The vectors $\langle 1+i,2+i,\dots,n-1+i,n+i\rangle$, for $0\leq i \leq n-1$ are not independent How to prove that the vectors ($n>2$):
$$
\langle 1+i,2+i,\dots,n-1+i,n+i\rangle,\qquad 0\leq i \leq n-1
$$
are not independent? If is possible suggest method except determinant of matrix of vectors. 
Thanks for any suggestion  
 A: It is so easy:
\begin{align}
v_0 &= \langle 1,2,\ldots,n \rangle \\
v_1 &= \langle 2,3,\ldots,n+1 \rangle \\
v_2 &= \langle 3,4,\ldots,n+2 \rangle \\
v_3 &= \langle 4,5,\ldots,n+3 \rangle
\end{align}
$$v_3 = v_2 + v_1 -v_0$$
A: All the vectors $v_i$ are in the subspace $V$ spanned by
$$v_0=\langle1,2,3,\ldots,n\rangle$$
and the vector
$$
u=\langle1,1,1\ldots,1\rangle.
$$
This is clear because $v_i=v_0+iu$.
Anyway, as $\dim V=2$, and your list has $n>2$ vectors, they are necessarily linearly dependent.
A: Consider the matrix having the given vectors as rows:
$$
\begin{bmatrix}
1 & 2 & 3 & \dots & n-1 & n \\
2 & 3 & 4 & \dots & n & n+1 \\
3 & 4 & 5 & \dots & n+1 & n+2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n & n+1 & n+2 & \dots & 2n-2 & 2n-1
\end{bmatrix}
$$
Subtract the first row from the second row, the third row and so on:
$$
\begin{bmatrix}
1 & 2 & 3 & \dots & n-1 & n \\
1 & 1 & 1 & \dots & 1 & 1 \\
2 & 2 & 2 & \dots & 2 & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-1 & n-1 & \dots & n-1 & n-1
\end{bmatrix}
$$
What can you say about the rank of this matrix?
