Let $n \geqslant 2$ be a positive integer. Write $v_i$ for the vector in $\mathbb{R}^n$ with entry $0$ at the $i$th coordinate and entry $1$ at every other coordinate. Is $\{v_i\}_{i=1}^n$ linearly independent?
It looks like, according to the matrix calculator, that:
\begin{equation*} \det [v_1^T, \cdots, v_n^T] = (-1)^{n-1}(n-1) \end{equation*}
But How do I prove it?
Let $e \in \mathbb{R}^n$ be a column vector of ones, and let $I$ be the $n\times n$ identity matrix. Then $$ \det(ee^T - I) = (-1)^n \det(I - ee^T) = (-1)^n(1+(-e)^TI^{-1}e)\det(I) = (-1)^n(1-n), $$
where the second equality uses the Matrix Determinant Lemma I've found in this answer.
Well, this is exactly what Theophile's comment said...
We can prove it by showing that these vectors span the whole space $\mathbb{R}^n$, because there is exactly $n$ of them so they would have to be the base, which means that they are also independent. Firstly, sum up all the vectors $v_i$ to get $(n-1)\cdot u$, where the vector $u$ has all coordinates equal to $1$. This shows that $u$ is spanned by your vectors. Now substract each of your vectors from $u$: $u-v_i$ to get a vector which has all coordinates equal to $0$ ecxept the $i$-th coordinate, which is $1$. This shows that $u-v_i=e_i$, and vectors $e_i$ (the base vectors) span the whole space.
The matrix $J_n$ with all entries equal to $1$ has one eigenvalue equal to $n$ and the other $n-1$ eigenvalues are zero. See this link, for example.
The matrix in question is $J_n-I_n$, where $I_n$ is the $n \times n$ identity matrix. The eigenvalues of $J_n$ are shifted by $1$, which leads to the determinant in question.
