Prove that a vector is in the span of a linearly independent set of vectors iff coefficients = 0 Here is the question I have been given.
Let S = {~v1, ~v2, ~v3) be a set of three vectors in R4.  Assume S is
linearly independent. Show that there are numbers a,b,c,d so that

~x=[x1,x2,x3,x4] is in the Span of S 
iff ax1+bx2+cx3+dx4=0

I know that because S is linearly independent, that means that the only solution to
 c1v1+c2v2+c3v3=~0 is if c1=c2=c3=0.

I am just wondering how I should apply this to the problem. It seems to simple to say that because S is linearly independent, and that if ~x spans S, then ~x must be a linear combination of S. I am not sure where to go from here.
The idea that I have is that because S is linearly independent, if ~x does span S, then the equation
c1~v1+c2~v2+c3~v3+c4~x=~0

must be true iff
c1=c2=c3=c4=0

 A: This question has a clean geometric interpretation. Consider the space $A$=span($S$). It is a $3$-dimensional subspace of $\mathbb{R}^4$. So the subspace orthogonal to $A$ is $1$-dimensional. Call it $P$ for perpendicular and suppose that it is spanned by the vector $p$. Now, it is "geometrically clear" that any vector orthogonal to $p$ is in $A$. That's the idea. And then you can take $$p=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$.
Now, how do you find $p$ given $A$? Simple, since $p$ is orthogonal to every vector in $A$, in be in the null space of $A^T$.
Now you have your $a,b,c,d$ and I will leave it up to you to work through the IFF logic!
A: If $(a,b,c,d)$ are such that the conditions are satisfied, then we can note that because each $\tilde v_i$ is in the span of $S$, we should have
$$
a\,\tilde v_i(1) + b\,\tilde v_i(2) + c\, \tilde v_i(3) + d\,\tilde v_i(4) = 0 \quad i = 1,2,3
$$
where $\tilde v_i(j)$ denotes the $j$th entry of the vector $\tilde v_i$.  This gives us a system of equations on $(a,b,c,d)$.  In particular, we have $Vy = 0$ where $y$ is the column vector $y = (a,b,c,d)$ and $V$ is the matrix whose rows are the vectors $\tilde v_i$.
Note that $V$ is a matrix with $3$ linearly independent rows, so there must exist a non-zero element $y$ of its nullspace.
Now, for such non-zero $y = (a,b,c,d)$, we want to show that the set of solutions of
$$
ax_1 + bx_2 + cx_3 + dx_4 = 0
$$
contains the entire $3$-dimensional subspace spanned by $S$, and nothing more.  To do so, it suffices to show that the nullspace of the (row-)matrix $[a\;b\;c\;d]$ (i.e. the solution space to the above equation) is $3$-dimensional; we already know that it contains the vectors $v_i$.
