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Thread: Determine if each of the following sets is a basis for the indicated subspace

  1. #1
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    Determine if each of the following sets is a basis for the indicated subspace

    {[1,2,3]T,[4,5,6]T,[7,8,9]T} in R3

    So, I've learned that for a set of vectors to be a basis it has to satisfy 2 of 3 of the following rules (if it meets two, the third is given):
    1.) Must be linearly independent
    2.) Must be a spanning set
    3.) the set has dimension(V) elements

    I've shown that the set is not linearly independent so it doesn't meet #1.
    The set does meet #3.

    I am running in to some confusion on how to test if it is a spanning set. Would I set up an augmented matrix and solve to test for consistency?
    Also, I had previously thought that if a set was LI, then it could not be a basis...
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  2. #2
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    Re: Determine if each of the following sets is a basis for the indicated subspace

    $\mathbb{R}^3 \text{ has 3 dimensions so we must have 3 linearly independent vectors of 3 elements each in order to span it.}$

    $\text{Since the 3 vectors are not linearly independent they cannot be a spanning set of }\mathbb{R}^3$

    A more interesting problem is when you have greater than 3, 3 dimensional vectors.

    Clearly they aren't all linearly independent but do they span $\mathbb{R}^3?$

    They do if a subset of 3 of them are linearly independent.
    Last edited by romsek; Oct 13th 2018 at 11:38 AM.
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    Re: Determine if each of the following sets is a basis for the indicated subspace

    Ok, then being that they are not linearly independent, and not a spanning set then it is not a basis?

    Also, I noticed that the middle vector is the average of the first two vectors.. does this mean I can remove that and get a basis?
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    Re: Determine if each of the following sets is a basis for the indicated subspace

    Quote Originally Posted by MrJank View Post
    Ok, then being that they are not linearly independent, and not a spanning set then it is not a basis?

    Also, I noticed that the middle vector is the average of the first two vectors.. does this mean I can remove that and get a basis?
    Yes, a basis both spans it's space and it's elements are linearly independent.

    yes but only for a 2 dimensional subspace of $\mathbb{R}^3$, i.e. a plane
    Last edited by romsek; Oct 13th 2018 at 01:25 PM.
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