# linear independance of a span

I might be going in circles, so would appreciate some clear input....As i understand, a span of a set of vectors can include dependent vectors. A span 'creates' a subspace, therefore a subspace can include depent vectors? So does any span alway include the zero vector and be closed for scalar muliplication & addition?

A set of vectors V is only a basis if they are L. Independent and spans V - isnt the spans part superfluous, every set of vectors has a span?

Ta.

• What's your definition of a span? – user2345215 Jan 13 '15 at 9:42

a span of a set of vectors can include dependent vectors.

Of course. If $v$ is in a span, $\alpha v$ is also there for any $\alpha\in K$.

A span 'creates' a subspace, therefore a subspace can include dependent vectors? So does any span alway include the zero vector ...

Again, yes. Use the same argument for $\alpha = 0$.

... and be closed for scalar multiplication & addition?

The span of a set is the intersection of all the subspaces that contain the set, hence it's a subspace, so it's closed under addition and multiplication.

A set of vectors is only a basis if they are linearly independent and spans $V$

Right.

isn't the span's part superfluous, every set of vectors has a span?

Yes, but it doesn't mean its span is the whole space! For example, $(1,0,0)$ and $(0,1,0)$ in $\mathbb{R}^3$ are linearly independent and they span the xy plane, but they don't form a basis of the space, since they can't generate vectors in the z axis.

Feel free to ask if you have more questions.