Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

blibber Group Title

Find the matrix B of the linear transformation T(x)=Ax with respect to the basis B = (v1, v2,...,vm): A= [0 1; 2 3] v1= [1; 2] v2= [1; 1]. I can make the [x]_B span...how do you get that final matrix????? (basis change)

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. swissgirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Jemurray3 Can u check this out?

    • 2 years ago
  2. blibber Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    To clarify, I can get to B = [ [ 2; 8 ]_B , [ 1; 5 ]_B ] but don't understand how the final B works out to be B = [ 6 4 ; -4 -3 ]

    • 2 years ago
  3. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Sorry, just lost a lot of text. Re-typing...

    • 2 years ago
  4. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Okay sorry. If I have a vector \[\vec{x} = \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) \] that I want to express as a vector y with respect to the basis vectors v1 and v2, then \[ \vec{x} = [\vec{v_1} \space \vec{v_2}]\vec{y} = V\vec{y} \] so, \[\vec{y} = V^{-1}\vec{x} \] So, the linear transformation \[T(x) = A\vec{x} \] can also be written as \[T(y) = AV\vec{y} \] But at the end of the day, this quantity \[ \left( AV\vec{y}\right) \] is a vector with respect to the basis V. To make it a vector with respect to x again, I must apply V inverse: \[T(x) = V^{-1}AV\vec{y} = B\vec{y} \] so in conclusion, long story short, \[B = V^{-1}AV \]

    • 2 years ago
  5. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Wait, that last part is wrong. Sorry, ,my explanation got kinda muddled.... the end result is right, but the last two steps weren't explained properly.

    • 2 years ago
  6. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Here: If T maps X to X', then \[T(x) = \vec{x'} = A\vec{x} \] but \[\vec{x'} = V\vec{y'}, \vec{x} = V \vec{y} \] so \[ V\vec{y'} = AV\vec{y} \] \[\vec{y'} = V^{-1}AV\vec{y} = B\vec{y}\]

    • 2 years ago
  7. blibber Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome awesome! very cool..thanks sooooo much!! :) one final question (sorry to be annoying..) but so I found V(inv) to be [-2 -4; -2 -5] and then when I go through V(inv)*A*V, I am dealing with [-2 -4; -2 -5] *[0 1; 2 3]*[1 1; 2 1] but I keep ending up with [-10 -6; -12 -7]. I'm pretty sure it's not my matrix multiplication or inverted matrix V that's the problem..do you notice if I'm using a vector out of place somewhere?

    • 2 years ago
  8. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    V(inv) should be [-1 1 ; 2 -1]

    • 2 years ago
  9. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    and V should obviously be [1 1 ; 2 1]

    • 2 years ago
  10. blibber Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    OH MY GOODNESS!!!!! haha okay, so I was a little ahead of myself in my notes..my ridic V(inv) above was actually my INCORRECT V(inv)*A already..that's what happens when your work gets all jumbled up, I guess. But my inverted V was wrong anyway..I guess I should take a step back and work on that now hahaha. THANK YOU SO MUCH!!! (exam in 15 hours..angsting so hard right now!) you're awesome! =D

    • 2 years ago
  11. blibber Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    and thanks, swissgirl for hooking me up with a math life-saver! (sorry, im new to the site and dont know how to do anything yet..) im off to study some more!

    • 2 years ago
  12. Jemurray3 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    No problem, best of luck.

    • 2 years ago
    • Attachments:

See more questions >>>

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.