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Math Help - Proof using linearly independent

  1. #1
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    Proof using linearly independent

    I'm not sure where to start with this: (these are all matrices -- are supposed to have bars over them, but couldn't figure out how to insert those)

    Prove that if {w, x, y, z} is a linearly dependent subset of V, then so is {w+z, x+y, y+z, w-z}.

    Any help or guidance would be much appreciated!!
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  2. #2
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    Re: Proof using linearly independent

    Hey widenerl194.

    Hint: Remember that {x,y,z,w} is independent iff for real numbers a,b,c,d that ax + by + cz + dw = 0 holds iff a,b,c,d are all 0. You can assume the statement for {w,x,y,z} now try adjusting this to get the next statement.
    Thanks from widenerl194
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  3. #3
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    Re: Proof using linearly independent

    Okay so this is what I did:

    Let {w, x, y, z} be a linearly independent subset of V. For a,b,c,d, in the reals, aw+bx+cy+dz=0 because a=b=c=d=0.
    We want to show {w+x, x+y, y+z, w-z} is linearly independent.
    So, a(w+x) + b(x+y) + c(y+z) + d(w-z) = 0.
    Then, (a+d)w + (a+b)x + (b+c)y + (c-d)z = 0.
    So then we have: a + d=0, a+b=0, b+c=0, c-d=0. I put that into an augmented matrix and row reduced getting
    [1 0 0 0]
    [0 1 0 0]
    [0 0 0 1]
    [0 0 1 0]
    So, a=b=c=d=0. Therefore {w+x, x+y, y+z, w-z} is linearly independent.
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  4. #4
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    Re: Proof using linearly independent

    Looks good to me.
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