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Question

Let $\mathcal{E}=\left\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\} \quad$ be the standard basis for $\mathbb{R}^{3}$ , $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\}$ be a basis for a vector space $V,$ and $T : \mathbb{R}^{3} \rightarrow V$ be a linear transformation with the property that
$$T\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{3}-x_{2}\right) \mathbf{b}_{1}-\left(x_{1}+x_{3}\right) \mathbf{b}_{2}+\left(x_{1}-x_{2}\right) \mathbf{b}_{3}$$
a. Compute $T\left(\mathbf{e}_{1}\right), T\left(\mathbf{e}_{2}\right),$ and $T\left(\mathbf{e}_{3}\right)$
b. Compute $\left[T\left(\mathbf{e}_{1}\right)\right]_{\mathcal{B}},\left[T\left(\mathbf{e}_{2}\right)\right]_{\mat cal{B}},$ and $\left[T\left(\mathbf{e}_{3}\right)\right]_{\mathcal{B}}$
c. Find the matrix for $T$ relative to $\mathcal{E}$ and $\mathcal{B}$

Chris Trentman · Accepted Answer

were given these standard basis for our three the basis for a vector space V in a linear transformation T from our three to V with the property that t of x one x two x three In our three, this is going to be equal to x three minus x to be one minus x one plus x three b two plus X one minus x to be three In part, they were asked to compute the image of each of the standard basis vectors. So we have that TV one. Well, this is the same as t of by definition 100 And so this is going to be x three minus x two is simply zero minus zero times be one minus x one plus x three. So negative the two then plus X one minus X to be three. So plus b three and likewise we have that the image of e two. This is the image of the vector 010 So we have zero minus one B once or negative one B one minus and then we have excellent. 63 is zero plus zero. So zero b two and finally we have X one minus x two b three. So this is one zero minus one. So negative B three. So if the image is negative, B one minus b three. And finally we have that the image of the three. This is the image of the vector 001 This is going to be one minus era. Be one simply be one minus zero plus one B two. So negative be to and zero minus zero B three c zero B three. So the images be one minus B two park. We were asked to compute the image with respect to the basis the projection under the basis be This is pretty easy. So we have that the image of e one projected on to be Well, we see that the coordinates you won, we have negative B two plus B three. So this is going to be the vector zero negative 11 Likewise, we have it t of e two sabee. This is going to be the coefficients of the basis. This is negative 10 negative one. And finally we have t of the three. So be this is one negative 10 and in part C. We&#x27;re has to find the Matrix for T relative to Epsilon the standard basis and be so to do this, we have that the Matrix for T relative depths on and be has the form The Matrix, whose column vectors are going to be TV ones of B TV, too, said the and TV three Sub and putting together what we found in the previous part. We obtain the Matrix with column vectors zero negative 11 one negative 10 negative one and one negative 10

Runpeng Li · Answer

Okay, so for problem for I r. Again given assumption we have t of x one, be one of those x two you two plus x three The three It&#x27;s the new vector, which is to x one minus four x two was five like three and ActiveX too. Plus three x three. So to find out the matrix for tea rented to be first need to find out the f B one from court in a C where C is Thea faces off off are too. So here we can just take X one to be one and next to x three to be zero. So our specter will be too zero. And similarly, we can find bee the F B two. So this will be negative for Nick to one and finally t f B three so T o b three Just be five and three now. So the Matrix TB, we&#x27;ll just be the span off these three vectors. Two negative four and five zero negative one and three

Runpeng Li · Answer

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&#x27;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&#x27;s t l Q. So that&#x27;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&#x27;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&#x27;re done for the first part because I our this relation and second relation. Then we can conclude that he&#x27;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&#x27;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&#x27;re plugging t which gives Dr Inactive three negative one, one and three. Then we&#x27;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&#x27;s it.

Runpeng Li · Answer

Okay, so for problem to we have the basis di at his d one and you too, and are so basis be, which is B one B two. So we we need to find out the matrix t relative to D and B. So by the giving permission ti of the one is to be one minus three b two and t of d to its inactive for B one plus five b two. So by using that relation by observing their coefficient, you can have a transformation off the one and we should have a deep coordinate. Sorry, the be coordinate here. So the first thing that the 100 people made by our first equation it&#x27;s okay to be one Sorry. Step two. We should be two and three and would be other. You too. And under ordinate b, this will be by our second equation, like a born by now the meet our metrics t I would just be the span of these two vectors to elective three next four and five

Runpeng Li · Answer

pick. So, in this problem, we&#x27;re already given the, um, Helene Alina transform Matrix relative to be, which is a three by three matrix. That is the real inductive. 61 05 elective one one negative two and seven. Now the only thing we need to do to find out a t of three of the one minus for B two is to take the coefficient off the given victor three V one minus four Be too. So it&#x27;s three off the one negative for P 20 for B three and enter the cleaner transformation off this matrix and we find that is 24. Order for century and active 20 for a second entry and three waas eight. You love it for the second entry. So that is actually the tea. Uh, three three b one. Honest for Pete to

Rakvi . · Answer

the fact that we wanted me to our basis off capitally and you using the information. How do you know creates on each one of them? You can write the Matrix Chorus wanting to be one. First. Problem is, just choose the coordinates off, do you? Everyone, which is one Goma minus one and second goal. A misty off we toe. Also it even off people and the court nettle. Those are 21 OK? Similarly, the Matrix course wanting to Tito Most well, Thomas stewed Wolf. Everyone, which is one and second column, is corresponds to the coordinates of day to Veto, which is three minus one. And now, if you want to find out, do you do off the one I just want to play the two matrices out Do you two off? Even Wilco&#x27;s Oneto 13 tu minus one. Their product. This is a quartile. So 22 off the one off me is equal to this matrix eight times we for any vector

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Let $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b…

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

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David C. Lay, Steven R. Lay, Judi J. McDonald

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