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anonymous
 4 years ago
2. Suppose that vectors v1 = (1, 2), v2 = (2,−1), and that the basis B is B=v1, v2 . (this is a list)
Let T be the linear transformation from R2 to R2 given by T(v1) = v1 and T(v2) = 0.
(a) Write down the matrix for T in the new basis B. (You should be able to do this
directly from the definition of T).
(b) Use this to write down the matrix for T in the standard basis.
anonymous
 4 years ago
2. Suppose that vectors v1 = (1, 2), v2 = (2,−1), and that the basis B is B=v1, v2 . (this is a list) Let T be the linear transformation from R2 to R2 given by T(v1) = v1 and T(v2) = 0. (a) Write down the matrix for T in the new basis B. (You should be able to do this directly from the definition of T). (b) Use this to write down the matrix for T in the standard basis.

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11 2 2 1 is B T(x) = Ax a c b d = A 1(a,b) + 2(c,d) = (1,2) 1a + 2c = 1 1b + 2d = 2 2(a,b)1(c,d) = (0,0) 2a 1c = 0 2b  1d = 0 Of course this inst according to the definition of a linear transform, but its how i do it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11a + 2c = 1 2a  1c = 0 *2 1a + 2c = 1 4a  2c = 0  5a = 1 ; a = 1/5 1a + 2c = 1 *2 2a  1c = 0 2a 4c = 2 2a  1c = 0  5c = 2; c = 2/5

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11b + 2d = 2 2b  1d = 0 2b 4d = 4 2b  1d = 0  5d = 4 ; d=4/5 1b + 2d = 2 4b  2d = 0  5b = 2; b = 2/5 \[A=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\] maybe might have to leave the 1/5 in there to work tho

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1B = v1, v2 newB = T(v1), T(v2) \[T\binom{1}{2}+ T\binom{2}{1}\] \[1\binom{1}{2}+2\binom{1}{1/2}\] ugh ... i just dont have the practice at doing it htis way :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so i'm confused, the matrix that I got for a was [1 0] [0 0]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1howd you get it? that might help me know if im completely off

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it because if T(v1) is (1, 2) and T(v2) is (0, 0), then the matrix that takes the vectors to that one is [1 0] [0 0]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i could be wrong, which is exactly why i posted the question here haha.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11. 1 2. 0 = 1+0 1 0 0 = 0+0 not equal 2 so the transformation matrix doesnt tramsform v1 into v1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lmao ok then i'm wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but that was my mind set for trying to get the right answer..

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[A=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{1}{2}=1\binom{1}{2}+2\binom{2}{4}=\binom{1+4=5}{2+8=10}/5=\binom{1}{2}\] right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lets see if it works for v2 :) \[A\vec{v_2}=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{2}{1}=2\binom{1}{2}1\binom{2}{4}=\binom{22=0}{44=0}/5=\binom{0}{0}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1how to get to that thru just using the definitions of linear transformations I cant tell ...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i take a generic A matrix, and create 2 sets of equations to solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, now do you know how i would find the stand basis matrix for that?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i think that is it, but im reading up on it at the moment to be sure

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the part "a" is spose to get you to the A matrix that I did; but thru a different technique i think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, oh. i dont think so, part a asks for the matrix in the basis B

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1B = 1 2 2 1 newB = 1 0 2 0 is that part a?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if i could see your textbook im sure id be like "doh!!"

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11.1 +0. 0 = 1 2 0 0 0.1 +1. 0 = 0 2 0 0 which is why you came up with the : 1 0 0 0 to begin with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wahhh i wish i knew. that makes sense.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ill read up on it tonight and see what I can remember :) good luck

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you ever end up figuring anymore of it out? i'm so stumped.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1nothing really that makes any sense to me .... B = v1 v2 B = 1 2 2 1 .......................................... newB = T(v1) T(v2) newB = 1 0 2 0 seems to be the way its looking from what I can gather. that should be part a its the part b thats got me frazzled

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so all of the stuff you put yesterday with everything over 5, that matrix, that has nothing to do with it? haha

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1from what ive gathered; it was not the manner in which they are wanting this accomplished

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1im not sure what newB and B have to do with each other rather than writing them up as baseses for vector spaces

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1what is the name of the chapter that you are working on?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's all about changing basis... idk if you gathered that. so basically its just asking for the matrix of the transformation with respect to basis b and then in part b its asking for that same matrix only with respect to the standard basis. they're homework questions assigned and not in the textbook.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so, part b is what: T(v1) = (1,0) ; T(v2) = (0,1)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not sure if that's how you write it ... because i dont know if it's neccessarily going to the standard matrix, but what would the standard basis matrix be of the transformation .. if that makes any sense.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i aint got a clue yet :) B = v1 v2 \[B\vec{x} =x_1v_1+x_2v_2\] \[B\vec{x} =x_1\binom 12+x_2\binom 2{1}\] \[B' = T(\vec{v_1})\ T(\vec{v_2})\] \[B'\ \vec{x} =x_1T(v_1)+x_2T(v_2)\] \[B'\ \vec{x} =x_1\binom 12+x_2\binom 00\] what this has to do with anything I cant tell

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lmao its all good, it's like the blind leading the blind here! thanks anyways!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, all in all, good luck with it ;)
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