1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Linear Algebra proof about linear transformations

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Given:
    T is a linear transformation from V -> W and the dim(V) = n and dim(W) = m

    Prove:
    If β = {v1, ..., vm} is a basis of V, then { T(v1), ..., T(vm) } spans the image of T.
    NOTE: because of bad hand writing I can't tell if the bold is suppose to be an 'm' or an 'n'. I think 'm' because that makes more sense to me.

    3. The attempt at a solution

    Let A = { T(v1), ..., T(vm) } .

    We must show that the vectors in A are L.I., and that the dim(A) is m.

    If we show the vectors of A are L.I. then since there are m vectors we know the dimension is m.

    Then there must be only the trivial solution to c1T(v1) + ... + cmT(vm) = 0 .
    Or, by linearity, T(c1v1 + ... + cmvm) = 0 .

    Note: I don't think I can jump straight to the next step. (in a similar thread of mine I did so by utilizing the fact that the null space is 0).

    Since the vectors v1...vm form a basis, they are L.I. and only the trivial solution exists.

    Therefore span(A) = Im(T).
     
  2. jcsd
  3. Oct 25, 2011 #2

    Mark44

    Staff: Mentor

    But it doesn't make sense to me. It's given that dim(V) = n, so any basis for V must have n vectors. So β = {v1, ..., vn}.

    From this, the set of image vectors must be { T(v1), ..., T(vn) }
     
  4. Oct 25, 2011 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The problem isn't asking you to show that A is a basis for W. It's asking you to show A spans Im(T). The vectors in A may not, in fact, be independent.
     
  5. Oct 25, 2011 #4
    Thanks Mark and vela. How does this look:


    Let A = { T(v1), ..., T(vm) } and β = {v1, ..., vn}.

    Since β is a basis of V, all vectors in V can be written as v = c1v1 + ... + cnvn.

    Then T(v) = T(c1v1 + ... + cnvn) = c1 T(v1) + ... + cn T(vn).

    Because the Im(T) = T(v), and T(v) can be written as a linear combination of the elements of A, A spans Im(T).
     
  6. Oct 25, 2011 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's the basic idea. I'd probably slightly reorder it: start the proof with "Let [itex]w\in \mathrm{Im}(T)[/itex]" and show that this implies that w can be written as a linear combination of the vectors in A.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Another Linear Algebra proof about linear transformations
Loading...