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Linear Independence: Writing vectors as linear combinations

  1. Mar 21, 2012 #1
    Forgive me for not writing in latex, but I searched this site for 10 minutes looking for a latex reference and could not find anything on matrices. Also, excuse for the excessive amount of info.

    1. The problem statement, all variables and given/known data

    Determine whether this list of 3 polynomials in P1:
    p1 = 1+3x
    p2 = 1+2x
    p3 = 2+3x
    is linearly independent. If not, write one of the pi in terms of the others.
    To test independence, let's see if the linear system
    x1 p1 + x2 p2 + x3 p3 = 0 has any non-trivial solutions.
    First, write the coefficient matrix A for a linear system representing the polynomial equation.

    2. Relevant equations

    3. The attempt at a solution

    I reduced the matrix
    [1 1 2]
    [3 2 3]

    to

    [1 0 -1]
    [0 1 -3]
    I set Ax=0.
    I found that the set {pi} is linearly dependent, and that a non-trivial solution to AX = 0 is (0,0,0) (Is this always the case? Pretty much all of the problems that I've come across have this as a solution for AX=0)

    Now, my the next part is my issue. The directions then read:
    "In particular, your solution to AX = 0 implies
    = 0
    Now use this linear dependence relation among the vectors { p1, p2, p3 } to write one of these vectors as a linear combination of the others."


    This is where I am totally lost. I'm solving problems via a website, and every attempt to this solution has returned wrong. I've guessed every combination of p1 = p2 + p3 and I am not sure how to approach this last question.

    There are numerous problems that ask the same question listed above. It would be greatly appreciated if someone could give an explanation and the proceeding steps to solving a problem like this as I will have to apply it to other similar problems.
     
    Last edited: Mar 21, 2012
  2. jcsd
  3. Mar 22, 2012 #2

    hunt_mat

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    Homework Helper

    I am not too sure exactly what you're doing here but your first step was fine, you wrote:
    [tex]
    a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}=0
    [/tex]
    And came up with an over determined system which shows that there is no solution and therefore the set is linearly dependent.

    Not what you do is write:
    [tex]
    p_{1}=b_{1}p_{2}+b_{2}p_{3}
    [/tex]
    and equate coefficinte of the powers of x, this should give the answer you want.
     
  4. Mar 22, 2012 #3

    Ray Vickson

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    Homework Helper

    Here is a matrix in LaTeX:

    [tex] A =\left[ \matrix{1 & 1 & 2 \\ 3 & 2 & 3} \right][/tex]

    The commands are "A = \left[ \matrix{ 1 & 1 & 2 \\ 3 & 2 & 3 } \right]". In a row the "&" is a column separator, and the "\\" is an end-of-row command. You get closed brackets by the "\left[ ... \right]" pair; you must *always* have a pair, not just one of them by itself. If you want rounded brackets, use "\left( ... \right)" instead; you can even have one square and one rounded:
    [tex] B = \left( \matrix{1&2&3\\4&5&6}\right) , \; C = \left( \matrix{a+b & c+d \\ e & f} \right]. [/tex]

    RGV
     
  5. Mar 22, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    One very quick point is that the space you are talking about is two dimensional. A basis is {1, x}. It is impossible to have three independent vectors in a two dimensional space.
     
  6. Mar 22, 2012 #5
    I have found it useful to remember that the dimension of a space is:

    1. The number of vectors in its basis
    2. The number of vectors in the largest possible set of linear independent vectors in the space
    3. The number of vectors in the smallest possible set of vectors that spans the space

    Thinking about how each one of these interacts with the others helped me a lot in understanding bases and dimension.
     
  7. Mar 22, 2012 #6

    Fredrik

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    Gold Member

  8. Mar 22, 2012 #7
    "To test independence, let's see if the linear system
    x1 p1 + x2 p2 + x3 p3 = 0
    has any non-trivial solutions."

    I would say ##(x1,x2,x3) = (0,0,0)## is a trivial solution, not a non-trivial solution.

    By contrast, by inspection, ##(x1,x2,x3) = (1,-3,1)## is a non-trivial solution.
     
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