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(a) Show that the set B =4H form basis for R?The matnx P;the "change of coordinatcs matrix" . and it satisfics:"-[; for any vector in R' whcn is...

Question

(a) Show that the set B =4H form basis for R?The matnx P;the "change of coordinatcs matrix" . and it satisfics:"-[; for any vector in R' whcn is thc coordinale Vcctor of relalive t0 the basts B.(b} C onsider the vecior {=-h 4h: (that is [1.-[4 Find the coordinate vector x rclative t0 the standard basis.(c) Find the inverse matrix P; t0 determine the coordinate !ector [F1 of the !ectorDraw sketch t0 illustrate the vcctor15 4 lincar combination ofJnd

(a) Show that the set B = 4H form basis for R? The matnx P; the "change of coordinatcs matrix" . and it satisfics: "-[; for any vector in R' whcn is thc coordinale Vcctor of relalive t0 the basts B. (b} C onsider the vecior {=-h 4h: (that is [1.-[4 Find the coordinate vector x rclative t0 the standard basis. (c) Find the inverse matrix P; t0 determine the coordinate !ector [F1 of the !ector Draw sketch t0 illustrate the vcctor 15 4 lincar combination of Jnd



Answers

Define $T : \mathbb{P}_{3} \rightarrow \mathbb{R}^{4}$ by $T(\mathbf{p})=\left[\begin{array}{c}{\mathbf{p}(-3)} \\ {\mathbf{p}(-1)} \\ {\mathbf{p}(1)} \\ {\mathbf{p}(3)}\end{array}\right]$
a. Show that $T$ is a linear transformation.
b. Find the matrix for $T$ relative to the basis $\left\{1, t, t^{2}, t^{3}\right\}$ for $\mathbb{P}_{3}$ and the standard basis for $\mathbb{R}^{4} .$

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.

First we have the Colleen Ami o Space and we first have the basis that is B, which is one minus three t squared, two plus T minus five. He's square and one plus two tea. Okay, so the standard days, that's just a key one. This one you do is two. Three. Is he squared? So if we transform those constant and call your patient and two squared by by a P want Peter MP three So we have basis be is you want minus three the three. Two p one us. You do minus why p three And he won t o p. Two. So that gives, uh, change. Accorded a matrix from E to see to see to be first. 10 negative. Three and 2165 and 120 It's so here's dollar matric. So the next thing we need to write t squared as a wiener combination, not the point. Normalcy B. So that is too. Find a p from E to see times record be e and record UFC and work. Even pfc to is the victor 001 and we will solve this system. 121 There were 12 and the three negative. Five and zero and 001 Here's the system we need to solve. So we apply the gushing in the nation two. Why don't we? I'll just write the result here. That is 100 They're all one deal 001 and have three negative to one. So that gives, um, Specter record. PB is three negative too. And what? So our we need a combination is that is three times the first in your question one minus three T squared, minus two times a second question two plus T minus five. T squared to us. T minus five teeth were or we had we say t squared. Where you either We can see three and plus one times off one of us to tea, and we can't even check. We have, uh, three minus two. Oh, wait. Um, yeah, we have three. Why is two times two is negative for and plus one. So that's countenance with zero. And we have that give 60 squared and and positive 10 t squared and has tools when that years. And yeah, we have, uh So then will leave us a T squared and we have to t cancel two by two t like duty here. So that is exactly two D. Oh, Sorry. Uh, second U T squared.

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

The answer to this problem is the embarrass off the metrics off in problem 17. But to solve it separately the fresh right, The first picture off BC's see Because to And then we need to find a better C one C two that been multiplied by the victors off B gives us, uh, first victor off. See, I mean, he every Red Sea nine to close see to for at three. So here this is X. This is why and then see one bye bye except the first si two foot by takes a second and keep this X. So she's not equations here. Nine c one plus for C to give this to on to C one plus series Negative self minus three C to you. Just one. Then we need to solve these. I just defying games. You can write c one musical, too one plus three c two divided by two and then replace see one here and then and gets you want a C two she's going to be write it as tell you netted two basis be to seventh and negative one. Yeah. So this is the first vector I need to find a second and we write the again. Victor off basis. See 83 in one. Go see one 92 plus C two for a negative three. I seem likely. Solve it. Um, like above. You're right. This is the equations. Nine C one plus for C two. Close to negative. Three to see Yvonne minus C three if you to you calls to one, then here. You can't just simplify as before. See? One used to call to one plus three C two you wanted by two on. Then replace. See one here, get C one and C two. The second victor will be as to be minus 1/7 on three minutes. So these two pictures are the columns off. Um, the change of metrics. Change of basis, Metrics. So hairy. Right? The the metrics from here. Noise from C to B. Easy call to the metrics. Seven. Maybe so just be joke metrics from C to B


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