IMC 2012 · Day 1 · P2

For $n = 1$ the only matrix is $(0)$ with rank 0. For $n = 2$ the determinant of such a matrix is negative, so the rank is 2. We show that for all $n \ge 3$ the minimal rank is 3.

Notice that the first three rows are linearly independent. Suppose that some linear combination of them, with coefficients $c_1, c_2, c_3$, vanishes. Observe that from the first column one deduces that $c_2$ and $c_3$ either have opposite signs or both zero. The same applies to the pairs $(c_1, c_2)$ and $(c_1, c_3)$. Hence they all must be zero.

It remains to give an example of a matrix of rank (at most) 3. For example, the matrix $$\begin{pmatrix} 0^2 & 1^2 & 2^2 & \dots & (n-1)^2 \\ (-1)^2 & 0^2 & 1^2 & \dots & (n-2)^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ (-n+1)^2 & (-n+2)^2 & (-n+3)^2 & \dots & 0^2 \end{pmatrix} = \Bigl( (i-j)^2 \Bigr)_{i,j=1}^{n} =$$ $$= \begin{pmatrix} 1^2 \\ 2^2 \\ \vdots \\ n^2 \end{pmatrix} (1, 1, \dots, 1) - 2 \begin{pmatrix} 1 \\ 2 \\ \vdots \\ n \end{pmatrix} (1, 2, \dots, n) + \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} (1^2, 2^2, \dots, n^2)$$ is the sum of three matrices of rank 1, so its rank cannot exceed 3.