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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 $\mathcal{B}$

a) $T\left(e_{3}\right)=1 b_{1}-1 b_{2}+0 b_{3}$b) $\begin{aligned}\left[T\left(e_{1}\right)\right]_{\mathrm{B}} &=\left[\begin{array}{c}{0} \\ {-1} \\ {1}\end{array}\right] \\\left[T\left(e_{2}\right)\right]_{\mathrm{B}} &=\left[\begin{array}{c}{-1} \\ {0} \\ {-1}\end{array}\right] \\ \text { and } \quad\left[T\left(e_{3}\right)\right]_{\mathrm{B}} &=\left[\begin{array}{c}{1} \\ {-1} \\ {0}\end{array}\right] \end{aligned}$c) $\left[\begin{array}{ccc}{0} & {-1} & {1} \\ {-1} & {0} & {-1} \\ {1} & {-1} & {0}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 4

Eigenvectors and Linear Transformations

Vectors

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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'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

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