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DETERMINE ThE ElGEWVALUES ANp A BAsis For EACx ElGENSMcE OF '3-8-3DETERNINE IF A 15 Diagonalizable...

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DETERMINE ThE ElGEWVALUES ANp A BAsis For EACx ElGENSMcE OF '3-8-3DETERNINE IF A 15 Diagonalizable

DETERMINE ThE ElGEWVALUES ANp A BAsis For EACx ElGENSMcE OF ' 3 -8 -3 DETERNINE IF A 15 Diagonalizable



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Let $T: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}$ be defined by $T(x, y, z)=(2 x+y-2 z, \quad 2 x+3 y-4 z, \quad x+y-z) .$ Find all eigenvalues of $T,$ and find a basis of each eigcnspace. Is $T$ diagonalizable? If so, find the basis $S$ of $\mathbf{R}^{3}$ that diagonalizes $T,$ and find its diagonal representation $D$

Okay, so for this exercise we got these matrix and we need to calculate the geometric and the break multiplicity of each of the Eigen values of these matrix. So let's calculate the Eigen values of the matrix. So that means that we need to consider the associated characteristic polynomial. That means taking the determinant of a minus from the And the identity matrix equals to zero. That is equal to take in the determinant of -1 minus lambda for minus two. Horrible mine islam 0.0 minus 313 And my nose this determinant is equal to from the cube -6 Lumber Square. The last 11 lambda minus six equals to zero. And actually we can uh characterize this as follows to islam -1 from the -2 and λ -3 equals to zero. So from this, it is really easy to see what are going to be the values. In this case our number one equals to one lumber two equals two And I'm the three equal to three. So this I can value the pier only once. That means that all these again Eigen values have geometric multiplicity equals to what all of them. Now we need to check the algebraic multiplicity and that is related to how many Eigen vectors associated to that. Again, value we have. So we need to consider each of these Eigen values separately. So what I mean is that we need to take this system a minus on the I the identity. We need to consider this system here and find the solutions. So let's start with number one equals to one. The case this system becomes yeah two minus four. 2, -30. 3 -1. 2 Equals to 000. Okay, so we got the system and we need to solve this. So what we need to do is reduce this to the action form looks like, so we need to reduce this to the asian form and we obtain the following system is 1010, -1000. Oh right, yeah. And here we can find a solution. So from this we got just one free variable that corresponds to this position, that is X three. So it's three, we are going to call T and X one, It's going to be equals two -T. The next two is going to be t. So this solution in this case the the Eigen vector associated to lumber one is equal to tee times -111. Um I'm I'm sorry, here is -1. So yeah, here is right, X one is equal to T. Extras equal city. So we got the vector 111 Right, So we got just one Eigen vector associated to them, The one. So that means that the Algebraic multiplicity of land, the one is equal to one. Oh Now let's continue. Let's go with λ two equals 2. We're going to repeat the procedure. We got a minus the two via entity matrix equals to your I'm sorry. Here is times X equals to zero. We need to find a solution for this system. So I am going to put great this this matrix. So that is minus 34 minus two minus 3 to 0 -3. 1 1 Here, x one x 2 x three Equals zero vector. And then we were we reduce these metrics to the Asian form And that give us one 0 -2/3 0, -1000 X- one x 2 X three equals zero factor. And in this case the solution is again, we got a free variable that is equal to that is X three, So X three is equal to some value in this case at T And X two is equal to T&X one is equals two, 2/3 of T. This means that the wagon the Eigen vector can be written in this way T 11 and two thirds. So we can choose actually entity. In the previous case I just equals to one. But in this case, in order to eliminate this fraction here, I'm going to say that T is equal to three. So the Eigen vector is 2-3 Tories. 3, 3, 2. Yeah, sorry, Is 3, 32 actually here. But in different order Here, the two thirds is first because it's X one on one, so that is 233 This is the Eigen vector associated to this Eigen value and here again the algebraic multiplicity is equal to one and the last Eigen value, It is equal to three. Again here the system London, three I X equals to the zero vector Is equal to -44 -2 -310. And here in -310. Yeah. Mhm. We reduces matrix deviation forum and we obtain 10 minus one half 1/4 0, -3. 4000 x one X two X three equals to the zero factory. So when we solved this again we got the X three is a free bearable is equal to T. X two is equal to oh home uh huh X two eZ equals two 3/4 T. And x one is equals two, 1/4 T. So in this case the Eigen vector is the cost of tee times uh one carter 3/4 and one again here we can choose any t to give the value to the Eigen vector. And that means that this I'm going to take the equals to four and that gave us 134 there's the third Eigen vector. Again it is only one Eigen vector, so that means that the algebraic multiplicity of these Eigen value equals to one. So all the Eigen values have the same algebraic and Geometric multiplicity. That is equals to one. Then we can find in me we can finalize this matrix. Yeah. Oh yeah yeah. And the answer is yes, because the algebraic and geometric multiplicity of each of the Eigen values is the same and how we're going to construct that people is just the um again factor. So but for constructive we need to consider the diagonal matrix. So the diagonal matrix are going to have the iron values. In this case, I'm going to choose 123 is the minimum metrics with the iron values. And in P. I need to put the associated Eigen vectors for each of the Eigen values in the corresponding. Cool. So here is 111 233 And the last one is 134. Um That's it.

Do you like? We're told to suppose that the matrix B in problem 9 11 represents a linear operator on complex space C. Two. Whereas to show that in this case B is in fact diagonal. Izabal by finding a basis S of C two consisting of Eigen vectors of B. Well from problem 9 11. The matrix B Has Elements 1, -1 2 -1. Air walks. Just as we found in the previous exercise. The characteristic polynomial of B is still felt of T equals T squared plus one. Yours as a polynomial over the complex number. It does factor as T -1 times T plus I using difference of squares. And therefore it follows that the roots landed equals I and landed equals negative. I are the Eigen values of our matrix B. Report like an accurate accurate tigers rsx. That was like the next. Okay, now for our first Eigen value, I'll subtract I down the main diagonal of B. So we get the matrix M which is d minus I times I, which is a matrix ah one minus I negative 12 negative one minus I. This corresponds to the homogeneous system, 1 -1 times x -Y equals zero and two X uh plus negative one minus I times Y equals zero. And this reduces to the single equation one minus I times x minus Y equals zero. Now the system has only one independent solution. Take, for example X to be one and why to be one minus I. And so the vector V one with coordinates 11 minus I is an icon vector. It's exactly uh that belongs to in fact spans the Eigen space of land. It equals I. Now for the other Eigen value negative, I we subtract negative by down the diagonal of be. So we get the matrix M which is B plus I times big guy, which is uh What are you doing a bit? All right, well there we go. It's back to normal. So this gives us the matrix one plus I negative 12 -1 Plus I. Which corresponds to the homogeneous system. I just live one plus i times x minus Y equals zero and two times X plus negative one plus I times Y equals zero. This reduces to the single equation one plus I times X minus Y equals zero. And we see that this system has only one independent solution. For example, take X equal to one and y equal to one plus I. Then it follows that the doctor the two is an Eigen vector belonging to and in fact also spanning the Eigen space of Eigen value. Lambda equals negative. I you said leave her now as a complex matrix, we see that D. Is now diagonal. Izabal. Typically we have a set S with vectors V. One V. Two or 11 minus I. And 11 plus I can't have it. This is a basis of c. two. Yeah. We're going to do a house and I am the mummy birthday hit in my ass carrying. And it went all the way up to the rubber bend consisting of Eigen vectors of matrix B. And therefore it follows that B is in fact diagonal. Izabal don't do cream cheese please as a complex matrix. And using this basis we can actually represent B. I love it's a fucking addiction. I can't help myself. Were traveling all this is also fat. Yeah, yeah. By the diagonal matrix D. Which is the diagonal matrix which entries I and negative I.

Hello there. Okay, so for this exercise we get this meat. We need to calculate these things here. So for each of the values of these metrics, we need to calculate the geometric multiplicity and that separate multiplicity. And based on that result we need to that remind if this matrix A. Is are not recognizable in case that it is a recognisable. We need to find that make XP that therein lies the matrix. So let's start first by calculating the I can values. Sorry. So that means that we need to calculate the determinant of the system. And here I is the identity matrix in this case is three by three matrix And these are termination shall be equals to zero. This is equivalent to calculate the determinant of this 25 -11- Lambda minus nine, 17 -9 and -4 this determinant. Give us the following uh polynomial. So λ -1 sq λ -2 equals to zero. Okay, so from this it is easy to calculate to obtain the roots and the roots are one and two. But he here is important to mention something. And is that the first I can value is the generator it has A and direct multiplicity equal institute. So for this case the algebraic multiplicity is equal to two and for the other one is just one. Okay, so we got now don't you break multiplicity of each of the Eigen values now? So for lambda equals to one. We got the algebraic multiplicity is equal to one. I'm going to copy here again. The matrix a real fast. Okay, so the point is that we need to get now the as of right d geometric multiplicity. And for that we need to see the we need to calculate the Eigen vectors, but that's not necessarily true if we just want to know what is the geometric multiplicity, I'm sorry. Here is to Yes, for these, I can value the geometric multiplicity is equal to two. And now we need to calculate the geometric multiplicity for that. We need to consider the system and this is equivalent to her to this 18 -9, 6, 25 -12 -9, 17 -9 -5 X one x two x three equals to the zero factor. Okay, so we need to reduce this system to the issue of the forum to see what happened with the dimensions of the new space. Why? I'm going to tell you why nature. So this is the result thing asian form of this system. The point is that the no space will give us the nontrivial solutions for this system. So that means that the annuity of these matrix here will be equals to the number of linearly independent Eigen vectors of these associated with these Eigen value. That means that this is the cost to the geometric multiplicity. And in this case the news, the majority of this matrix is just one because after reducing the matrix to the action form, we got only one roll full of zeros. So that means that the geometric multiplicity in this case is just one. Another way to calculate to obtain the geometric multiplicity is actually solving this system and you will find that there is only one Eigen vector generated by some value T. And here is um one, one and four thirds. Okay, so this is the only Eigen value to generate the Eigen space associated to λ one. So That means that the geometric multiplicity of the second value is one. So that's important to take into account. No, let's see what happened with the other, invited. So for for the other Eigen value that is exhausted too. The matrix a minus lambda i is equal two right, 17 -9 -6, 25 -13 minus nine, 17 -9 and -6. Okay, So if we reduce this matrix radiation for what we obtain is 10 -33/4 0, -3, Okay, so again he applies the metric for the majority of this, of this uh matrix will be related with the geometric multiplicity of this Eigen value here. And in this case is again what? Because we only have one roll full of zeros. So the geometric multiplicity is one. But in this case for lambda to the algebraic multiplicity and the geometric multiplicity, both R. to one. However, for the previous Eigen value I'm the one equals to one. For But here's to In this case the algebraic multiplicity was equal to two and the geometric multiplicity was equal to one. That means that they are different. And because of the serum 5.4.5 five .35 point two 0.4 if a E s diagonal sizable if and only if for all the Eigen values I, the algebraic multiplicity is equal to the geometric multiplicity. And in this case for one of them, this is not satisfied. So that implies immediately that A is not diagonal eyes.

Hello there. Okay, so in this exercise we got this matrix A. And we need to uh first for all the values of these matrix, we need to calculate the algebraic and geometric multiplicity. Okay, so let's start in this case, the asian values is really easy to calculate because this is a lower triangular matrix and story here. I got them uh A copy from here. This here is one, so is a lower triangular matrix. That means that the Eigen values are located at the diagonal of these matrix. Okay, so it is clear that the Eigen values are going to be zero with algebraic multiplicity equals to two because appeared two times and λ two equals to one with algebraic Multiplicity equals to one. Great. Now we need to get the geometric multiplicity and for that we need to consider the associated the Eigen space for each of the Eigen values. That means that for example, for lambda, one equals to zero. We need to consider this system and find all the non trivial solutions. So that system is equivalent to this case is just a. X equals two. So we got this system Extracts three equals 2, And the solution of this is that we got to free variables X two and X three. So that means that X two is going to be equal to t. For example, an X three is going to be equal to us. And the Based on this equation here, we got that X one, it's equal two minus s over three. So this gives us the general solution of the of this system, so is equals two, T 010 plus s -1, 01. So what happened is that in this case the Eigen space associated to this Eigen value here is generated by two vectors. Is the span of these two vectors. So these two our Eigen vectors. So actually we can choose C equals to one. To obtain the first Eigen vector, that's going to be 010 and as equals to three. And the second Eigen vector it's going to be minus 103 Great. So for number one equals to zero, we got to Eigen vectors. One is equal to 1010 and the second one is equal two -103. So we got two linearly independent Eigen vectors are associated to love the one. That means that the geometric multiplicity in this case is also equal to two. And as we know from what we calculate before the age of like multiplicity is also constitute. So both multiplicity hours are the same. So we are in the in a good way we can probably, we can analyze actually we can and the second I can value is he close to one? So we must find one Eigen vector. So we repeat the procedure we considered the system and we find the non trivial solutions. In this case this system is equivalent to -1000 -10300 x one X two X three equals to the zero vector. Mhm. The solutions of this system, well this first were reduced to the echelon form And we obtain 1000 1000 X three. Yeah. And the solutions is that X one is the cost to zero X two is also a cost to zero and here X three because we don't have any pilot is a free variable, So X three is equal to some constant value. And that means that the general solution is generated by just one vector is the span of this vector, Which means that the geometric multiplicity is one. Okay, so the number of Eigen vectors associated to each Eigen value will give you the geometric multiplicity. So in this case for lambda too, the geometric multiplicity and the algebraic multiplicity are the same. Great. So that's the first part of this exercise. No. Based on this result, we have checked that for for all the Eigen values, the geometric multiplicity is equal to the anti black multiplicity. And by the theorem 5.2.4, it says that eight is diagonal. Izabal if and only if for all the Eigen values of a the geometric multiplicity is close to the edge of wreck multiplicity, which is this case. So we can organize these matrix and the the process is not that hard. So we need to find the the matrix speed and the matrix speed is generated by the it has in the columns, the corresponding Eigen vectors. Okay, so that means that in this case I'm going to great here again. The Eigen vectors or I N vectors are 010 -103. And the third that we just computed previously Is 001. So you can put in any other actually, but in this case I'm going to put in the other. They obtained them. So that means 010 -103 And 001. Right? So we got P. And then we need to calculate um to complete the organization, the members of the so the embers of B Is equal to 301 -100 and 01. Great. Um yeah, I'm going to to put this I'm going to shift this to because otherwise this doesn't make sense. Yes, here is 010 and 001 It doesn't matter how the order that you put the iron vectors on the metric speed. Okay, so here we got P and P inverse. So I'm going to copy them here again. So P is able to 001 -103 010 And the inverse of this matrix is 301 -100 and 010. And what happened is that we are going to use this to calculate the diagonal matrix. That is supposed to be obtained after a plane PM burst A P um after playing this where A is equal 2000000301 This diagonal matrix is equal to 00000001. Is that they have no metrics, as you can see. But the most important thing that they want you to notice here is that this, the econometrics is actually formed by the Eigen values of the matrix A. Then we competed before, and that's it. Thank you.


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