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Linear Algebra - showing sets are linearly independent/dependent

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Using the fact that a set S is linearly dependent if and only if at least one of the vectors, vj, can be expressed as a linear combination of the remaining vectors, obtain necessary and sufficient conditions for a set {u,v} of 2 vectors to be linearly independent. Determine by inspection whether each of the following sets is linearly independent or linearly dependent.

    a.) {1+x, x2}
    b.) {x, ex}
    c.) {x, 3x}

    the following parts d.) and e.) are both matrices, which i'm not sure how to properly represent, I don't know if there is a way to do it while posting this or not so I'll just do my best to say what I mean. All of the matrices are 2x2, there are 2 in part d.) and 2 in part e.), I'll write them in the form [a, b, c, d], where a is the top left entry in the matrix, b is top right, c is bottom left, and d is bottom right.

    d.) {[-1, 2, 1, 3], [2, -4, -2, -6]}
    e.) {[0, 0, 0, 0], [1, 0, 0, 1]}



    2. Relevant equations

    none



    3. The attempt at a solution

    I'm getting all of them as linearly dependent but I'm just looking for some reassurance on whether or not what I'm doing to achieve these results is allowed.

    a.) if we take (1+x) and multiply it by itself we get (1+ 2x + x2), and if we take -2 [itex]\times[/itex](1+x) we get (-2-2x), and combining this with (1+x+x2) would just give us x2. so I guess what I'm saying is x2 can be written as a linear combo of (1+x) by following those steps.

    b.) I claim x can be written as a linear combo of ex by taking the natural log of ex which would then just give us x. however I am not sure if you are allowed to use natural log in this context.

    c.) i think this one is obvious. 3x is just 3 times x, so they are clearly linearly dependent.

    d.) this one i think is also obvious. if we multiply the first matrix by -2 we get the second matrix, so they are linearly dependent.

    e.) if we multiple the 2nd matrix by 0 we get the 1st matrix which is all 0's, so in this way i think it is linearly dependent. however this is just the trivial solution so i'm not 100% sure, seeing as when the only solution to something is the trivial solution it is linearly independent, but i'm not sure if that applies in this situation.

    any help would be greatly appreciated
     
  2. jcsd
  3. Nov 11, 2011 #2
    If A is a linear combination of B then A = cB for some constant c. If you think about it you cant write x^2 = c(1+x) or x = ce^x, but you can write x = c3x. just set c = 1/3
     
    Last edited: Nov 11, 2011
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