Any linearly independent set in a vector space is a basis for that space? Any linearly independent set in a vector space is a basis for that space? Is that true or false in general?
I would think it would be true because the fact that is it a linearly independent set would force the set of contain a pivot position in every row in echelon form. Since it has a pivot pos in every row, the set can span the space. 
 A: A basis has to have 2 properties: It has to be linearly independent and spanning. There are linearly independent sets of vectors that are not spanning, for example the set containing only one vector $\{(1,0)\}$ in the vector space $\mathbb{R}^2$ over $\mathbb{R}$. Thus, the statement is false in general.
A: It's hard to be a basis -- it's a delicate balance of power and parsimony. You have to have just the right amount of fatness. If you're too skinny, you won't be able to span the space. If you're too fat, you'll have some redundant extra baggage, and you'll lose linear independence. 
So "linearly independent" just means "free from redundancy". It doesn't say anything about being fat enough to span the space.
All of this is far from rigorous, of course. Don't let your math professor hear you talking this way. But it should help you develop some intuition about these notions.
A: If a set of vectors is linearly independent, any of its subsets is, too. But whenever you remove an element from a linearly independent set, the removed vector is not in the span of the remaining ones.
So, removing an element from a basis gives a linearly independent set which is not a basis.
A: $\langle 1,0,0 \rangle, \langle 0,1,0 \rangle$ is a linearly independent subset of $\mathbb{R}^3$.
A: In a $n$-dimensional vetor space, any independent set containing exactly $n$ vectors is a basis. Otherwise, definitely in a $n$-dimensional vetor space, an independent set containing $k$ vectors, $k<n$, is not a basis.
A: The thing is that if we are considering a Vector space of dimension n(say,V),then any set of n linearly independent vectors will form the basis for V.
It can be easily shown using Replacement Theorem which states that if b belongs to the space V,it can be incorporated in trivial basis set formed by n unit vectors,replacing any one of the n unit vectors.
we can continue doing this n times to get a completely new set of n vectors,which are linearly independent.
