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Thread: Generalizing the Replacement Theorem

  1. #1
    Senior Member I-Think's Avatar
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    Generalizing the Replacement Theorem

    Let $\displaystyle \beta$ be a basis for a vector space $\displaystyle V$ and let $\displaystyle S$ be a linear independent subset of $\displaystyle V$. Prove there exists a subset of $\displaystyle \beta$,$\displaystyle S_1$, such that $\displaystyle S\cup{S_1}$ is a basis for $\displaystyle V$

    Proof
    Let $\displaystyle S_1$ be the set of all elements in $\displaystyle \beta$ such that $\displaystyle S\union{S_1}$ is linearly independent. Now consider $\displaystyle S\cup{S_1}$

    Note that every vector in $\displaystyle S$ can be expressed as a linear combination of vectors in $\displaystyle \beta$. So it follows that every vector in $\displaystyle V$ can be expressed as a linear combination of vectors in $\displaystyle S\cup{S_1}$
    Thus $\displaystyle S\cup{S_1}$ is linearly independent and generates $\displaystyle V$, so it is a basis.

    Is this proof 100% correct?
    Last edited by I-Think; Jun 5th 2011 at 05:41 PM.
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  2. #2
    Senior Member Tinyboss's Avatar
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    Can you clarify the first line after "proof"? I'm not sure if $\displaystyle SS_1$ is meant to be $\displaystyle S\cup S_1$, or what. Also, (and I'm going with a different interpretation here) it's entirely possible (in fact overwhelmingly likely) that every element of B is linearly independent (I would say "not in the span of") S. In that case, you will get a set that's too big to be a basis, at least in finite dimension. Even in infinite dimension, it won't be linearly independent.
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