$S$ is a set of $m$ vectors in a vector space $V$ of dimension $n$. If $m>n$, then is $V = \operatorname{Span}(S)$? Let $S=\{v_1, v_2, \ldots,v_m\}\subset V$, where $V$ is a vector space of dimension $n$.
If $m>n$, is $V = \operatorname{Span}(S)$?
 A: This is not true in general; for example, if $V=\Bbb{R}^2$ and $v_i=(i,0)$. Then $\dim\operatorname{span}(\{v_1,\ldots,v_m\})=1$ no matter what $m$ is.
A: As others have said, this is not true. Consider two dimensional space $\mathbb{R^{2}}$ and consider three vectors that are multiples of $e_1$ say $ \{ e_1, 2e_1 3e_1 \} $. Clearly adding these just would keep you on a one dimensional axis. Hence giving a counter example
However, it is important to know that if you have a set with $m>n$ vectors, where $n$ is the dim. of your vector space. Then you must have vectors in the set which are NOT linearly independent. Based on the fact vectors are closed under addition and scalar multiplication, hence their span must be fully in contained in $V$.
A: Already many answers have been given. For simplicity I assume the vector space is either over the real numbers or complex numbers.
In this case any vector subspace of dimension at least 1 will have infinitely many elements.
Now take  your vector space $V$ to be of dimension $n\geq 2$ 
and take a subspace $W$ of dimension at least 1.  Then one can find infinite number of vectors inside $W$. SO the span of any finite number of vectors from $W$ will be contained in the proper subspace $E$ and can not be whole of $V$.
(of course this argument can also be applied to vector spaces over finite fields provided the number of elements of that field is more than the dimension of the subspace $W$. )
