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Need the vectors in a basis be elements of the vector space? watch

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    Do the things in a basis need to be in the vector space?

    Find a basis of the following vector spaces V over the given field K:

    V is the set of all polynomials over \mathbb{R} of degree at most n, in which the sum of the coefficients is 0; K=\mathbb{R}
    If we take the set \begin{Bmatrix} x^n, x^{n-1},..., x, 1 \end{Bmatrix} then this spans V and they're all linearly independent, so I think they should work, but I've not yet encountered a basis that ain't in the vector space, so I'm just checking it's kosher. Thanks.
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    'A basis B of a vector space V over a field F is a linearly independent subset of V that spans (or generates) V.' says wikipedia, which agrees with my thinking that the basis elements do need to be in the space.
    I'd probably wait for someone else's opinion though, I'm not certain.
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    Yes.

    I've forgotten how to write Latex in this forum but...
    Suppose B = {e_i} for 0<i<=n
    is a basis for V (finite dim'd VS).
    Then
     v \in V if and only if  \exists scalars a_i st v can be written v=\sum _1 ^n a_i e_i.
    Choose a_i =1 and run through all i to get each basis vector a member of V.

    This is basically how you prove anything that somehow seems obvious about vector spaces.
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    They have to be part of the vector space.
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    Wiki Support Team
    The term "linearly independent" is shorthand for "linearly independent in V", which doesn't make any sense if they're not in V.

    (Original post by Hathlan)
    this spans V
    Yes, in the sense that {(1,0,0), (0,1,0), (0,0,1)} spans any embedding of R into R^3, but its span is actually something much bigger. Think about this: you've given a basis for W = {K-polynomials with degree at most n}, which seems to suggest that dim W = n+1, and V is a subspace of W (which isn't equal to W itself), so we should have dim V at most equal to n. So your proposed basis is too big. In fact, since you've taken W and added one extra restriction, I personally would automatically guess that dim V = n.

    (Another way of thinking about this. Take a polynomial a_nx^n + \dots + a_1x + a_0 in V. Then obviously a_0 is completely determined by a_1, \dots, a_n, so we only really have n coefficients to play with...)

    Can you think of a basis now?
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    (Original post by generalebriety)
    The term "linearly independent" is shorthand for "linearly independent in V", which doesn't make any sense if they're not in V.


    Yes, in the sense that {(1,0,0), (0,1,0), (0,0,1)} spans any embedding of R into R^3, but its span is actually something much bigger. Think about this: you've given a basis for W = {K-polynomials with degree at most n}, which seems to suggest that dim W = n+1, and V is a subspace of W (which isn't equal to W itself), so we should have dim V at most equal to n. So your proposed basis is too big. In fact, since you've taken W and added one extra restriction, I personally would automatically guess that dim V = n.

    (Another way of thinking about this. Take a polynomial a_nx^n + \dots + a_1x + a_0 in V. Then obviously a_0 is completely determined by a_1, \dots, a_n, so we only really have n coefficients to play with...)

    Can you think of a basis now?
    x^n-1, x^{n-1}-1,...x-1?
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    Aye.
 
 
 
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