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Question about vector independence

  1. Dec 11, 2014 #1
    So I was reading my textbook and I confused myself about a theorem

    Where if S={v,v2,.....,vr} and in ℝn then if r>n, then it is linearly dependent

    It doesnt make sense to me because if we look at 2 vectors in ℝ3 (lets say u and v)

    we have u=(u1,u2,u3) and v=(v1,v2,v3)
    So i do k1(u1,u2,u3)+k2(v1,v2,v3)=0
    If i use a matrix:

    u1 v1 0
    u2 v2 0
    u3 v3 0

    Then it would seem to me as both vectors would be linearly dependent, no?
     
  2. jcsd
  3. Dec 11, 2014 #2

    Stephen Tashi

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    The theorem doesn't say that 2 vectors in R^3 cannot be dependent. It doesn't say that there is dependence "if and only if r > n".
     
  4. Dec 11, 2014 #3
    But if I solve for a matrix of a 2x3, I would end up with a free variable no? ( 3rd line, after I REF or RREF the matrix) so that would make it linearly dependent
     
  5. Dec 11, 2014 #4
    The theorem does not say that 2 vectors in [itex]\mathbb{R}^3[/itex] must be independent. There is no contradiction.
     
  6. Dec 11, 2014 #5
    Well I don't know what im trying to say is that I see a contradiction, more like " what is wrong in my reasoning" type of thing. I'll give a concrete example I just did, maybe my question will be more clear

    Is this set of vector linearly dependent

    (8,-1,3),(4,0,1)

    8 4 0 --> ( interchange row 2 with row 1, then row 3 with "new" row 2)
    -1 0 0
    3 1 0

    -1 0 0
    3 1 0
    8 4 0
    We end up with 2 rows of leading one, and one row that I can reduce to 0 0 0 ( the third one )easily, so how isn't that linearly dependent?
     
  7. Dec 11, 2014 #6

    Stephen Tashi

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    The two vectors (8,-1,3),(4,0,1) in your example are not linearly dependent.

    To determine if two 3-dimensional vectors [itex](u1,u2,u3)[/itex] and [itex] (v1,v2,v3) [/itex] are dependent, you need to find non-zero solutions [itex] x [/itex] and [itex] y [/itex] to the vector equation:

    [itex] x ( u1,u2,u3) + y (v1,v2,v3) = 0 [/itex]

    Written in matrix form, this can be expressed as:

    [itex] \begin{pmatrix} u1&v1 \\ u2 & v2\\ u3&v3\end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\0\end{pmatrix} [/itex].

    Apparently you analyze this equation by writing the "augmented matrix" for it ( where the constants on the right-hand side are put in the matrix, we are dealing with:

    \begin{pmatrix} u1 & v1 & 0\\ u2&v2& 0\\u3&v3 & 0\end{pmatrix}

    If we row-reduce the matrix so the bottom row is zeroes, what does that tell us?

    If it reduces to [itex] \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix} [/itex] , this implies that [itex] x = 0, \ y = 0 [/itex] is the unique solution to the equations, so the two vectors are linearly independent.

    If it reduces to something like [itex] \begin{pmatrix} 1 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} [/itex] , this implies there are solutions like [itex] x = 2, y = 1 [/itex] and [itex] x = 4, y = 2 [/itex] etc., so the two vectors are dependent.
     
  8. Dec 11, 2014 #7
    Oh I didn't know you guys had a matrix format that I could use... sorry about that

    So, trivial solutions are only determined whether or not a x or y ( in this case) variable would have a leading one or not? just seems kinda odd that one row could be 0 without affecting the depedence of the vector
     
  9. Dec 11, 2014 #8

    Stephen Tashi

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    You might be confusing two different techiques for testing independence. Another (and more common technique) for testing the indepdence for a set of n-dimensional vectors is to write the vectors as rows in a matrix and try to row reduce the matrix toward the identity. If you get a row of zeros in that technique you have managed to express the zero vector as a linear combination of other rows, so the vectors are dependent.

    If you row reduce [itex]\begin{pmatrix} u1 & u2 & u3 \\ v1 & v2 & v3 \end{pmatrix}[/itex] to something like [itex] \begin{pmatrix} 1 &-2 &0 \\ 0 & 0 & 0 \end{pmatrix} [/itex] then the vectors are dependent.

    If you row reduce it something like [itex] \begin {pmatrix} 1 & 0 & -2 \\ 0 & 1 & 7 \end{pmatrix}[/itex] the vectors are independent.
     
  10. Dec 12, 2014 #9

    Mark44

    Staff: Mentor

    If you have two vectors to consider, it's very easy to determine whether they are linearly independent. If neither one is a constant multiple of the other, the two vectors are linearly independent.

    It's a lot harder when you have three or more vectors. Even if no one vector is a multiple of any of the others, the set can still be linearly dependent. For example, consider S = {<1, 1, 0>, <1, -1, 0>, <2, 0, 0>}. No one vector is a multiple of any of the others, but the equation c1v1 + c2v2 + c3v3 = 0, has a solution where not all of the constants are zero.
     
  11. Dec 12, 2014 #10

    FactChecker

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    That is not always possible: If u=(1,0,0) and v=(0,1,0) it is not possible to find nonzero k1 and k2 that will give 0.
    But it can certainly be true: If u=(1,0,0) and v = (-1,0,0) clearly k1=1 and k2=1 would give 0.
    So as long as the number of vectors is less than or equal to the dimension of the space, it can go either way.

    However, for R3, it is impossible to find 4 vectors where no set of them are linearly dependent. That is what the theorem says.
     
  12. Dec 13, 2014 #11

    Fredrik

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    The theorem says that for all positive integers ##n##, every subset of ##\mathbb R^n## with cardinality (=number of elements) greater than ##n## is linearly independent. So in particular, it says that every subset of ##\mathbb R^3## with at least four elements is linearly dependent. It says nothing about sets with only two elements, like your ##\{u,v\}##.

    Some subsets of ##\mathbb R^3## with two elements are linearly dependent, and some are linearly independent. ##\{(1,0,0),(2,0,0)\}## is an example of the former, and ##\{(1,0,0),(0,1,0)\}## is an example of the latter.
     
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