Linear dependence and determinant of submatrices 
We have two column vectors $x,y \in \mathbb{R^n}$. Prove that these are linearly dependent iff every $2 \times 2$ submatrix $M$ of $[x,y]$ has a determinent which equals $0$. Here $M$ is defined as follows (for an $n \times m$ matrix):
$\begin{bmatrix}
    a_{ij}       & a_{ik} \\
    a_{lj}       & a_{lk} \\
\end{bmatrix}$
with $1 \leq i, l \leq n, 1 \leq j, k \leq m, i \neq l, j \neq k$.

So we have $x =\begin{bmatrix}
    x_{1}\\
    x_{2} \\
 \dots \\
 x_{n}
\end{bmatrix} $ and $y = \begin {bmatrix}
    y_{1}\\
    y_{2} \\
 \dots \\
 y_{n}
\end{bmatrix} $
They are linearly dependent, so there exists $c_1, c_2 \neq 0$ such that $c_1 \begin{bmatrix}
    x_{1}\\
    x_{2} \\
 \dots \\
 x_{n}
\end{bmatrix} + c_2 \begin {bmatrix}
    y_{1}\\
    y_{2} \\
 \dots \\
 y_{n}
\end{bmatrix} = 0 $
so we have $$ \begin {bmatrix}
    x_{1} & y_{1}\\
    x_{2} & y_{2} \\
 \dots & \dots \\
 x_{n} & y_{n}
\end{bmatrix}  \begin {bmatrix}
    c_1 \\
    c_2 \\
\end{bmatrix} = 0$$
We can turn this into an augmented matrix and if the rank is less than $n$ the vectors are linearly dependent. But I don't know how to prove what's asked. Can anybody give a hint? 
 A: HINT: You have already shown that
$$
\begin {bmatrix}
    x_{1} & y_{1}\\
    x_{2} & y_{2} \\
 \dots & \dots \\
 x_{n} & y_{n}
\end{bmatrix}  \begin {bmatrix}
    c_1 \\
    c_2 \\
\end{bmatrix} = 0,
$$
must hold for some nonzero constants $c_1$ and $c_2$. So in particular
$$
\begin {bmatrix}
    x_{i} & y_{i}\\
    x_{j} & y_{j} \\
\end{bmatrix}  \begin {bmatrix}
    c_1 \\
    c_2 \\
\end{bmatrix} = 0
$$
for all $i$ and $j$.
A: This follows at once from the fact that $\;x,\,y\;$ are linearly dependent iff one of them is a multiple scalar of the other one, i.e. iff $\;y=kx\;,\;\;k\in\Bbb R\;$ , and thus
$$x=(x_1,x_2,...,x_n)\implies y=(kx_1,kx_2,...,kx_n)$$
so that each $\;2\times2\;$ submatrix of $\;M\;$ is of the form
$$M_{im}:=\begin{pmatrix}x_i&kx_i\\x_m&kx_m\end{pmatrix}\implies \det M=0$$
A: Consider the linear map $f_{ij}\colon \mathbb{R}^n\to\mathbb{R}^2$ defined by
$$
\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}
\mapsto
\begin{bmatrix}x_i\\x_j\end{bmatrix}
$$
where $1\le i<j\le n$. This map is clearly linear. Thus, if $x$ and $y$ are linearly dependent, then so are $f(x)$ and $f(y)$.
