Let $A = (a_{i j})\in M_n(\mathbb F),n\geq 3$,s. t. $a_{ij} = i − j,\;\forall i,j\in\{1,\ldots,n\}$. Prove that $\det(A) = 0$.
I was thinking that I can prove that $\det(A)$ is $0$, by showing that the last row is $0$. But not really sure how to go about this.
Let's denote $C_i$ the $i$th column of $A$.
We can make $C_1\leftarrow C_1-C_2$ and $C_2\leftarrow C_2-C_3$
The two first columns are equal so the determinant is 0.
Notice that given matrix is skew symetric, so $$A=-A^{\top}$$
so if $n$ is odd $$\det A = det(-A^{\top}) = (-1)^n\det(A^{\top}) = -\det A$$ and thus a conclusion.
Let $x = (1,\dots,1)^T$ and $y = (1,2,\dots,n)^T$. It suffices to note that $$ A = yx^T - x^Ty = \pmatrix{x & y} \pmatrix{0&-1\\1&0} \pmatrix{x & y}^T. $$ It follows that $A$ has rank at most $2$.
