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Math Help - Vectors forming a basis,spanning

  1. #1
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    Vectors forming a basis,spanning

    Do the following vectors form a basis of R^3, span R^3, or neither?

    a1 = (1, 2, 1), a2 = (-1, 0, -1), a3 = (0, 0, 1)

    Do I just check to see if they're linearly independent?
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  2. #2
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    Do I just check to see if they're linearly independent?
    Correct. Since the dimension of the space is 3, and you have 3 vectors, it follows that being a basis is equivalent to spanning the space.
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    Ok, so can I say this is true:

    If the number of vectors equals the n in R^n, then they are spanning? So, in this case, if there are two vectors, they do not span? What if I have more than 3 vectors?

    I can also say that if they span AND they are linearly independent, then the vectors form a basis?
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    If the number of vectors equals the n in R^n, then they are spanning?
    Only if they are linearly independent. In that case, they are also a basis.

    What if I have more than 3 vectors?
    In three-dimensional space, more than 3 vectors could span the space, but they definitely won't be a basis, because they won't be linearly independent.

    I can also say that if they span AND they are linearly independent, then the vectors form a basis?
    Yes.
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  5. #5
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    Thank you for your quick help.

    For clarification:

    If I have two vectors, for a 3-dimensional space, these vectors can not span or be a basis. This is true?

    If I have four vectors, for a 3-d space, these vectors CAN span, but can not be a basis, since they will not be linearly independent. This is true?

    How would you show that the 4 vectors span?
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    If I have two vectors, for a 3-dimensional space, these vectors can not span or be a basis. This is true?
    True.

    If I have four vectors, for a 3-d space, these vectors CAN span, but can not be a basis, since they will not be linearly independent. This is true?
    True.

    How would you show that the 4 vectors span?
    Write an arbitrary vector x in the space as a linear combination of the 4 vectors. That will generate a linear system of equations. If there are any solutions, then the vectors span. If there are not solutions, the vectors do not span. Make sense?
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    With the linear system of equations, what is the right hand equal to for the equations?
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    With the linear system of equations, what is the right hand equal to for the equations?
    The components of your arbitrary vector x.
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    Hmm, I don't quite get it. I have four vectors: (-1, 2, 3), (0, 1, 0), (1, 2, 3), (-3, 2, 4).

    Would I be solving for:

    -1*X_1 + X_3 - 3*X_4 = Y_1
    2*X_1 + X_2 + 2*X_3 + 2*X_4 = Y_2
    3*X_1 + 3*X_3 + 4*X_4 = Y_3

    I'm sure I didn't set that up right. There are 7 unknowns and 3 equations.
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  10. #10
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    Your system is correct, but you're interpreting it incorrectly. In the context of solving that system, you don't treat your Y_{j} as unknowns for which to solve. They are just arbitrary numbers. Treat them like you would treat the RHS of any system. You need to solve for the X_{k}'s. Make sense?
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  11. #11
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    I can't solve this to get each X in terms of only Y's.

    Does this mean it does not span?
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  12. #12
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    You can't solve it exactly to get a unique solution. However, you're not after a unique solution (which, incidentally, would correspond with being a basis). You're after any solutions at all. Since the system is under-determined, if you have any solutions, you're going to have infinitely many solutions. You're checking to see whether you have zero solutions, or infinitely many solutions. Zero solutions means you don't have a spanning set. Infinitely many solutions means you have a spanning set, but it's not a basis. Exactly one solution means you have yourself a basis. Make sense?
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  13. #13
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    Very much so. Thank you for your help.
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  14. #14
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    Great. You're welcome. Have a good one!
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