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I'm terrible with proofs linear algebra problem

  1. Nov 23, 2009 #1
    let { v_1, v_2, v_3, ... v_m } (m > 3) be linearly independent vectors in a vector space V. Prove that the set { v_1 - 2v_2 + v_3, v_2, v_3, ..., v_m } is also Linearly.







    I did this:

    a_1( v_1 - 2v_2 + v_3) + a_2(v_2) + a_3(v_3) + ... + a_m(v_m)
    = a_1(v_1) - a_1(2v_2) + a_1(v_3) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m)
    = -a_1(2v_2) + a_1(v_3) + a_1(v_1) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m)

    and since we know the first part, we know that a_1(v_1) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m) is Linearly independent so we know that a_1 = a_2 = a_3 = ... = a_m = 0. so we can see now that the first two terms have scalars a_1 and a_1 which equal zero to the second set of vectors is also Linearly independent.
     
  2. jcsd
  3. Nov 23, 2009 #2

    Office_Shredder

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    You have to combine your like vectors here

    -a_1(2v_2) + a_1(v_3) + a_1(v_1) + a_2(v_2) + a_3(v_3) + ... +a_m(v_m)

    Becomes

    [tex] a_1v_1 + (a_2-2a_1)v_2 + (a_3+a_1)v_3 + a_4v_4...[/tex]

    and then each coefficient here comes out to zero
     
  4. Nov 23, 2009 #3

    Mark44

    Staff: Mentor

    To prove that a set of vectors {v1, v2, ..., vn} is linearly independent, you have to show that this equation has EXACTLY ONE SOLUTION for the constants c1, c2, ..., cn:
    c1v1 + c2v2 + ... + cnvn = 0, namely, the solution where c1 = c2 = ... cn = 0.

    The part about EXACTLY ONE SOLUTION is significant here, and is something that beginning linear algebra students miss. The definition for a set of linearly dependent vectors has the same equation as above, but there will be a solution to the equation where at least one of the constants isn't zero. That's too subtle for many linear algebra students.
     
  5. Nov 24, 2009 #4
    although it may not be apparent in my attempt at a proof, i do know the definitions of linearly independent and linearly dependent. are you saying that what i have done does not necessarily prove that there is only one possible (trivial) solution? i can understand how a right answer works, but what is wrong with my answer?
     
  6. Nov 24, 2009 #5

    Mark44

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    I wasn't sure that you understood the definitions, since you didn't even have an equation. You had an expression, a_1( v_1 - 2v_2 + v_3) + a_2(v_2) + a_3(v_3) + ... + a_m(v_m), which you were manipulating.

    The key thing is that you need to show that the equation a_1*v_1 + ... + a_m*v_m = 0 has one and only one solution.

    For example, let v_1 = (1, 0, 0), v_2 = (0, 1, 0, and v_3 = (1, 1, 0)
    If I form the equation c_1*v1 + c_2*v2 + c_3*v_3 = 0, I can see that c_1 = 0, c_2= 0, and c_3 = 0 makes this equation a true statement. Can I conclude that the three vectors are linearly independent?

    In fact, I can't, because there is another solution, c_1 = 1, c_2 = 1, and c_3 = -1. So my solution where all three constants were 0 is not the single, solitary, sole, unique, all by itsef solution to this equation.
     
  7. Nov 24, 2009 #6
    what i meant was a_1( v_1 - 2v_2 + v_3) + a_2(v_2) + a_3(v_3) + ... + a_m(v_m) = 0

    and i was trying to prove that a_i = 0. and then using the other set of vector's linear independence showed that the terms have only the solution of a_1 = a_2 = a_3 = 0.
     
  8. Nov 24, 2009 #7

    Mark44

    Staff: Mentor

    OK, that's the ticket.
     
  9. Nov 24, 2009 #8
    but still, that is incorrect? i thought that since the a_i from the first set of vectors were trivial, then if that same linear independent set of vectors is in the second set of vectors, then the a_i that it had in common would be zero....?

    i know i am being annoying probably, but i need to get this stuff...
     
  10. Nov 24, 2009 #9

    Office_Shredder

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    Like I said in my post.

    You haven't written that expression out in the form [tex] \lambda_1v_1 + \lambda_2v_2 +...[/tex] and asked when it eqals zero, so you can't say anything about the coefficients yet,

    Suppose the sum is equal to zero. Re-arrange it to get this

    [tex]
    a_1v_1 + (a_2-2a_1)v_2 + (a_3+a_1)v_3 + a_4v_4...[/tex]


    Now what, exactly, can you say must be zero?
     
  11. Nov 24, 2009 #10

    Mark44

    Staff: Mentor

    I think that you get the idea, but some of the stuff you said in your OP is incorrect or vague. No, I don't think you're being annoying. If you don't understand, it is not being annoying to ask questions to try to clear things up.

    Linearly what? independent? dependent?
    There's not a zero in sight here anywhere.
    The sum of the vectors is not linearly independent. It's the vectors themselves that are or are not linearly independent. The fact that a_1 = a_2 = a_3 = ... = a_m = 0 is immaterial. That happens whether the vectors are linearly independent or linearly dependent. What you have to establish is that there is no other solution for these constants (to prove linear independence).

    You are given that the vectors v_1, ..., v_m are linearly independent. This means that a_1*v_1 _ a_2*v_2 + a_3*v_3 + ... + am_v_m = 0 has only the trivial solutions for the constants a_1, a_2, ..., a_m.

    Let's look at this equation...
    c_1*(v_1 - 2v_2 + v_3) + c_2*v_2 + c_3*v_3, ..., + c_m*v_m = 0 [Note: I can always do this for any set of vectors from the same vector space.]
    <==> c_1*v_1 + (-2c_1 + c_2)*v_2 + (c_1 + c_3)*v_3 + c_4*v_4 + ... + c_m*v_m = 0

    Now since {v_1, v_2, v_3, ..., v_m} is a linearly independent set of vectors, the only solution to the equation above is the trivial solution: c_1 = 0, -2c_1 + c_2 = 0, c_1 + c_3 = 0, c_4 = 0, ..., c_m = 0.
    c_1 = 0 ==> c_2 = 0.
    c_1 = 0 ==> c_3 = 0.
    So all the constants are zero, and there is no other solution.
    Therefore the set of vectors {v_1 - 2v_2 + v_3, v_2, v_3, ..., v_m} is linearly independent.
     
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