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Linear Independent vector

  1. May 3, 2005 #1
    Hi

    I'm presented with a set of vectors in [itex]\mathbb{R}^{3}[/itex]

    [itex]v_1 = (a_1,a_2,a_3)[/itex] [itex]v_2 = (b_1,b_2,b_3)[/itex] and [itex]v_3 = (c_1,c_2,c_3)[/itex]


    I'm suppose to determain if these vectors are linear independent.

    My question is: The most simple way of proving the above is that written the vectors as a set of linear combinations ???

    [itex] a_1 x + a_2 y + a_3 z = 0 [/itex]
    [itex] b_1 x + b_2 y + b_3 z = 0 [/itex]
    [itex] c_1 x + c_2 y + c_3 z = 0 [/itex]

    /Fred
     
  2. jcsd
  3. May 3, 2005 #2

    SpaceTiger

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    What do you suppose linear independence would imply for the solution to those equations?
     
  4. May 3, 2005 #3

    OlderDan

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    This looks like you are setting your three vectors to zero. That is not correct. What you need to show is that there is no linear combination of the three vectors that will add to zero unless the coefficients multiplying the three vectors (not their internal components) are individually zero.

    [itex]Av_1 + Bv_2 + Cv_3 = 0[/itex]

    if and only if A = B = C = 0
     
  5. May 3, 2005 #4
    Hi and thanks for Your answer.

    How I have understood Linear Independence is that in order to prove if a set of vectors are linear linear independent, then one has to write this set as linear combinations, where each lin.comb is set to equal cero.

    Then if there exist a solution for the set of linear combinations then the previously mentioned vectors are linear independent.

    Is that wrong?

    /Fred
     
  6. May 3, 2005 #5

    SpaceTiger

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    It's right, if you alter a bit to say that if there exists a non-trivial solution to the set of equations, then they're linearly independent. The trival solution would involve just setting x, y, and z to 0.

    So do you know how you can you figure out if there are non-trivial solutions?
     
  7. May 3, 2005 #6
    Hi

    [itex]x,y,z[/itex] are not zero. Its each of the linear combinations who I have set to equal zero.

    /Fred

     
  8. May 3, 2005 #7

    SpaceTiger

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    It's being assumed that you already know ax, bx, and cx. When you set it up in equation form like that, it implies that you are indeed solving for x, y, and z. One solution to the equations is that they are all equal to zero, but there may be other solutions. Whether or not there are depends on the values of ax, bx, and cx.
     
  9. May 3, 2005 #8

    OlderDan

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    I think this explains your confusion. You are thinking of x, y, and z, as possibly linearly independent and trying to show that a linear combination of x, y, and z can only be zero if the coefficients are zero. That is true, but that is not the issue here. Take it as given that x, y, and z ARE linearly independent. Now you are forming three other vectors using x, y, and z as a linearly independent basis for representing those three vectors. By non-trivial, what Space Tiger means is that those three vectors are NOT themselves zero. At least one of the coefficients in each vector is non-zero. That is given.

    You could construct three vectors by taking linear combinations of x, y, and z that are not linearly independent. For example, if you added some multiples of your first two vectors together to make the third, you would have three vectors, but they would not be independent. If you took the negative of the third vector and added it to those same multiples of the first two vectors you would get zero, and that would prove that your three vectors are not independent.

    It might help you to reduce the problem to two dimensions and look at

    [itex]Av_1 + Bv_2 [/itex]

    where

    [itex]v_1 = a_1x + a_2y [/itex] , [itex]v_2 = b_1x + b_2y [/itex]

    Draw some vectors in two dimensions that are not parallel to one another, and see if there is any way multiples of those vectors can add to zero. It cannot be done, unless both vectors are multiplied by zero. Now try to make a third vector

    [itex]v_3 = c_1x + c_2y [/itex]

    that cannot be formed by adding multiples of the first two vectors. That also cannot be done. There is no way you can have three linearly independent vectors in two dimensions, but there are infinitely many pairs of two linearly independent vectors in two dimensions.

    Now look again at your three dimensional problem. You have three non-zero vectors. They might be linearly independent, and they might not. They could all be multiples of the same vector. What you are trying to show here is that there is no way one of your three vectors can be written as a linear combination of the other two. None of your vectors are zero, and there is no combination of those vactors that is zero UNLESS all three coefficients in the combination of those three vectors is zero.
     
  10. May 5, 2005 #9

    xanthym

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    The following summarizes and clarifies info already presented or referenced in this thread.

    Define three vectors {V1, V2, V3 [tex]\mathbb{R}^{3}[/tex]} by the following:
    (Note: This definition differs from OP's.)

    [tex] 1: \ \ \ \ \vec{V}_{1} \ = \ \left [
    \begin{array}{ccc}
    a_{1} \\
    b_{1} \\
    c_{1} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{2} \ = \ \left [
    \begin{array}{ccc}
    a_{2} \\
    b_{2} \\
    c_{2} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{3} \ = \ \left [
    \begin{array}{ccc}
    a_{3} \\
    b_{3} \\
    c_{3} \\
    \end{array}
    \right ] [/tex]

    These vectors are Linearly Independent if and only if the scalar triple (x, y, z)=(0, 0, 0) is the SINGLE UNIQUE solution to:

    [tex] 2: \ \ \ \ x\vec{V}_{1} \ + \ y\vec{V}_{2} \ + \ z\vec{V}_{3} \, \ = \ \, \vec{\mathbf{0}} [/tex]

    Or equivalently, the single unique solution to:

    [tex] 3: \ \ \ \ \ x\left [
    \begin{array}{ccc}
    a_{1} \\
    b_{1} \\
    c_{1} \\
    \end{array}
    \right ]
    \ + \ y\left [
    \begin{array}{ccc}
    a_{2} \\
    b_{2} \\
    c_{2} \\
    \end{array}
    \right ]
    \ + \ z\left [
    \begin{array}{ccc}
    a_{3} \\
    b_{3} \\
    c_{3} \\
    \end{array}
    \right ]
    \ \ = \ \
    \left [
    \begin{array}{ccc}
    0 \\
    0 \\
    0 \\
    \end{array}
    \right ]
    [/tex]

    Or equivalently, that (x, y, z)=(0, 0, 0) is the single unique solution to:

    [tex] 4: \ \ \ \
    \begin{array}{ccccccc}
    a_1 x & + & a_2 y & + & a_3 z & \, = \, & 0 \\
    b_1 x & + & b_2 y & + & b_3 z & \, = \, & 0 \\
    c_1 x & + & c_2 y & + & c_3 z & \, = \, & 0 \\
    [/tex]

    The triple (x, y, z) will be the single unique solution to Eq #4 if and only if the determinant of its coefficients is NON-zero:

    [tex] 5: \color{red}\ \ \ \ \ \ \left |
    \begin{array}{ccc}
    a_{1} & a_{2} & a_{3} \\
    b_{1} & b_{2} & b_{3} \\
    c_{1} & c_{2} & c_{3} \\
    \end{array}
    \right | \ \ \neq \ \ \textsf{0} \ \ \ \color{black} \Longleftrightarrow \ \ \ \color{red} \{\vec{V}_{1}, \ \vec{V}_{2}, \ \vec{V}_{3} \ \ \ \textsf{Linearly Independent} \} [/tex]

    Eq #5 probably provides the easiest method for determining Linear Independence in [itex]\mathbb{R}^{3}[/itex].
    For convenience, Left Side of Eq #5 is rapidly determined from:

    [tex] 6: \ \ \ \ \left [
    \begin{array}{ccc}
    a_{1} & a_{2} & a_{3} \\
    b_{1} & b_{2} & b_{3} \\
    c_{1} & c_{2} & c_{3} \\
    \end{array}
    \right ] \ \ = \ \ \left [ \ \vec{V}_{1} \ \ \vec{V}_{2} \ \ \vec{V}_{3} \ \right ]
    [/tex]


    Note that only 3 possibilities exist for Eq #4 solutions:
    1) Single, Unique Solution
    2) Infinite # of Solutions
    3) No Solutions
    Eq #5 indicates situation #1.



    ~~
     
    Last edited: May 6, 2005
  11. May 5, 2005 #10
    I think you mean the oppsite! If there are non-trivial solutions, the vectors are dependent. The converse is also true.
     
  12. May 6, 2005 #11

    xanthym

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    For the homogeneous case, if 1 non-trivial solution exists, then an infinite number exist (and of course the vectors are dependent).

    It should also be noted that Linear Independence for given vectors {V1, V2, V3 [tex]\mathbb{R}^{3}[/tex]} is usually stated in terms of:

    [tex] 2: \ \ \ \, \ \ \ \ \ \ x\vec{V}_{1} \ \ + \ \ y\vec{V}_{2} \ \ + \ \ z\vec{V}_{3} \ \ \, = \, \ \ \vec{\mathbf{0}} \ \ \ \ \ \ \ \ \textrm{ (solve for Real scalars x, y, z) } [/tex]

    In fact, any Real CONSTANT vector can appear on the Right Side:

    [tex] 2a: \ \ \ \ x\vec{V}_{1} \ + \ y\vec{V}_{2} \ + \ z\vec{V}_{3} \ \ = \ \ \left [
    \begin{array}{c}
    261 \\
    -38.2 \\
    7 \\
    \end{array} \right ] [/tex]

    The necessary and sufficient condition for Linear Independence is that Eq #2a (like that for Eq #2) has a SINGLE UNIQUE solution triple (x, y, z). (There's no need to actually find it.)

    Again, the necessary and sufficient condition for a single unique solution of Eq #2a (like that for Eq #2) is that the determinant of its coefficients be NON-zero:
    (See Msg #9 for more details.)

    [tex] 5: \color{red}\ \ \ \ \ \ \left |
    \begin{array}{ccc}
    a_{1} & a_{2} & a_{3} \\
    b_{1} & b_{2} & b_{3} \\
    c_{1} & c_{2} & c_{3} \\
    \end{array}
    \right | \ \ \neq \ \ \textsf{0} \ \ \ \color{black} \Longleftrightarrow \ \ \ \color{red} \{\vec{V}_{1}, \ \vec{V}_{2}, \ \vec{V}_{3} \ \ \ \textsf{Linearly Independent} \} [/tex]

    where:

    [tex] 1: \ \ \ \ \vec{V}_{1} \ = \ \left [
    \begin{array}{ccc}
    a_{1} \\
    b_{1} \\
    c_{1} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{2} \ = \ \left [
    \begin{array}{ccc}
    a_{2} \\
    b_{2} \\
    c_{2} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{3} \ = \ \left [
    \begin{array}{ccc}
    a_{3} \\
    b_{3} \\
    c_{3} \\
    \end{array}
    \right ] [/tex]


    ~~
     
    Last edited: May 6, 2005
  13. May 6, 2005 #12
    Hi and thanks for Your answer,

    If I have three vectors in [tex]\mathbb{R}^2[/tex]

    [tex]\vec{v_1} = \left( \begin{array}{ccc} a_1 \\ a_2\\ \end{array} \right )[/tex]

    [tex]\vec{v_2} = \left( \begin{array}{ccc} b_1 \\ b_2\\ \end{array} \right )[/tex]

    [tex]\vec{v_3} = \left( \begin{array}{ccc} c_1 \\ c_2\\ \end{array} \right )[/tex]

    What is the most commen way of proving if these vectors are linear independent or not?

    I have heard that if the vectors [tex]\mathrm{span} \{ \mathbb{R}^2 \}[/tex] then these vectors [tex]\vec{v_1}, \vec{v_2}, \vec{v_3}[/tex] are linear independent.

    Is that correct ???

    Sincerely
    Fred
     
    Last edited: May 6, 2005
  14. May 6, 2005 #13

    xanthym

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    Three (3) vectors in [tex] \displaystyle \mathbb{R}^2 [/tex] are NEVER Linearly Independent.


    ~~
     
    Last edited: May 6, 2005
  15. May 6, 2005 #14
    Sorry,

    I'm beginning to get tired. :zzz:

    Hope You can forgive me ?? :blushing:

    A set of vectors in [tex]\mathbb{R}^2[/tex] are NEVER linear independent because they are situated in the same plane.

    Futhermore in my given situation there are three vectors with only two entries each. Therefore these vector are linear dependent.

    /Fred

     
  16. May 6, 2005 #15

    xanthym

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    .
    For Mathman23:


    *** POP QUIZ ***

    Given two (2) vectors {V1, V2 [tex]\mathbb{R}^{3}[/tex]}, state a necessary and sufficient condition that these vectors are Linearly Independent.



    CLICK BETWEEN DASHED LINES BELOW To Reveal Answer In Pop-Up Window
    ---------------------
    [tex]\color{white} ANSWER: V1 X V2 NotEqualTo 0 [/tex]
    ---------------------
    CLICK BETWEEN DASHED LINES ABOVE To Reveal Answer In Pop-Up Window


    ~~
     
    Last edited: May 6, 2005
  17. May 6, 2005 #16

    OlderDan

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    If the number of vetors in your set were equal to the dimensionality of the space, then if the vectors spaned the space they would be linearly independent. You just have too many vectors in your set for this condition to be meaningful.

    I'm wondering if you are able to follow some of the other responses that have been given here. It seems to me the notation has been substantially changed around from the problem you originally stated. As I understood your early posts, you were talking about vectors in a function space and had chosen as a basis the linearly independent functions x, y, and z with coefficients given by the defining equations

    [itex]v_1 = (a_1,a_2,a_3)[/itex] [itex]v_2 = (b_1,b_2,b_3)[/itex] and [itex]v_3 = (c_1,c_2,c_3)[/itex]

    Mapping the problem into a 3 dimensional Euclidean vector space is certainly a good way to approach the problem, but the vectors

    [tex]\ \ \ \ \vec{V}_{1} \ = \ \left [ \begin{array}{ccc}a_{1} \\b_{1} \\c_{1} \\\end{array}\right ]\ \ \ \ \ \ \vec{V}_{2} \ = \ \left [ \begin{array}{ccc}a_{2} \\b_{2} \\c_{2} \\\end{array}\right ]\ \ \ \ \ \ \vec{V}_{3} \ = \ \left [ \begin{array}{ccc}a_{3} \\b_{3} \\c_{3} \\\end{array}\right ] [/tex]

    have redefined the coefficients, and the linear combination

    [tex]\ \ \ \ x\vec{V}_{1} \ + \ y\vec{V}_{2} \ + \ z\vec{V}_{3} \, \ = \ \, \vec{\mathbf{0}} [/tex]

    has given new meaning to x, y, and z. They are analogous to the coefficients I wrote earlier as A, B, and C. In the function space spanned by x, y, z, the condition x = y = z = 0 would be forcing your original function vectors to [itex] \vec{\mathbf{0}} [/itex] There is nothing wrong with different representations of vectors in [tex]\mathbb{R}^3[/tex] , but remembering back to my first encounter with function spaces, I think I would have been confused by these changes. But that's just me. If you have it sorted out don't worry about it. If it is confusing to you, I'm sure the notation can be sorted out.
     
  18. May 6, 2005 #17

    xanthym

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    O-Dan:

    You're certainly correct that there might be some confusion regarding notation in this thread. Nevertheless, the objective of Msg #9 (and subsequently of Msg #11) was to present this concept:

    [tex] 5: \color{red}\ \ \ \ \ \ \left |
    \begin{array}{ccc}
    a_{1} & a_{2} & a_{3} \\
    b_{1} & b_{2} & b_{3} \\
    c_{1} & c_{2} & c_{3} \\
    \end{array}
    \right | \ \ \neq \ \ \textsf{0} \ \ \ \color{black} \Longleftrightarrow \ \ \ \color{red} \{\vec{V}_{1}, \ \vec{V}_{2}, \ \vec{V}_{3} \ \ \ \textsf{Linearly Independent} \} [/tex]

    where:

    [tex] 1: \ \ \ \ \vec{V}_{1} \ = \ \left [
    \begin{array}{ccc}
    a_{1} \\
    b_{1} \\
    c_{1} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{2} \ = \ \left [
    \begin{array}{ccc}
    a_{2} \\
    b_{2} \\
    c_{2} \\
    \end{array}
    \right ]
    \ \ \ \ \ \ \vec{V}_{3} \ = \ \left [
    \begin{array}{ccc}
    a_{3} \\
    b_{3} \\
    c_{3} \\
    \end{array}
    \right ] [/tex]

    The above stands on its own and should be fairly clear.
    (Above Eq #'s reference original #'s in Msg #9.)


    ~~
     
    Last edited: May 7, 2005
  19. May 7, 2005 #18

    OlderDan

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    Standing on its own it is clear. My only concern is that the OP, who seems to be wrestling a bit with the concepts, presented the same three equations you presented in your derivation

    From the OP in post #1

    [itex] a_1 x + a_2 y + a_3 z = 0 [/itex]
    [itex] b_1 x + b_2 y + b_3 z = 0 [/itex]
    [itex] c_1 x + c_2 y + c_3 z = 0 [/itex]

    From your post #9

    [tex] 4: \ \ \ \ \begin{array}{ccccccc}a_1 x & + & a_2 y & + & a_3 z & \, = \, & 0 \\b_1 x & + & b_2 y & + & b_3 z & \, = \, & 0 \\c_1 x & + & c_2 y & + & c_3 z & \, = \, & 0 \\[/tex]

    A careful look at the starting vectors and the linear combinations in the two cases shows that these sets of equations are not equivalent, even though they look identical. I wouldn't want the OP to conclude that the original three equations were the path to the proof.

    My recollection of my own conceptual difficulties with function spaces is no doubt a contributing factor to my concern. I thought I knew everything there was to know about vector algebra, and probably did when the representation was in terms of three dimensional spacial vectors with an i, j, and k unit vector basis, or a set of basis vectors that could be expressed in terms of i, j, k. The first time I encountered the idea that "vectors" can be written in the form

    [itex] a_1 + a_2 x + a_3 x^2 [/itex]

    or

    [itex] a_1 x + a_2 y + a_3 z [/itex]

    because 1, [itex] x [/itex] and [itex] x^2 [/itex] are linearly independent functions, and [itex] x [/itex] , [itex] y [/itex] and [itex] z [/itex] are linearly independent functions, it kinda blew my mind.
     
  20. May 9, 2005 #19
    Hello again,

    From the later stuff You Guys have written regarding my initial question I have become a bit unsure.

    Is the method where you take the determinant of the coefficients for the six linear equation, and if the derterminant is different from zero , then the three vectors in [tex]\mathbb{R}^3[/tex] linear independent.

    Is this the most commen method of dertermining if a set of vectors in [tex]\mathbb{R}^3[/tex] linear independent ??

    Sincerely

    Fred

     
  21. May 9, 2005 #20

    OlderDan

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    I am certainly not trying to confuse you, but my sense is that you have been unsure from the beginning, which is why I thought it good to point out that the 3 equations you originally posted were not equivalent to the three equations posted by xanthym. I am going to take the liberty of using xanthym's post, exactly as it was structured, but changing the symbols to match your original presentation of the problem so that you can see the difference.

    So here is xanthym's post, with changes needed to match the three vectors as you defined them. I will comment on the changes I make

    The three equations for three unknowns in #4 can be solved by various techniques. The evaluation of the determinant as a test for the category of solution may be the easiest way for a 3 dimensional case. For higher dimensions, calculating a determinant gets a bit messy, and other matrix reduction methods might be more efficient.

    I hope this makes it clear how to put your problem into the form of the solution xanthym presented, and why the x, y, and z you used in your original post are not the same as the x, y, and z that xanthym used.
     
    Last edited: May 9, 2005
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