Generalizing the Replacement Theorem

• Jun 5th 2011, 05:17 PM
I-Think
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?
• Jun 5th 2011, 06:18 PM
Tinyboss
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.