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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 $\mathbb{R}^{4} .$

a) $T$ is a linear transformation.b) $=\left[\begin{array}{cccc}{1} & {-3} & {9} & {-27} \\ {1} & {-1} & {1} & {-1} \\ {1} & {1} & {1} & {1} \\ {1} & {3} & {9} & {27}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 4

Eigenvectors and Linear Transformations

Vectors

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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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