Do n + 1 vectors, assuming they are not collinear, always span n? I'm trying to wrap my head around linear independence/dependence and I'm hoping that I'm understanding this correctly, but I'm not certain.
So to my understanding, by what my linear algebra prof calls the "too many vectors" theorem, $n + 1$ vectors in $R_n$ , assuming they are not collinear, will always be linearly dependent as one vector will be a linear combination of the others.
So does this mean that as long as this condition is fulfilled, the vectors will always span $R_n$? 
 A: Here is an example in $\mathbb{R}^3$: 
$$\{(1,1,0),(1,2,0),(1,3,0),(1,4,0)\}$$
These clearly do not span $\mathbb{R}^3$, as they all have a zero in the third coordinate.  They are also no two collinear, as they all have a one in the first coordinate.
As pointed out in the comments, "not collinear" only guarantees that the dimension of what they do span is at least 2, not necessarily the dimension of the entire space.
A: Generally, the following are equivalent for a set $S$ of vectors in a vector space:


*

*$S$ is linearly independent. (Namely, every non-trivial linear combination from $S$ is non-zero. Contrapositively, if the zero vector is expressed as a linear combination $\mathbf{0} = \sum_{k} x_{k} \mathbf{v}_{k}$ from $S$, then $x_{k} = 0$ for all $k$.)

*Every vector in the span of $S$ can be expressed uniquely as a linear combination from $S$.

*Every subset of $S$ is linearly independent. (Consequently, if $S$ is linearly independent then the zero vector is not in $S$; no two vectors in $S$ are collinear; no three vectors are coplanar; etc.)

*No proper subset of $S$ spans the same subspace as $S$.
A: No. They span at least a two-dimensional subspace of $\mathbb{R}_n$. Consider $(1,0,0)$, $(2,0,0)$, $(3,0,1)$ and $(4,0,0)$. These are 4 vectors in $\mathbb{R}_3$ so at least one of them is a linear combination of the others, but they do not span $\mathbb{R}_3$. They span a 2D plane instead.
