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From Wikipedia:

"A basis $B$ of a vector space $V$ over a field $K$ is a linearly independent subset of $V$ that spans (or generates) $V$.(1)

$B$ is a minimal generating set of $V$, i.e., it is a generating set and no proper subset of B is also a generating set.(2)

$B$ is a maximal set of linearly independent vectors, i.e., it is a linearly independent set but no other linearly independent set contains it as a proper subset."(3)

I tried to prove (1) => (3) => (2), to see that these are equivalent definitions, can you tell me if my proof is correct:

(1) => (3): Let $B$ be linearly independent and spanning. Then $B$ is maximal: Let $v$ be any vector in $V$. Then since $B$ is spanning, $\exists b_i \in B, k_i \in K: \sum_{i=1}^n b_i k_i = v$. Hence $v - \sum_{i=1}^n b_i k_i = 0$ and hence $B \cup \{v\}$ is linearly dependent. So $B$ is maximal since $v$ was arbitrary.

(3) => (2): Let $B$ be maximal and linearly independent. Then $B$ is minimal and spanning:

spanning: Let $v \in V$ be arbitrary. $B$ is maximal hence $B \cup \{v\}$ is linearly dependent. i.e. $\exists b_i \in B , k_i \in K : \sum_i b_i k_i = v$, i.e. $B$ is spanning.

minimal: $B$ is linearly independent. Let $b \in B$. Then $b \notin span( B \setminus \{b\})$ hence $B$ is minimal.

(2) => (1): Let $B$ be minimal and spanning. Then $B$ is linearly independent: Assume $B$ not linearly independent then $\exists b_i \in B, k_i \in K: b = \sum_i b_i k_i$. Then $B \setminus \{b\}$ is spanning which contradicts that $B$ is minimal.

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There is a mix up in your numerotation. For instance your proof of $1$ implies $2$ is actually a proof of $1$ implies $3$. – Olivier Bégassat May 3 '12 at 12:08
Another way to define (3) is to say that B spans $V$. or, for all vectors $v\in V$, $v$ is also in $Span{B}$ – Joel Cornett May 3 '12 at 12:30
up vote 1 down vote accepted

The proof looks good (appart form the obvious mix up in the numbering). One thing which is not totally precise: In your second proof you write Let $v\in V$ be arbitrary. $B$ is maximal hence $B\cup\{v\}$ is linearly dependent. i.e. $\exists b_i\in B,k_i\in K: \sum_i b_ik_i=v$, i.e. $B$ is spanning. To be precise you have $k_i\in K$ not all vanishing such that $k_0v+\sum_i k_ib_i=0$. Since $B$ is linearly independent $k_0=0$ implies $k_i=0$ for all $i$, therefore $k_0\neq 0$ and $v$ is in the span of $B$.

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