Feedback on answer regarding Linear Algebra I am new to Linear Algebra, and have no teacher at present. I am just looking for some feedback re my answer to the following question:
Let $v_1,..,v_k,u,w$ be vectors in linear space V. It is given that the equation $x_1v_1+…+x_kv_k=u$ has a single solution and that the equation $x_1v_1+..+x_kv_k=w$ has no solution. Where $x_1,\cdots,x_k$ are scalars in the underlying field.


*

*Prove that the set ${v_1,…,v_k}$ is linearly independent

*Find the dimension of Sp ${v_1,…,v_k,w}$
I reasoned the following:
1.
It is given that $x_1v_1+…+x_kv_k=u$ has a single solution, so each variable must have a fixed value for that single solution to hold. If u equals the zero vector, that single solution must be the trivial solution, as the trivial solution always exists with homogenous systems. Therefore, ${v_1,…,v_k}$ is linearly independent.
2.
As ${v_1,…,v_k}$ was shown to be linearly independent, its dimension must be $k$. It was given that $v_1,..,v_k=w$ has no solution, therefore $w$ is not a linear combination of $v_1,..,v_k$. As such, ${v_1,…,v_k,w}$ is a basis of Sp ${v_1,…,v_k,w}$, and its dimension is $k+1$.
 A: Your idea is correct, but you need to modify the statement of the problem.
In a vector space $V$ vectors $\{v_1,v_2,....,v_k\}$ are such that for every $u$ in $V$, the equation $$c_1 v_1 + c_2 v_2 +....+c_k v_k = u $$ has a  unique solution, and for every $w$ in $V$, the equation  $$c_1 v_1 + c_2 v_2 +....+c_k v_k = w $$ has no solution, then  $\{v_1,v_2,....,v_k\}$ is linearly independent and the  dimension of Span $\{v_1,v_2,....,v_k,w\}$ is $k+1$.
A: I'm going to try to restate the problem (a la Mohammad Riazi-Kermani)
Suppose that in a vector space $V$ we have vectors $\{v_1,v_2,....,v_k\}$, $u$, and $w$ such that the equation $$c_1 v_1 + c_2 v_2 +....+c_k v_k = u \tag{1}$$ has a  unique solution, and yet the equation  $$d_1 v_1 + d_2 v_2 +....+d_k v_k = w \tag{2}$$ has no solution. 


*

*Prove that the set ${v_1,\ldots,v_k}$ is linearly independent.

*Find the dimension of Sp ${v_1,\ldots,v_k,w}$.
The argument for the first part is this:  Let $c_1, \ldots, c_k$ be the scalars solving equation 1. 
Now suppose the set ${v_1,\ldots,v_k}$ were dependent.
Then there are not-all-zero constants $a_1, \ldots, a_k$ such that 
$$
a_1 v_1 + a_2 v_2 +....+a_k v_k = 0.
$$
But now consider
\begin{align}
(a_1+c_1) v_1 + (a_2+c_2) v_2 +....+(a_k+c_k) v_k 
&= (a_1 v_1 + a_2 v_2 +....+a_k v_k) + (c_1 v_1 + c_2 v_2 +....+c_k v_k) \\
&= 0 + u\\
&= u.
\end{align}
Thus this new set of coefficients is a second solution to equation $1$ (because not all the $a_i$ are zero), which contradicts the hypothesis that equation 1 has a single solution. Hence the assumption of dependence must be false, so the set is independent. 
For part 2, because we know that $v_1, \ldots, v_k$ are independent, they span a subspace of dimension $k$; the vector $w$ is not in this subspace (because Equation $2$ has no solutions). I claim that $\{v_1, \ldots, v_k, w\}$ is independent. Suppose not. Then suppose that 
$$
b_1 v_1 + \ldots + b_k v_k + b_{k+1}w = 0\tag{3}
$$
is a nontrivial linear combination of the elements. 
Case 1: $b_{k+1} \ne 0$. Then we can move $w$ to the other side and divide through by $b_{k+1}$ to get
$$
-\frac{b_1}{b_{k+1}}v_1 - \ldots -  -\frac{b_k}{b_{k+1}}v_k = w,
$$
which would be a solution to equation 2, which has no solutions, so this is impossible. This leaves us in case 2, where $b_{k+1} = 0$. 
Case 2: $b_{k+1} = 0$. In this case, equation 3 becomes
$$
b_1 v_1 + \ldots + b_k v_k = 0
$$
which implies that the vectors $v_1, \ldots, v_k$ are linearly dependent, which is a contradiction. 
Hence our original assumption, that the set $v_1, \ldots, v_k, w$ is dependent, is false, so the set is independent. The dimension of the span is therefore the number of elements in the set, i.e., $k+1$. 
