Generalizing the Replacement Theorem

**I-Think** · June 5th 2011, 05:17 PM

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

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

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

Is this proof 100% correct?

**Tinyboss** · June 5th 2011, 06:18 PM

Can you clarify the first line after "proof"? I'm not sure if $SS_1$ is meant to be $S\cup S_1$ , 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.

Thread: Generalizing the Replacement Theorem

LinkBack

Thread Tools

Display

Generalizing the Replacement Theorem

Similar Math Help Forum Discussions

Generalizing with Variables

[SOLVED] Generalizing the n-th power of a matrix

Hartog's Theorem and the Axiom of Replacement

generalizing pictorial growth pattern

Matrices and Inverses - Generalizing?

Search Tags