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Math Help - Proof of linear dependence of coplanar vectors

  1. #1
    Senior Member OReilly's Avatar
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    Proof of linear dependence of coplanar vectors

    There is a theorem in my math book which proof I don't understand.

    Theorem is: Vectors x,y,z are linear dependent if they are coplanar and two of them are collinear.

    Proof:
    1) Vectors x,y,z are in plane alpha.
    2) Vectors x and y are collinear.
    3) There is real number k != 0 such that y = kx
    4) From 3) follows that y - kx + 0z = 0 which means that they are linear dependent.


    I don't understand step 4). Where did come from 0z?
    Can someone explain me step 4)?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OReilly View Post
    There is a theorem in my math book which proof I don't understand.

    Theorem is: Vectors x,y,z are linear dependent if they are coplanar and two of them are collinear.

    Proof:
    1) Vectors x,y,z are in plane alpha.
    2) Vectors x and y are collinear.
    3) There is real number k != 0 such that y = kx
    4) From 3) follows that y - kx + 0z = 0 which means that they are linear dependent.


    I don't understand step 4). Where did come from 0z?
    Can someone explain me step 4)?
    Recall what it means to be linearly dependent:

    A subset S of vector space V is called linearly dependent if there exist a finite number of distinct vectors v1, v2, ..., vn in S and scalars a1, a2, ..., an, not all zero, such that
    Note that the zero on the right is the zero vector, not the number zero.



    Here your distinct vectors are x, y, z as opposed to v_1, V_2,...,v_n. since you are considering only 3 vectors, you can prove they are linearly independent by showing there are distinct constants (that's why it says not all zero) that you can multiply each vector by and get the zero vector. this proof simply chose those three constants to be 1, -k, and 0.



    (1)y + (-k)x + (0)z = 0


    You get all the other steps right?
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Jhevon View Post
    Recall what it means to be linearly dependent:

    A subset S of vector space V is called linearly dependent if there exist a finite number of distinct vectors v1, v2, ..., vn in S and scalars a1, a2, ..., an, not all zero, such that
    Note that the zero on the right is the zero vector, not the number zero.



    Here your distinct vectors are x, y, z as opposed to v_1, V_2,...,v_n. since you are considering only 3 vectors, you can prove they are linearly independent by showing there are distinct constants (that's why it says not all zero) that you can multiply each vector by and get the zero vector. this proof simply chose those three constants to be 1, -k, and 0.



    (1)y + (-k)x + (0)z = 0


    You get all the other steps right?
    Yes, I understand now all.
    In order to prove linear dependency of coplanar vectors x,y,z we must prove that there is at least one constant k != 0 so that k_1x + k_2y + k_3z = 0.
    Since we have kx - y = 0 then it's k_1 = k, k_2 = -1 and chosing for k_3 = 0 isn't problem because all others are not zero so linear dependency is proven.
    In other words we found that they aren't all zero which is condition for linear independency.

    Thanks for help!
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