in-exercises-13-20-find-an-invertible-matrix-p-and-a-matrix-c-of-the-form-leftbeginarrayrra-b-b-ae-6

Question

In Exercises $13-20,$ find an invertible matrix $P$ and a matrix $C$ of the form $\left[\begin{array}{rr}{a} & {-b} \ {b} & {a}\end{array}ight]$ such that the given matrix has the form $A=P C P^{-1} .$ For Exercises $13-16,$ use information from Exercises $1-4$ .
$$
\left[\begin{array}{rr}{1} & {-1} \ {.4} & {.6}\end{array}ight]
$$

Runpeng Li · Accepted Answer

So for problem 18 were given the matrix A to B one negative. One 14.6. Okay, so first thing, we need to find out the Yagan vectors, uh, good manners and taken pictures. So we apply the determinant of a minus land. I I was keep the calculation here. So zero, and that implies our Aiken values are or 50 plus or minus 3 50 Bye. You just take take take London to be 4/5 minus You came by and the corresponding liken vector will be one have minus three. If I and what get so bye bye. Observing this psychic Victor, we take a to B 4/5 and B two b 3/5. So our a trick See before fifth connective three and 3 to 4. Fifth and our metrics P first, the real the real part of this again. Victor one have one, then the imaginary part Collective 3/2 asthma, zero

Michael Jacobsen · Answer

for this example, we&#x27;re going to be illustrating the the&#x27;re, um that&#x27;s provided here with our particular matrix A. And for this matrix, we know that Lambda 12 is two. Plus, I is an Eigen value, and the corresponding Eigen vector is V one with this, the&#x27;re, um and the Given matrix say our goal in this problem is to determine the Matrix c indicated here as well as the Matrix P that gives this factory ization for a Let&#x27;s start off with the Matrix p. To do that, we&#x27;re going to focus on our Aiken. Vector at right is as follows. We can say that V one is equal to the rial part of E plus the imaginary part of E. And that would be first the real part. Let&#x27;s put a night I hear is well, for the real part. We focus on these values that are not coefficients to the complex. You know, I So the real part is negative one and one that we add to that the imaginary parts of the complex number. For the imaginary part, we pull the coefficients of the complex unit I, which is a one here and from here we could say that the zero there is a zero times I to give us a 10 Then don&#x27;t forget to copy the complex unit I. So this is what V one is equal to. And then fire the&#x27;re, um here. We see that p is going to be first, the real part of E which is this vector? Negative 11 then the imaginary part of E. But that&#x27;s the specter here at 10 So we found that Matrix P and what we can do to find the Matrix C is used this theory and provided here so or that portion of here, Um, we take C, and we&#x27;ll also go to the Eigen value That&#x27;s provided Lambda One is of the form of a plus B I. But in other words, we have a is to and B is one, and those values go inside here, so we&#x27;ll have a 21 than a negative one and a two. So this is the Matrix P and the matric See that allows this matrix factories, ation, toe work

Runpeng Li · Answer

Okay, So the matrix in problems 16 is given from all our problem or yeah, home for So it is five naked too. 13 Now, remember, our reviews result did There is a There&#x27;s Agon value, which is for minus I and the corresponding I get victory would be to one, plus I to one of us. I So our matrix p will be theirs to re apart to one that he measuring heart 01 and our metric see will be so here, Ace. For these one, we have four next, 11 and four soldiers.

Michael Jacobsen · Answer

in this video. We&#x27;re going to be illustrating the present there, missy. Here for when we have a complex Eigen value of the form A plus B I, and it gives us a matrix factories ation in this way. So our goal here is to determine the Matrix C as well as the matrix p. For this given matrix, this matrix has Eigen value negative 0.6 minus point ai. And that tells us from the Aiken value portion the A is negative 0.6 and B equals negative 0.8 the imaginary part. So with this information, we&#x27;re ready to determine the Matrix C See, for column one involves just and be, which is negative 0.6 and negative 0.8. There for the second column, we take negative B and then a soul have positive 0.8 and a negative 0.6 below. So that&#x27;s her matrix C. And now to find the Matrix P, we need the rial and imagine a part of our given vector Aiken Vector V here. So the real part of this Eigen vector V is equal to first Just take the real value. The rial part of each of the entries which is 0.37 and 0.91 Then for the imaginary part, if you like, we can write down plus zero times I just to help out, and we find that the imagine a part of the We just take the coefficients, which are 0.18 and zero. So that&#x27;s going to be this column vector here. But now P once again involves a real parts in the imaginary part. So we have found that P is equal to first the real part, which is 0.37 and 0.91 then put in a second calm with the imaginary part, which is 0.18 and zero. So that&#x27;s the set of matrices, p and C such that a factors in this specific way.

Michael Jacobsen · Answer

in this video, we&#x27;re going to be illustrating the serum that&#x27;s provided here, which gives us a very useful matrix factory ization. Arkle here is then to determine the Matrix P as well as the matrix C in that factory ization where we have a matrix A already provided for this matrix. Say we have Eigen value here as well. And given the form of a plus B, I, we see that a is 0.96 and B is 0.28 This is already enough to tell us what the Matrix C should be. From this information, C is equal to first 0.96 and 0.28 for the first column. Then take negative 0.28 the negative B divide by 0.96 and this forms are matrix C that is here in the factory ization. Now let&#x27;s turn our attention to the Matrix P. We first need the real part of the Eiken Vector V that&#x27;s associate toe lambda and the real part involves thes two coefficients. So say that the real part of E is equal to 0.75 and 0.60 likewise, for the imaginary part of of V we can put in a plus zero times I and we just pulled this coefficient and this coefficient along with a negative sign for its imaginary part so we can write imaginary part of the vector V is zero negative 00.30 Then, from that information, we combine the two vectors and we see that P must be equal 2.75 point six. So then zero negative 0.30 And these are the two required maitresse ease CNP to obtain this factory ization.

Michael Jacobsen · Answer

for this example, we&#x27;re going to be illustrating the the&#x27;re, um that&#x27;s provided here with our particular matrix A. And for this matrix, we know that Lambda 12 is two. Plus, I is an Eigen value, and the corresponding Eigen vector is V one with this, the&#x27;re, um and the Given matrix say our goal in this problem is to determine the Matrix c indicated here as well as the Matrix P that gives this factory ization for a Let&#x27;s start off with the Matrix p. To do that, we&#x27;re going to focus on our Aiken. Vector at right is as follows. We can say that V one is equal to the rial part of E plus the imaginary part of E. And that would be first the real part. Let&#x27;s put a night I hear is well, for the real part. We focus on these values that are not coefficients to the complex. You know, I So the real part is negative one and one that we add to that the imaginary parts of the complex number. For the imaginary part, we pull the coefficients of the complex unit I, which is a one here and from here we could say that the zero there is a zero times I to give us a 10 Then don&#x27;t forget to copy the complex unit I. So this is what V one is equal to. And then fire the&#x27;re, um here. We see that p is going to be first, the real part of E which is this vector? Negative 11 then the imaginary part of E. But that&#x27;s the specter here at 10 So we found that Matrix P and what we can do to find the Matrix C is used this theory and provided here so or that portion of here, Um, we take C, and we&#x27;ll also go to the Eigen value That&#x27;s provided Lambda One is of the form of a plus B I. But in other words, we have a is to and B is one, and those values go inside here, so we&#x27;ll have a 21 than a negative one and a two. So this is the Matrix P and the matric See that allows this matrix factories, ation, toe work

Problem

In Exercises $13-20,$ find an invertible matrix $…

Problem 18 Medium Difficulty

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

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