A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 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

  2. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1a + 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

  3. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1b + 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

  4. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    B = 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 :/

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so i'm confused, the matrix that I got for a was [1 0] [0 0]

  6. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    howd you get it? that might help me know if im completely off

  7. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I 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]

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but i could be wrong, which is exactly why i posted the question here haha.

  9. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1. 1 2. 0 = 1+0 1 0 0 = 0+0 not equal 2 so the transformation matrix doesnt tramsform v1 into v1

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lmao ok then i'm wrong.

  11. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but that was my mind set for trying to get the right answer..

  12. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  13. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  14. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lets see if it works for v2 :) \[A\vec{v_2}=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{2}{-1}=2\binom{1}{2}-1\binom{2}{4}=\binom{2-2=0}{4-4=0}/5=\binom{0}{0}\]

  15. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how to get to that thru just using the definitions of linear transformations I cant tell ...

  16. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i take a generic A matrix, and create 2 sets of equations to solve

  17. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, now do you know how i would find the stand basis matrix for that?

  18. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think that is it, but im reading up on it at the moment to be sure

  19. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the part "a" is spose to get you to the A matrix that I did; but thru a different technique i think

  20. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm, oh. i dont think so, part a asks for the matrix in the basis B

  21. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    B = 1 2 2 -1 newB = 1 0 2 0 is that part a?

  22. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if i could see your textbook im sure id be like "doh!!"

  23. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1.1 +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

  24. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wahhh i wish i knew. that makes sense.

  25. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ill read up on it tonight and see what I can remember :) good luck

  26. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks!

  27. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    did you ever end up figuring anymore of it out? i'm so stumped.

  28. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    nothing 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

  29. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so all of the stuff you put yesterday with everything over 5, that matrix, that has nothing to do with it? haha

  30. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    from what ive gathered; it was not the manner in which they are wanting this accomplished

  31. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    im not sure what newB and B have to do with each other rather than writing them up as baseses for vector spaces

  32. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what is the name of the chapter that you are working on?

  33. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it'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.

  34. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so, part b is what: T(v1) = (1,0) ; T(v2) = (0,1)?

  35. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'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.

  36. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i 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

  37. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lmao its all good, it's like the blind leading the blind here! thanks anyways!

  38. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, all in all, good luck with it ;)

  39. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.