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Find an invertible matrix P and a diagonal matrix D such that P-I _ AP-D_ -14 -12 -8 A = 14 24 18...

Question

Find an invertible matrix P and a diagonal matrix D such that P-I _ AP-D_ -14 -12 -8 A = 14 24 18

Find an invertible matrix P and a diagonal matrix D such that P-I _ AP-D_ -14 -12 -8 A = 14 24 18



Answers

find an invertible matrix $P$ and a matrix $C$ of form (15) such that $A=P C P^{-1}$. $$A=\left[\begin{array}{rr} 8 & 6 \\ -3 & 2 \end{array}\right]$$

Hello there. So for this exercise we got this matrix A. And we need to find the matrix B. That diagonal eyes these matrix. So that means that we can find some matrix P. Such that we can grade a diagonal matrix equals two. Be in verse A and B. Of course beans should be in vertical. So just to remind you how you should proceed in this case is when you require a diagonal ization of matrix. First step is to calculate the icon values. And for that I mean we got this system where X is a vector and lambda is just some real number. Okay, so from this system we obtain this here. I note the identity matrix times X equals to zero. So we should find a solution for the system and we should find the non trivial solutions. So that means that we need the determinant of this matrix A minus lambda I. And should be close to zero. So that actually corresponds to the the characteristic polynomial of these matrix. So the characteristic polynomial is just Okay, there is some notation here I'm going to use is this this triangle? And it's a function of lambda. So how to construct this? This polynomial will is just the determinant of the system. Okay, so in this case is just taking the determinant of -14- Lambda 12 -20 and 17- Lambda. These are terminated is equal to lambda plus 14 times lambda minus 17 Plus 214 And they should be equal to zero. Okay, so that characteristic Polynomial is equals two, λ -1 times λ -2. After we fact arises and we need to find the rid of this point of view. So the druids means just taking equating this to zero and you can see it here. We got already the solution. So lambda one is equal to one and λ two is Equal Institute. So these are the Eigen values. These are going to be the matrix that we're going to put in the diagonal. So here we are going to put on the one the to this corresponds to the diagonal matrix. Now we need to find the matrix speak and the matrix B. It's going to be the matrix composed by the Eigen vectors. Uh each kilometer into corresponds to one Eigen factor. So these I can vector here is correspondent to this idea in fact, so let's go to it. These these Eigen vectors. And so let's start with λ one. So this is the second part calculating the Eigen vectors. So we consider one of the alien values. In this case, I'm going to consider this lumber one. So we got this system a minus lambda one, Her identity. And hear some vector x equals to 00 vector. So we need to find a solution for this system. So this in this case is equivalent two 15 minus 12 minus 15 20 and 16 Times X- one X 2. That's going to corresponds to the solution. This equals to 00. Okay. And then we need to stop the system. You can. Well we make these systems singular because it determines going to be zero. So the Israel or these, we're going to be a linear combination of the other one. And we obtain the following. You obtain here five minus 400 times X one X two equals to the zero factor. And the solution for this system is actually is a set of solution because we got one free viable. And the solution is written in this way. Some free variable tee times four and five. Okay. If we take any any any value for T. Here and we multiply this vector here and we put this X onto the system. You're going to obtain the solution 00 So this corresponds to the first Eigen value for far. Okay, now let's continue. So just to put the things here for lambda, one equals to one. We got the Eigen vector 45 and now we need to calculate the second Eigen value. Eigen vector. So for λ two equals to two. What's going to be the second factor? Well, we repeat the process, we get a minus in this case slammed to the entity here. The vector X equals to the zero vector. This becomes okay minus 16, 12 -20 and 15. And here X one X two equals to zero. Who is it? So let's find a solution for the system. The system reduced to four minus 300 And here X one X two zero. So here again we got a free variable so we can put any any and a free variable. And you can see that the solution for this system is going to be t 3 4. So this is the second solution. Okay, So for land to the Eigen vector corresponds to three four. So great. So what we got so far is that For the metrics a equals 2 -14. Yeah. Right. Oh, 21 minus 20 17. The wagon values are one to the associated Eigen vectors are for yeah, equipment five. And the second one is 3 4. So with this quantities here we can construct these matrix speak. So the metrics be, as I mentioned, corresponds to the Eigen vectors, uh putting in the same order that appear the Eigen values in the diagram. So if I put the diagonal number one, number two, The first column, I should I should put the first Eigen vector, so that's 45. And for the 2nd 1 that is in the second column, Then I could hear 34 in the second call and there's a matrix speak. Now we should check that. Indeed We can grade the equals two, be the members of B times A and B again. And that is actually the same, is equivalent to taking P times Z equals two A. B. So that's how we're going to check this. So be D is equal to 4, 4 Times the econometric, that is 1002. And this is equal to 46 58 on on the right hand side. We got that A. B. And that is equal to -14, -2017 time four, Uh huh. And these multiplication give us Again for 658. So yes, they are equal. So that means that we can organize these matrix A. Using these matrix here.

Hello there. Okay. So in this exercise we have this matrix A Equals 2 -1 -5, 4 and seven. And we know that these kind of matrices can be greeted as the multiplication of an orthogonal matrix B times and a matrix of this structure A minus B. B. And A. Um the members of N P is given us the real part of the one Eigen value of this matrix A. With well this this matrix B has as columns the real part of one of the wagon vectors of A. And on the other column we have the the imaginary part of that wagon factory. So we need to construct this and these uh these matrix in the middle that has coefficients A. And B corresponds to the coefficients that appear on the Eigen values of A. So let's suppose that lambda is the Eigen value of A. Then is it has this structure eight plus minus I B. So from from here is that we obtained these matrix. So the first step is to calculate the Eigen values of the matrix hey, and to calculate the Eigen values of A. We need to compute the determinant of lambda. The identity matrix in this case is uh identity matrix two by two minus the matrix A equals to zero, but is equivalent to taking the determinant of the matrix lambda plus one, five minus four And λ -7. This determinant give us the characteristic pulling a milk for these matrix and is equals to love the square minus six. Lambda plus 13 equals to zero. And you can observe that here we have a quadratic equation and we can find the roots of this, this polynomial and the roots corresponds to Eigen values. There are complex. So the iron values in this case are three Plus 2 i. And of course the the conjugated, I can value that is 3 -2 i. Right? So from here we we we can observe the A. It will take the will take the value of three and B. It's going to take the value of -2. If we choose this, I can value of course the first one. Okay, so we have the value of A is equal to three. Vz equals two minus two. And from that we can construct the middle uh metrics that I'm going to label A. C. And this matrix C is equals to a -7. B. A. Which in this case for that particular agent value is equal to three to minus two and three. Great. And now we need to construct the metric speed. So for the second part we need to construct the metrics be but for that we need the Eigen vectors or at least one Eigen vector. Okay, so one Eigen vector of a. So let's choose the value of lambda equals 23 plus two. I. So to obtain the again the Eigen vector, we need to find a nontrivial solution for the system, Lunda identity matrix minus A. X equals to the zero vector. But there is equivalent. Two, solve the system four blasts to I five and -4 minus four plus two. I times the X factor. That's going to be X one X two equals to the zero factor here. Okay, so we have this system, this linear system and we need to find a nontrivial solution. And we we reduce um if we try to put here a zero, following the Belgian elimination procedure, we obtain the matrix four Plus two, I 5, 00 And extended matrix will look like this. And from this we obtained that the Eigen vector is Equals 2 -2 plus I and two. Okay, so there's the Eigen vector associated to this Eigen value under three plus two. I now we know that at the beginning I write that the matrix speed is equal to the columns given by the real part of the Eigen vector and the imaginary part of the Eigen vector of A. So that means that in this case this matrix B is going to be minus 2120 And the last thing that we need is the members of this matrix. So the inverse of B in this case is equal two. The matrix zero. One, half 11 Okay, so there's a matrix, the inverse of P. And now we have all the required. So matrix A is equal To the multiplication of the Matrix B -2, 1, 20. The matrix C, That is the matrix given by the coefficients of the Eigen value that is three 2 -2, 3. And the embers of being That is equals to 01/2 11

Hello there. In this case we have these matrix Equals to 4 -5 1 and zero. And we need to write this matrix A. As a multiplication of three matrices P. Which is an orthogonal matrix formed by the real part of the Eigen Eigen vector of X. One of the Eigen vectors of X. And imaginary part of the same Eigen vector the matrix B inverse which is just the inverse of the matrix peak. And these matrix in the middle that has the coefficients that constitute the I can value of the matrix A. So there's the structure of the organ values of A. So we're going to choose the coefficients A. And B are going to organized in this way on this matrix. Okay so the first thing that we need to calculate are the Eigen values of A. So for that we need to obtain the characteristic polynomial that is equal to lamb, the I minus a equals to zero. But this is just taking the determinant of the matrix, lambda minus four. 5 -1. London equals to zero. But this is equal to the characteristic polynomial. That in this case is lambda square minus. For lambda plus five equals to zero. The two solutions. Our response to the Eigen values of these matrix and the Eigen values are complex. So in this case two plus I. And of course the conjugated two minus I. So if we consider here the The same value λ equals to two plus I. Then we obtain that A. Is Equal Institute and B is equal to -1. And from this we can construct the middle matrix that I'm going to label it, see that is equals two A minus B. B. A. So if we replace the values here we obtain the matrix to -1, 1, -1 and two here. Great. So we have the middle metrics. Now we need to construct the matrix B. So for the second part we need to obtain the Eigen vectors. Okay? We only need one, but the other one is obtained just by taking the conjugated. So let's consider the Eigen vector associated to the golden value. Lambda equals to two plus I. So that means that we need to find a nontrivial solution for the system. Lambda, I minus A X equals to the zero factor and homogeneous system from this, we obtain the linear system with matrix minus two plus I five minus one to plus I hear the vector X one X two and equals to +00 But this a week. Right? This as a extended matrix, we obtain two plus minus two plus I five years 00 If we try to eliminate the element on the position to one, we obtain 00 here, zero and zero. Okay, Otherwise we're going to obtain a trivial solution and that means that we calculated wrong. Diegan ph values. And the solution for this is that the Eigen vector Is given by two plus i. and one. Now how to construct this? This matrix B. Well, the matrix B is equal to the real part of one of the Eigen values. Again, vectors Together with the imaginary part of the picture. Okay, so the the real part of the second vector is 2, 1 And the imaginary part is just 10. So there's a matrix p. And the last thing is just taking the members of this matrix and the inverse in this case Is a Matrix 0, 1, 1 -2. So great. We have all the matrices that we need. So that means that we can write or matrix A. As the multiplication of the Matrix 2110. That is the Matrix B. The matrix, see that was the one that is in the middle. That is given by the the coefficients of the asian values that R 2 1 -1, 2 and the members of P. Okay, so it's 01, -2. This beginners, and there's a matrix seat.

So in order to see if this matrix a equals 13 minus 9 25 minus 17 is diagonal Izabal. We need to find that I can rallies of a But in order to do that, we need to find the determinant of a minus lambda times. The identity matrix said that equal to zero and sulfur lampa. So the determinant will be 13 minus lambda minus 9 25 minus 17 minus lander. So what set this equal to zero? So taking this determinant, we have 30 and minus lampa times minus 17 minus lambda minus negative. Nine times 25 equal standard. So multiplying this out, we have minus 2 21 minus 30 in Lambda Plus 17 Lambda plus Lambda Square plus 2 25 equals zero. So combining like terms, we have slammed a squared plus four lambda plus four equals zero. Now we factor this will get lambda plus two squared equals zero. Therefore, our Aiken value will be lambda equals negative to with the multiplicity of to now, in order to see if the matrix is diagonal sizable. What? We need to find a basis for the Eiken space for Lambda equals minus two and we can do that by taking the null space of the Matrix a minus minus two times the Identity Matron's, which is just no a plus to ah, so we need to find an all space of the matrix 13 plus two minus nine 25 minus 17 plus two. So this will be 15 minus nine 25 minus 50. Now we divide the top of I 15 We'll get one minus 3/5, 25 minus 15. And now we multiply the top row by 25 subtract this second row from it. We'll get one minus 3/5 00 Now we're called up the notes. Face of a matrix is the set of solutions to the equation. A X equals zero. So the north space of this matrix we'll just be exit to times the vector 3/5 want. Now we see that the Eiken space of the matrix has a dimension of what? Since there's only one factor, however, we found one I can value with the multiplicity of to That means that in order for the matrix to be diagonal Izabal, that Eigen space had to have a dimension of tube, which means that we needed to have two factors


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