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[points-2 5]Let TR? 7 R? be the linear transformation satisfying T ([H) - and T ([H)- [- Ilo 8 Then;whichof the following statements is false?Select one: T ([21)- [...

Question

[points-2 5]Let TR? 7 R? be the linear transformation satisfying T ([H) - and T ([H)- [- Ilo 8 Then;whichof the following statements is false?Select one: T ([21)- [5 @ b The eigenvalues of A are 2 andThematrix A expressing T in terms of the standard basis vectors oiland [%] 300 d Thematrix A expressing Tlinferms onithe standard basis vectors] and S 01 lis16 | 51

[points-2 5]Let TR? 7 R? be the linear transformation satisfying T ([H) - and T ([H)- [- Ilo 8 Then;whichof the following statements is false? Select one: T ([21)- [5 @ b The eigenvalues of A are 2 and Thematrix A expressing T in terms of the standard basis vectors oiland [%] 30 0 d Thematrix A expressing Tlinferms onithe standard basis vectors] and S 01 lis 16 | 51



Answers

Let $T_{1}$ be the linear transformation from Problem 10 and let $T_{2}$ be the linear transformation from Problem 5 (a) Find the matrix representation of $T_{2} T_{1}$ relative to the standard bases. (b) Verify Theorem 6.5 .7 by comparing part (a) with the product of the matrices in Problems 5 and $10 .$ (c) Use the matrix representation found in (a) to determine $\left(T_{2} T_{1}\right)(-3+8 x) .$ Verify your answer by computing this directly.

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Were given a transformation from, he said of polynomial of degree. At most two to our three i t. F p equals column Vector, whose entries are p of negative one p of zero in P of one. In part, they were asked to find the image under T of a certain polynomial. P f t equals five plus three t. So first of all noticed that P of tea, which is five plus three t, is in fact a polynomial of degree less than a record two. So this doesn't make sense. Thanks. Well, we have that. The image is going to be a column vector whose components are P of negative one p of zero and P of one. Then we see after plugging in these values, he of negative one is five minus three, which is to P of zero is five plus zero, which is five and p of one is five plus three, which is eight. So we get to column Vector 258 in Part B. We were asked to show that this transformation is in fact, a linear transformation, so we'll let P of tea and Q of t the polynomial of degree at most two and well, let's see be any scaler. Then we have that image of T plus or P plus. Q. Well, we have the both P and cure applying the moves of degree at most two. Therefore, the degree of P plus Q is also at most two. So this makes sense, and this is going to be the column. Vector, whose entries are polynomial P plus Q. Evaluated at negative one polynomial p plus Q. Evaluated at zero and P plus Q. Evaluated at one. Now we know that the addition of Polynomial is that a point is the same as Plano. Meals at a point added together. So this is the same as P of negative one plus Q of negative one P of zero plus Q of zero and then p of one plus Q of one. This can, of course, be written as thesis, um, of column vectors. He is negative. One p of zero p of one, plus the column. Vector que of negative one que of zero q of one we see it. This is the same as T of P plus T F. Q. Now we have the t of c times p well, because P is a polynomial of degree at most two, it follows that the scaler see Times P is also a polynomial of degree at most two. So this makes sense, and we have by definition, this is the column. Vector components. See Time's p evaluated at negative one. See Time's P evaluated at zero and c times p evaluated at one. And we have that because see is your scaler. You could be right. This as see Time's P evaluated at negative one. I think this is called quasi associative Lee. Maybe then we have C Times p of zero and C times p of one and can factor at a C from this column vector to give us simply c Times the column Vector. You have negative one p of zero p of one, which is the same as C times T of Pete is what we wanted to show and therefore follows. The tea is in fact, a linear transformation. Finally, in part, C were asked to find the Matrix for T relative to the basis won t t squared for p two in the standard basis for our three. So we have the basis p two won t he squared. So called be the basis See for our three, which is simply e one e two and e three. So to find this matrix, I want to compute each of the basis vectors of P two in terms off the basis vectors of our three. So we have that t of one. This is going to be column vector whose entries are mhm one evaluated at one just simply one verse. Everyone value at negative one, which is 11 evaluated at zero, which is also one on one of value. That positive one, which is also one. And this is the same. Has column Vector 100 plus the column. Vector 010 plus the column. Vector +001 which is the same has E one plus e two plus e three. Likewise, we have the t of tea. This is going to the column vector with entries. Well, t of negative one is negative. One T f 00 in the city of one is one, and this is equal to the sum of column victors. We have negative 100 plus 001 and this is the same as negative e one plus e three. Finally, we have that tea of T squared. This is going to be the column vector with entries. He squared value that negative one, which is simply one, then zero, then a positive one. And this is the same as the sum of column vectors 100 and column Vector 001 which is the same as E one plus e three. So going back, we have that the image of one with respect to basis. See, this is going to be the column vector with entries 11 and one coefficients of you. Only 23. We have that the image of tea with respect to the basis. See, this is the column vector of entries negative 10 one. And we have that. The image of t squared with respect to see is e column vector with entries 101 And so we have that The Matrix 40 relative to these bases is going to be the Matrix with column vectors the image of one with respect to see image of tea with respect to see and the image of t squared with respect to see which we calculated is the Matrix. With column vectors 111 negative 101 and 101

Were given a transformation. T Well, we're giving a Matrix A. Which is the standard matrix for a transformation. T and were asked to find the basis for our four with the property that the bee matrix of tea his diagonal So the Matrix were given a as column vectors. 15 0 12 Negative 66 13 Negative 15 18 Negative 44 21 Negative. 21 22 Negative. 33 Negative 15 12 8. So it suggested that to solve this problem, we use some computer algebra software like Maple. While it is technically possible to solve this problem by hand, it would require a lot of calculation which could be done by a computer in much faster time. So the first step will want to dio is you want to find the Eigen values of this matrix so this could be done using computers software. I'm gonna calculate the Eigen values here. Now you may want to pause if you want to calculate them on your own. After calculating I obtained the Eigen values of this matrix are lambda equals two with the multiplicity of one lambda equals four with multiplicity of to and Lambda equals five with multiplicity of one. So with these Eigen values now, I want to find the no basis for the Matrix minus each Eigen value times the identity matrix. So you want to find matrix a minus, starting with our 1st 2nd value to I times X. In fact, I actually start with the last I can tell you first. So minus five by times X equals zero. You want to solve this equation for X and find a basis for it. You see that again? This is not impossible to do by hand, but the calculations would be greatly simplified. Thes computer. So here again, I'm going to use Wolfram Alfa to find Eigen vectors. So you may want to pause here like calculate this So we see that the Eigen vector corresponding to the Eigen value five in the simplest notation is going to be the egg in Victor V one, which is 11 negative 344 Now I want to do the same thing. But with you, I can value of four. And in fact, the Zion Victor isn't just a Nikon Victor, but it's a basis vector for this wagon space. Okay, now we're considering the egg in space for having value four. So a minus four i times the vector X equals the vector zero. We want to find no basis. Once again, this is not impossible to do by hand. However, we greatly simplified if we use the computer and they went to pause here. So after using Wolfram Alfa, I find that to Eigen vectors which form a basis for this wagon space R V two, this is the Vector 39 503 and the three This is the vector negative 30 negative 730 So this these two vectors form a basis for this Eigen space. Finally want to consider the Eigen space corresponding to Eigen value Lambda equals two. So I want to find a no basis for a minus to I to do so again. It's not impossible to do by hand, but a computer making much faster. So I'm going to do that. They went to pause here. After using Wolfram Alpha, I find that an Eigen vector from this Eigen space is before which is the Vector zero negative 332 In this vector forms basis for this Eigen space since it has Dimension one. So now you have a complete set of Eigen vectors. Therefore, we have that the set B. This is going to be the set of vectors. Do you want through the four we have that This is a basis for our four such that the bee matrix of tea is diagonal.

Okay for problem Problem 10. We have the leaner, faster mission maps from P three to Arthur are poor. Such that t of pulling nominal is equal to by taking different input to the Polo Meo. Just pee of connective three first p A connected one and p of one and p up three. Hey, So the first thing we need to show that is that, um to show this transformation is linear. Okay, so do that. My transformation. We can first check tee up. People ask you where panic you are all clean on yourself. Order to be so then, by our assumption, we can have a p rescue while we take the include to be elected three. And he ask you with the input elective born people ask you with include one and people ask you with your food three. Now, because we are adding these two point nominees so we can we can calculate the number Sweet that with this input on dad them together just just by by the rules, our tradition. So that's he, uh, 93. I trust you collected three and PR. Negative one. Uh, you, uh, negative one and p of one less queue of one. And here, three askew on three. So by now we can separate this, uh, this vector with by a vector off Minami O P and the vector polynomial Q. So it turns out to be thio he that's t l Q. So that's the first part here. So the second part is considered skater Linda times people. So again, by our assumption, we still have Lunda cons. P off negative three Lambda SPF connective one, huh? Times p of one and Linda Times p of three. Excuse me. So we can take out the skater from our rector's. So we have loved a in front of our specter and P uh, negative three. He, uh, connected one he won and ps three. So that turns out to be Lunda time. So? So we're done for the first part because I our this relation and second relation. Then we can conclude that he's a renewed transformation. Now, Part B. We need to find the matrix for tea relative to the basis won t t square t t cubed for p three and standard basis are for Excuse me. So now let's, um defined business P start this is B to be given basis in our assumption, which is one two he squared and t Q. So we first calculate, uh, actually the tea with the input up won t t square and teeth cute separately. So first we have to be one which gives our vector all once and then we're plugging t which gives Dr Inactive three negative one, one and three. Then we're plugging our t squared. This gives Excuse me. Um 911 night, and at last we plugging our thank you. So we had a vector collective 27 negative. 11 and 27. So our metrics will be this band of these four vectors. So this will be he of this is B should be one collective three nine Next 27 one defective. 111 1111 and 139 27. And that's it.


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