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Can a matrix be a vector?

  1. Apr 3, 2010 #1
    This question makes no since to me are these matrixes v1, v2 ,v3 actually vectors as stated in the problem attached?

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  2. jcsd
  3. Apr 3, 2010 #2


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    They are not vectors in the form that you usually encounter them (i.e. as a string of coordinates in n-dimensional space).
    However, if you call any element of a vector space (that is, a set that satisfies certain axioms) a vector, then indeed they are.

    Note by the way that there is always a bijection between n x m matrices and vectors in Rnm.
  4. Apr 3, 2010 #3
    How would I go about finding the determinate of them or would I have to go about another method to test for independence?

    I can tell the vector {4,4;4,4} would not be in the span because there is no multiple of d for({a, b;c,d} that would ever equal 4 because d is always 0.
  5. Apr 3, 2010 #4

    D H

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    You don't need the determinant to solve this problem. You are just looking for some a, b such that [itex]a\,\mathbf v_1 + b\,\mathbf v_2 = \mathbf v_3[/itex] -- or show that no such a, b exist.
  6. Apr 3, 2010 #5
    Thanks so a = -7, b =1;

    Span(v1,v2,v3) = {<a, -a; c , 0> : a, c belong to R} ?
  7. Apr 3, 2010 #6


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    Nope. Edit: Oops, my bad. I checked it again after seeing DH's post, and I'm the one who messed up.

    I suggest you use the standard definition of linear independence. [itex]\{v_1,\dots,v_k\}[/itex] is linearly independent if

    [tex]\sum a_i v_i=0\Rightarrow a_i=0 \text{ for all }i[/tex]

    (What DH said works too).
    Last edited: Apr 3, 2010
  8. Apr 3, 2010 #7

    D H

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    Good. So what does that mean in terms of the problem? (You need to determine whether v1, v2, v3 are dependent or independent.)

    That looks good.
  9. Apr 3, 2010 #8
    That means they are dependent? v3 depends on v1 and v2
  10. Apr 3, 2010 #9

    D H

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  11. Apr 3, 2010 #10
    Thank you
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