Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

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 $B$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.