How can you show that if $B$ is a maximal linearly independent set of $V$, then this implies that $B$ is a basis of $V$?
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You want to prove that an element $x \in V$ can be written as a linear combination of elements of $B$.
Take such an $x$, then $B \cup \{x\}$ is not linearly independant by maximality of $B$. So there exist $\alpha_1,...,\alpha_{n+1}$ (if $B = \{b_1,\ldots,b_n\}$ ) such that $\alpha_{n+1} x + \sum_i^{n} \alpha_i b_i = 0$. $\alpha_{n+1}$ can't be null, else it would implies that $B$ is not linearly independant. So you can write $x = \frac{1}{\alpha_{n+1}} \sum_i^{n} \alpha_i b_i$.
Hagen von Eitzen
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yago
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8Remark: In the last step we divide by $\alpha_{n+1}$ of whic we only know that it is nonzero. This is where the difference between vector spaces over fields and modules over rings arises (with respect to existence of bases). – Hagen von Eitzen Apr 07 '14 at 17:54
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@HagenvonEitzen how do we know $\alpha_{n+1}$ is non-zero. We only know that since vectors {$x,b_1,b_2...$} are linearly independent, atleast one of $\alpha_i$ with $1\leq i\leq n+1$ is non-zero. But how is $\alpha_{n+1}$ non-zero? – zed111 Apr 22 '15 at 19:03
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2@zed111 if $\alpha_{n+1}=0$ then $\alpha_{n+1}x$ can be ignored and the dependence relation is really among the $b_i$ alone, but $B$ was assumed to be linearly independent. – Hagen von Eitzen Apr 22 '15 at 22:04
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Suppose not. Then there is some vector $w$ which is not in the span of the vectors in $B$. But then the set $B \cup \{w\}$ is a linearly independent set in $V$ which contains $B$, which is a contradiction.
Jeff
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Since B is a maximally linearly independent set, would B union {w} still be linearly independent? – Mike Richards Apr 07 '14 at 19:44
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That is the contradiction. ($B \cup {w}$ is linearly independent by assumption.) – Jeff Apr 07 '14 at 20:38
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Yes. I don't understand your line of questioning. Are you unfamiliar with proofs by contradiction? – Jeff Apr 08 '14 at 02:49
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I understand it, I was just dotting the i's and umlauts when I asked about the vector w. – Mike Richards Apr 08 '14 at 03:27
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Let $v \in V$ be any vector. Form the linear combination $c v + c_1 b_1 + \dots = 0$ where $b_i \in B$. If there is no solution $C = (c, c_1, c_2, \dots)$ other than $C = 0$, then we've shown that $B$ can be expanded with $v$ and thus is not maxmimally linearly indep. So there exists a solution $C$ such that not all of the $c$'s are zero. Within any such solution we must have $c \neq 0$ (the coeff of $v$) or else we'd be left with a linear combin. strictly in $B$ and so all $c_i = 0$. Thus there exists a solution $C$ with $c \neq 0$. From here prove that $B$ spans $V$.
Daniel Donnelly
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