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Reducing a Matrix of Variables

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the following vectors: (don't give values to variables)

    V0 = (1 1 1 1) V1 = (x1 x2 x3 x4) V2 = (x1^2 x2^2 x3^2 x4^2)

    V3 = (x1^3 x2^3 x3^3 x4^3)

    a) prove that the 4x4 matrix A, whose columns are the vectors above, row reduces to the 4x4 identity matrix.

    b) Do the Vectors above span IR^4? Are these vectors linearly independent?


    2. Relevant equations

    No equations, just have to know how to row reduce, etc...

    If I'm not mistaken, an identity matrix is one in which the leading entries are all 1's and descend diagonally.

    3. The attempt at a solution

    It's hard to show via keyboard what I did, but I first created the matrix (so the first column is all ones, etc...). Then I row reduced to get rid of the last 3 ones of the first coloumn by the following operation:

    (-1)R1 + R2 -- (-1) R1 + R3 -- (-1)R1 + R4

    So now my first column has a leading one (in terms of rows) followed by all zeros below it. However, I'm left with a mix of variables in the other columns (like -x1 + x2, etc...) that I can't figure out how to get rid of! I keep going in circles.

    Can any one help me? I realize it's hard to describe this in words and can't attach a windows journal file for reference...

    Thank you!
     
  2. jcsd
  3. Feb 15, 2012 #2

    HallsofIvy

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    (Click on the formulas to see the LaTeX code used.)
    So your matrix is
    [tex]\begin{bmatrix}1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3\end{bmatrix}[/tex]

    Your first row reductions gives
    [tex]\begin{bmatrix}1 & x_1 & x_1^2 & x_1^3 \\ 0 & x_2- x_1 & x_2^2- x_1^2 & x_2^3- x_1^3\\ 0 & x_3- x_1 & x_3^2-x_2^3 & x_3^3- x_1^3 \\ 0 & x_4- x_1 & x_4^2- x_1^2 & x_4^3- x_1^3\end{bmatrix}[/tex]

    Now the obvious next step is to divide the second row by [itex]x_2- x_1[/itex]- and that is not as complicated as you might think. Remember that [itex]x_2^2- x_1^2= (x_2- x_1)(x_2+x_1)[/itex] and that [itex]x_2^3- x_1^3= (x_2- x_1)(x_2^2+ x_2x_1+ x_1^3[/itex].
     
    Last edited: Feb 15, 2012
  4. Feb 15, 2012 #3
    Great! thank you, I did that, now I have a leading one in the second row. I;ve tried the next step two ways,

    1) eliminate second entry of first row
    2) make all entries below leading one of second row 1's

    But neither helps me move forward
     
  5. Feb 15, 2012 #4
    I also tried first obtaining the diagonal 1's that I need, but it gets too messy!
     
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