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Linear independence

  1. Feb 15, 2009 #1
    Determine whether the set {[1,2,-1,6], [3,8,9,10],[2,-1,2,-2]} is linearly independent.

    3. The attempt at a solution

    I construct

    [tex]A = \left[\begin{array}{ccccc} 1 & 2 & -1 & 6 \\ 3 & 8 & 9 & 10 \\ 2 & -1 & 2 & -2 \end{array}\right][/tex]

    The row echelon form is

    [tex]A = \left[\begin{array}{ccccc} 1 & 2 & -1 & 6 \\ 0 & 1 & 6 & -4 \\ 0 & 0 & 1 & -1 \end{array}\right][/tex]

    Now there is a theorem saying that if the "rank" of V is smaller than the number of vectors in the set under consideration (i.e., number of rows of V) then the vectors are linearly dependent; otherwise they're independent.

    I can't understand this step, how do we determine the "rank" of V?


    Furthermore, I have another question;
    There is a property that states: "if the set contains more vectors than the dimension of its member vectors, the vectors are linearly dependent." They are thus NOT linearly independent.
    So, what if the set contains fewer vectors than the dimension of its member vectors?? Here in my problem I have 3 vectors which are of the 4th dimensions, what does that tell us?

    Thanks!

     
  2. jcsd
  3. Feb 15, 2009 #2

    Office_Shredder

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    It tells you you need to solve the problem another way (if the number of vectors you have is less than the dimension of the vector space they're in, you can't immediately decide if they're linearly independent or not).

    You're in a bind here, since the rank is the dimension of the span of the row (or column, they will always be equal) vectors of the matrix. Fortunately row operations (or column operations if you're using the column definition) preserve rank, so what you have is useful. The problem boils down to finding if the vectors

    (1,2,-1,6), (0,1,6,-4), (0,0,1,-1) are linearly independent. Use the basic definition of linear independence: If a*(1,2,-1,6) + b(0,1,6,-4) + c(0,0,1,-1) = (0,0,0,0)

    Look at the first coordinate, then the second coordinate, then the third coordinate
     
  4. Feb 15, 2009 #3
    I understand, thanks.


    I just started reading this topic yesterday from the book and I’m a little confused atm.
    I don’t see any good explanation in my book, it only says: “by inspection the rank of A is…” it doesn’t elaborate on how to "inspect". I appreciate that if you could please demonstrate how to find the rank of A.

    Well, if the vectors are linearly dependent then the condition a(1,2,-1,6) + b(0,1,6,-4) + c(0,0,1,-1) = (0,0,0,0) must hold (they're not zero) & the equation can be rewritten as

    (1,2,-1,6) = -(b/a) (0,1,6,-4) – (c/a) (0,0,1,-1)

    (0,1,6,-4) = -(a/b)(1,2,-1,6) - (c/b) (0,0,1,-1)

    (0,0,1,-1) = -(a/c) (1,2,-1,6) - (b/c) (0,1,6,-4)

    The first form being possible if [tex]a \neq 0[/tex] and the second if [tex]b \neq 0[/tex] and so on.
     
  5. Feb 15, 2009 #4

    Office_Shredder

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    a(1,2,-1,6) + b(0,1,6,-4) + c(0,0,1,-1) = (0,0,0,0)

    Ok, so the first coordinate gives us the equation:
    a = 0

    That's pretty easy. So we look at the second coordinate
    2a + b = 0

    But a=0 (from the first coordinate). So we get b=0

    Then we see c=0 also. So if a(1,2,-1,6) + b(0,1,6,-4) + c(0,0,1,-1) = (0,0,0,0), then a=b=c=0
     
  6. Feb 17, 2009 #5
    Thank you, I get it now & I see that it's linearly independent. But I want to know how to find the rank of A. The book says: "by inspection, the rank of A is 3". The book doesn't explain how they got it. I know that if the rank is 3 (which is equal to the number of the vectors in the given set) makes the set linearly independent.
     
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