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Linearly dependent sets

  1. Feb 14, 2010 #1
    If A is a 3x3 Matrix and {v1, v2, v3} is a linearly dependent set of vectors in R^3, then {Av1, Av2, Av3} is also a linearly dependent set???

    Is this true? Can someone please explain why or why not??

    What I think: I think it is true because I read that a linear transformation preserves the operations of vector addition and scalar multiplication.
  2. jcsd
  3. Feb 14, 2010 #2
    If v1,v2,v3 are linearly dependent you can find constants a1, a2, a3 not all 0 such that
    [tex]a_1v_1 + a_2v_2 + a_3v_3 = 0[/tex]
    Now left-multiplying this by A you get:
    [tex]A(a_1v_1+a_2v_2+a_3v_3) = A0=0[/tex]
    Now use your rules for matrix arithmetic to derive:
    (HINT: Ak = kA for constants k, and A(v+w) = Av+Aw for vectors v, w where the expression makes sense).
  4. Feb 14, 2010 #3
    That makes sense. So if we have a1(Av1) + a2(Av2) + a3(Av3) = 0, then at least one of the constants could be zero and that will definitely result to a linearly dependent set.

    That leads me to a related theory: Let's assume we are talking about {v1, v2, v3} being a linearly INDEPENDENT set now. If we multiply the vectors by the matrix A, how does it affect the independece? Would it make a differerence if the matrix A is invertible?
  5. Feb 14, 2010 #4
    This is actually a quite interesting little question (well in my opinion anyway). First for fixed A, v1,v2,v3 note that if we take the contrapositive of your initial result we get:
    If Av1, Av2, Av3 are linearly independent, then v1,v2,v3 are linearly independent.
    so for linear independence it goes backwards. For an arbitrary matrix A we can not prove your new statement since we can just let A be the 0 matrix. However if A is invertible, then we can just go backwards by noting that if,
    [tex]a_1Av_1+a_2Av_2+a_3Av_3 = 0[/tex]
    Then we can left-multiply by [itex]A^{-1}[/itex] to get,
    [tex]a_1v_1+a_2v_2+a_3v_3 = 0[/tex]
    so if v1,v2,v3 are linearly independent and A is invertible, then Av1, Av2, Av3 are linearly independent.
  6. Feb 15, 2010 #5
    Thanks rasmhop... I did think "A" being an invertible matrix could make a difference but I didn't know how to prove it.
    That was a very good help from you.
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