Linear Algebra - Linear transformation question Let $
b \in \mathbb{R}^4, \space A\in M_{4\times4} (\mathbb{R}). 
$ Suppose
$$
\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix}$$
are all solutions of the equation $Ax=b$. Prove that $$b = \begin{bmatrix}  0  \\ 0 \\ 0 \\ 0\end{bmatrix}.$$
This question appeared at my recent algebra exam. I had an intuiton that the given vectors are linear-dependent since there are five of them and the dim of the vector space is four. The professor solved that question after the exam using linear transformation which seemed much easier and clean solution however he didn't explain the intuition behind. I  would like to get an explanation regarding that. Thanks
 A: One thing that stands out is that if you sum the first four vectors together you get
$\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$
If we write those vectors as $x_1, x_2, \ldots, x_5$ in the order they're given, we can just write $x_1 + x_2 + x_3 + x_4 = 3 x_5$.
So it's very tempting to see what happens when you multiply that equality by $A$:
$\begin{eqnarray} x_1 + x_2 + x_3 + x_4 & = & 3 x_5 \\
A(x_1 + x_2 + x_3 + x_4) & = & A(3 x_5) \\
A x_1 + A x_2 + A x_3 + A x_4 & = & 3 A x_5 \\
b + b + b + b & = & 3 b \\
4 b & = & 3 b \\
\implies b & = & \mathbf{0}
\end{eqnarray}$
You can do this more generally by taking your observation that there are 5 vectors in a 4-dimensional vector space, and hence there must be a linear dependency between them, except that you need to be able to span $\mathbb{R}^4$ with subsets of the vectors since you'll need to be in a situation where you have something like $(\lambda_1 + \lambda_2 + \ldots + \lambda_4 - 1) b = \mathbf{0}$ and you need to be able to escape the possibility that the scalar part is equal to zero.
A: Call $e_i$ your vectors in the order you listed them, and note that $e_1+e_2+e_3+e_4 = 3e_5$. Then $A(e_1+e_2+e_3+e_4) = 4b = 3Ae_5 = 3b$. Hence $b=0$.
A: $$\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix} - 3\begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}  0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$A\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix} - 3A\begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix} = A\begin{bmatrix}  0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$b + b + b + b -3b = 0$$
$$b=0$$
