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I tried finding the solution by assuming i and j are both n, but I'm not sure if this is the proper direction to go.

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closed as off-topic by Leucippus, Shailesh, Cesareo, Lee David Chung Lin, Saad Mar 15 at 0:48

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    $\begingroup$ It looks like you may have a fundamental misunderstanding of what the matrix looks like. Try writing out the matrix for, say, a $3 \times 3$ example and see if that helps you understand what's going on. $\endgroup$ – Robert Shore Mar 14 at 23:18
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Hint: the first column is $\mathbf v=(2,3,\dots,n)^T$. Furthermore, the $i$th column is given by $\mathbf v+(i-1)\mathbf 1$ where $\mathbf 1\in\mathbb R^n$ is the vector whose entries are all $1$s. What does this tell you about the number of column vectors of the matrix that are linearly independent?

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  • $\begingroup$ Not that it changes the argument at all, but the entries in the first column are $2,3,....,n,n+1$. $\endgroup$ – Ned Mar 14 at 23:35
  • $\begingroup$ @Ned you're right, I will update the answer. $\endgroup$ – YiFan Mar 15 at 2:00

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