Vector Subspaces Proof I'm having a hard time trying to prove the following question:

I tried to use de definition of Span, but it just got redundant and got me running in circles.. 
Here`s what I've got so far...
Thanks in advance

 A: For (a) consider the vectors $v_1,v_2,\cdots,v_k$ as the rows of a $k \times n$ matrix $M$. Then consider the homogeneous system of linear equations associated to $M$ i.e. each row of $M$ contains the coefficients of a equation. Namely, in matrix notation the homogeneous system is $M x = 0$ where $x = (x_1 , \cdots, x_n)^{T}$. Since $k < n$ there are more unknows than equations. So the system $M x = 0$ has a non trivial solution $x \neq 0$.
Observe that $x \notin B$ otherwise $x = 0$. Indeed, the vector $x$ is perpendicular to all vectors in $B$ since $x$ is perpendicular to all generators $v_j$'s because the system $M x = 0$ is the same to say so. So if $x \in B$ then $x$ is going to be perpendicular to itself i.e. $x=0$. 
For $(b)$ consider now the matrix $M$ whose columns are the vectors $v_1,v_2,\cdots,v_k$ and consider the homogeneous system $M x=0$. Then again since $k > n$ there are more unknowns than equations hence there is a non trivial solution $x = (x_1, \cdots, x_n)^{T}$. Namely, there is $x \neq 0$ such that $M x = 0$. The coefficients of $x$ give the linear relation between the vectors since this is exactly the meaning of $Mx=0$.
After this I think it is better you try by yourself to solve $(c)$.   
