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Linear algebra questions

  1. Dec 4, 2007 #1
    1. Let V and W be finite dimensional vector spaces with dim(v) = dim(w). Let {v1,v2,...,vn} be a basis for V. If T:V->W is a one to one linear transformation, determine if {T(v1), T(v2), ... , T(vn)} is a basis for W.

    2. How do i get a matrix out of this: Let A be an 8x5 matrix with columns a1, a2, a3, a4, a5, where a1, a3, and a5 form a linearly independent set and a2=2*a1+3*a5, and a4=a1-a3+2*a5.

    I have looked all over, and I have starts to each of these problems, any help would be received with much thanks.

    So far, for 1. I know that it is true by a theorem I found, but I am really unsure how to prove it.

    on 2. I made a matrix like this

    1 2 0 1 0
    0 0 1 -1 0
    0 3 0 2 1

    I reduced it and came up with the answer that the dimension of NulA is 2 because it reduces to having 2 free variables.

    If this is the wrong way to get the matrix A, how do I do it?

    Thanks.
     
  2. jcsd
  3. Dec 4, 2007 #2
    Here's a hint for #1:

    You will need to use the fact that f is one-to-one. You want to show that any w in W can be written as a linear comibination of the { T(v1), ... , T(vn) }. Well, if f is one-to-one, what can you say about [tex]f^{-1}(w)[/tex]? Moreover, any v (such that f(v) = w) can be written as a linear combination of v1 through vn. What happens when you evaluate f(v) ?
     
  4. Dec 4, 2007 #3

    JasonRox

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    You're constructing it right, but that's not an 8x5! It's a 3x5 (sometimes I get confused on which are rows/columns).

    For question 1, show that { T(v1), ... , T(vn) } is linearly independent and then you're done.
     
  5. Dec 4, 2007 #4
    How would I go about doing that?

    Right now, I'm trying to find an answer using the invertible matrix theorem.
     
  6. Dec 6, 2007 #5
    for 1 - if T is linear from V-->W, linearly independent subsets of V always map to linearly independent subsets of W. Since you have a basis for V, it's linearly independent. All you really have to show is that the LI subset you get in W is actually a basis for W. But since the dim(V)=dim(W)=n, you mapped n LI vectors to an LI subset with n vectors. Therefore, you have a basis.

    If you have to, make a lemma for the part that LI subsets of V map to LI subsets of W (you need the injectivity (1-1) of T for this part). It's pretty easy to show with a proof by contradiction if you get stuck.
     
  7. Dec 7, 2007 #6

    HallsofIvy

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    Given a vector space U with basis [itex]\{u_1, u_2, ..., u_n}[/itex], a vector space V with basis [itex]\{v_1, v_2, ..., v_n\}[/itex], and a linear transformation, T, from U to V, a standard way of writing T as a matrix (with respect to those bases) is to apply T to each basis vector,[itex]u_1, u_2, ..., u_n[/itex] in turn. Applying T to [itex]u_i[/itex] will give a vector in V which can be written as a linear combination , [itex]a_1v_1+ a_2v_2+ ...+ a_nv_n[/itex]. Those coefficients, [itex]a_1, a_2, ... a_n[/itex] are the numbers in the i row of the matrix.
     
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