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Find the matrix of T corresponding to the bases B and D and use it to compute [itex] C_{D}[T(v)] [/itex] and hence T(v)

T; P2 - > R2

T(a + bx + cx^2) = (a+b,c)

B={1,x,x^2}

D={(1,-1),(1,1)}

v = a + bx + cx^2

ok i cna find Cd no problem it is

[tex] C_{D}[T(v)] = \frac{1}{2} \left(\begin{array}{ccc} 1&1&-1 \\ 1&1&1 \end{array}\right) [/tex]

now am i supposed to solve for X where

[tex] C_{D}[T(v)] X = T(v) [/tex]

but textbook doesnt do that...

it does [tex] C_{D}[T(v)] \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = X [/tex]

It doesnt make sense... doesnt the question ask to compute Cd[T(v)] and T(v) from it??

Next question[itex] M_{EB} (ST) = M_{ED} (S) \bullet M_{DB} (T) [/itex]

Verify this theorem for the given transformation and uses the standard basis in Rn

Theorem:

T;R3 -> R4

S: R4->R2

T(a,b,c) = (a+b,b+c,c+a,b-a)

S(a,b,c,d) = (a+b,c-d)

[tex] M_{EB} (S) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \end{array}\right) \

C_{D}\left(\begin{array}{c} 0 \\ -1 \end{array}\right)\right] [/tex]

[tex] M_{EB} (S) = \left[\begin{array}{cc} 1&0 \\ 1&0 \\ 0&1 \\ 0&-1 \end{array}\right] [/tex]

[tex] M_{DB} (T) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \\1 \\ -1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array}\right)\right] [/tex]

[tex] M_{DB} (T) = \left[\begin{array}{cccc} 1&0&1&-1 \\ 1&1&0&1 \\ 0&1&1&0 \end{array}\right] [/tex]

the dimensions of the matrices are not correct though..

is there something wrong in the way i am forming the matrices?

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# Homework Help: Matrix of a linear transformation

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