If $B$ is a maximal linearly independent set in $V$ then $B$ is a basis for $V$ How can you show that if $B$ is a maximal linearly independent set of $V$, then this implies that $B$ is a basis of $V$?
 A: 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 $x = \frac{1}{\alpha_{n+1}} \sum_i^{n} \alpha_i b_i$.
A: 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.
A: 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 $V$.
