2. Consider the matrix2A2Find the eigenvalues and eigenvectors_ Confirm the Sum and Product formulas of eigenvalues: A1 + A2 = tr A and A1A2 det A_ Diagonalize A Fi...

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 $B^{2}=A$

This question covers topics relating to the linear and running system of linear equations. Okay, But we will start from forming a new matrix. Vice of attracting lander for the from the organo interests of a given matrix. Right? So if we do you do that. You are going to get two minus lambda one and one negative lambda minus two. Okay. Right. And then you are going to find determinant of the obtained matrix. Right? So determinant of this matrix and they noted by a minor lumber I write. So this name is, it's called a a man Atlanta. I is going to be λ Square -5. Right? And then you solve the quadratic equation and then you get to judge right, London one equal to negative credit. Fine and London two equals the square root of five. So these are the Eigen values right? These ah I can values Okay, next, we are going to find the Eigen vector. Right? All right. So how do we do that? So for first I got valued the negative square to the five. You should lock in to have the matrix. Right? So the matrix is two plus quarter to five. One and one negative two plus quarter to five. Right? And you compute a noon space of this matrix and the noon of this matrix is So after computation, you are going to get this right And similarly you do for the second Eigen value. And finally you will get the new the new spy of the agon Matric? S the after two plus square root of 5 1. That's it

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.