Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Denote: $[A]$ as the span of $A$.

Theorem: Every linearly independent subset of a vector space is a subset of a basis of a space.

Proof: Let $A$ be a linearly independent subset of a vector space $V$ and let $\,B\,$ be a basis of $\,V$.

Case 1. $B\subseteq [A]\,$. Then $\,[B]\subseteq [A]\,$. But $\,[B]=V\,$ , hence, $\,A=B\,$.

Case 2. $B\not\subseteq [A]$. Then there exists $\alpha \in B$ such that $\alpha \not\in [A]$. Thus $A\cup \{\alpha\}$ is linearly independent. Repeat the argument until an enlarged set is produced that spans $V$.

My question is: In case 2, how is it that $A$ is contained in a basis $B$? I just can't get the idea in the repeating the argument part. Thanks for your help.

share|cite|improve this question
up vote 2 down vote accepted

The argument as shown is a bit suspicious. Firstly, you need the existence of a basis $B$ in the first place. Some authors use the theorem to be shown here, i.e. that every linearly independent family can be extended to a basis, to prove just that (namely by extending the empty family). Secondly, "repeat until" may not work as easily as it sounds if the vector space is infinite-dimensional.

However, what happens here is that a linear independent faimily is extended (possibly repeatedly) to a bigger linear independent family, thus spanning larger and larger portions of the vector space and especiallycontaining more and more members of the basis assumed given. At least with finite dimensional vector spaces this process must end after finitely many steps, aand that means we have found an extended linear independent family that also spans the whole vetor space, i.e. a basis.

share|cite|improve this answer
That explains it very well. Thanks Sir Hagen von Eitzen. – Philip Benj Marcoby Eragon Jan 6 '13 at 11:01

The first case is incorrect, take any two different bases $A,B$.

Since $A$ is a basis $B$ is contained in $[A]$, but $A\neq B$.

In case $B$ you are adding one linearly independent vector to $A$, when $|A|=dim(V)$ you will be done.

share|cite|improve this answer

Your Answer


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.