now look at our function F. We see that f of one is equal to one minus three minus six plus seven plus three, which is positive two and F of four on the other hand, well this is negative one and therefore we have that. They once again, this is the same as F of P times D times p inverse, which is the same as P times F of D comes p inverse which is equal to P, which is the matrix to one negative 11 times F of D. Which is well, it's a diagonal matrix. We simply evaluate the diagonals that. So F of one which is 200 and F of four which is negative one times P inverse, which we found to be one third negative one third, one third and two thirds. And this simplifies after matrix multiplication, 2, 1, 2 negative, 1, 0. Shit. I've seen it happen literally three or 4 times that's F. Of a. Finally in part D were has to find a real cube root quote unquote of B. Which is well defined to be a matrix B. Such that be cubicles A. And B has real item values list about a half a dozen. Mhm. Put it out for I want them so much. He keeps saying birthday boy. Yeah. Happy birthday. Well we see that the Matrix 100 Hubert or four. This is a by our definition real Q. Root of our matrix D. Since clearly the diagonal matrices is just the matrix whose diagonal entries or products. And it has really good values. And therefore by rearranging a real key brute of A. E. Is equal to our matrix P times uh Eight times p inverse that were taking the key brute of it. So this is key times again distributing the power, the cube root. Sorry? So initially D so cube root of D tends to the universe. And so this is equal to matrix P, which is 21 negative 11 times 100 cubed root of four times one third negative one third one third two thirds. And if you evaluate this matrix multiplication helps if you pull out a third, if you get one third times two plus the cube root of four negative two plus two times the cube root of four negative one plus the cube root of four and one plus two times the cube root of four came up with. And and this is the real cube root of a or a real keeper today.

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.


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