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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

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  • $\begingroup$ What is the definition of "independent set"? $\endgroup$ – Bernard Massé Jun 10 '17 at 20:25
  • $\begingroup$ @BernardMassé I mean,for every $n>2$ there are non-zero linear combination of sets that is zero. It was better to wrote the vectors instead of the sets. Thanks for comment. $\endgroup$ – Amin235 Jun 10 '17 at 20:30
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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}

$$v_3 = v_2 + v_1 -v_0$$

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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.

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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?

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  • $\begingroup$ The rank is $2$, am I right? $\endgroup$ – Amin235 Jun 10 '17 at 21:00
  • $\begingroup$ @Amin235 Yes, you are. $\endgroup$ – egreg Jun 10 '17 at 21:01
  • $\begingroup$ I got 10 minutes left before close down $\endgroup$ – user441532 Jun 14 '17 at 20:50

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