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How do I begin this linear algebra problem?

  1. Feb 5, 2013 #1
    http://i.imgur.com/MIazUji.png
    "Describe the matrix A so that Ax = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1]



    Ax = b



    I feel like after getting the ball rolling I'll actually be able to work on this problem, but for the time being I haven't the slightest idea how to begin it.

    Recently, we've been working on inverses, so I assume that you need to take the inverse of x, and then multiply both sides by the inverse of x so that you're left with A = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1](x^-1), but I don't know exactly how I'd even begin that since it's all variables. Hopefully this counts as "an attempt".

    Thank you very much in advance.
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    Mark44

    Staff: Mentor

    It makes no sense to talk about the inverse of the vector x, since it's not a square matrix. In any case, you don't have to find the inverse of A -- just a representation for A.

    Write this:
    $$ \begin{bmatrix} x_1 - x_2 \\ x_2 - x_3 \\ . \\ . \\ . \\ x_n - x_1\end{bmatrix}$$
    as an n X n matrix, and see where that takes you.
     
  4. Feb 6, 2013 #3
    I must be further behind than I thought, how can I write that as an n X n matrix if it is only an n X 1 matrix to begin with?
     
  5. Feb 6, 2013 #4
    Gosh, I feel completely daft. Thanks for your help Mark, I figured it out.
     
  6. Feb 6, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In general, you can find the matrix representing linear transformation T:U-> V, using ordered basis {u1, u2, ..., un} for U and ordered basis {v1, v[/sub]2[/sub], ..., vm} for V by:

    Apply T to u1. Write the result as a linear combination of the basis vectors for V. The coefficients will be the first column. Do the same thing for Tu2 to get the second column and so on.

    For example, using the standard basis of Rn for both domain and range spaces, T(1, 0, 0, ..., 0)= (1- 0, 0- 0, 0- 0, ..., 0- 1)= (1, 0, 0, ..., -1) so the first column has 1 in the first row, -1 in the last row and 0 in all other rows.
     
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