Suppose that $F$ is a finite collection of vectors from a vector space. Prove that every maximal subset of $F$ that is linearly independent is a basis of $\langle F\rangle$, which designates the span of $F$.
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Use https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference in future posts. – Arbuja Apr 27 '17 at 13:03
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Could you clarify what $\langle F \rangle$ is? Is it the span of the vectors in $F$ or something else? – michaelhowes Apr 27 '17 at 13:19
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1⟨F⟩ is the span of F – Charles Apr 27 '17 at 13:44
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Hint: Suppose a maximal linearly independent subset didn't span $\langle F \rangle$. Then find a contradiction. – Ken Duna Apr 27 '17 at 14:55
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We have to show that any maximal linearly independent subset $A\subset F$ generates $\langle F\rangle$. We begin by showing that each $f\in F$ is in $\langle A\rangle$.
Proof. Let $A=\{a_1,a_2,\ldots, a_r\}$, and consider an $f\in F$. The maximality of $A$ implies that the set $A\cup\{f\}$ is linearly dependent; hence there are scalars $\lambda_i$, not all $=0$, such that $$\lambda_0 f+\sum_{i=1}^r\lambda_i\,a_i=0\ .$$ As the $a_i$ are linearly independent we cannot have $\lambda_0=0$ here. This allows to conclude that $f=\sum_{i=1}^r{-\lambda_i\over\lambda_0}\,a_i\in \langle A\rangle$.$\qquad\square$
Now $F\subset\langle A\rangle$ immediately implies $\langle F\rangle\subset\langle A\rangle$.
Christian Blatter
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