Let $M_n$ be the $(2n+1) \times (2n+1)$ for which \begin{equation} (M_n)_{ij} = \begin{cases} 0, & i = j\\ 1, & i = j \equiv 1,...,n \mod 2n+1\\ -1, &i - j \equiv n+1,...,2n \mod 2n+1 \end{cases} \end{equation}
Find the rank of $M_n$.
After computing some small cases I'm quite convinced that the answer is $2n$, but I'm not sure how to prove it.
First, note that $M_n$ cannot be full rank because the vector $x = (1,\dots,1)$ satisfies $M_n x = 0$.
On the other hand, we can show that the rank of this matrix is at least $2n$ by considering the leading $(2n)\times(2n)$ submatrix. By this post, this submatrix must be invertible.
An alternative approach: if we use the formula for the eigenvalues of a circulant matrix, we find that they are given by $$ \lambda_j = \sum_{k=1}^n \omega^{jk} - \sum_{k={n+1}}^{2n} \omega^{jk}, \quad j = 0,1,\dots,2n $$ where $\omega = e^{\theta_n i}$, and $\theta_n = \frac{2 \pi i}{n}$. Using the formula for the sum of a geometric sequence, we can rewrite this sum for all $j \neq 0$ as $$ \lambda_j = \frac{\omega^{(n+1)j} - 1}{\omega^j - 1} - \frac{\omega^{j(2n + 1)} - \omega^{(n+1)j}}{\omega^j - 1} \\ = -\frac{\omega^{j(2n+1)} - 2 \omega^{j(n+1)} + 1}{\omega^j - 1} \\ = -\frac{2 - 2 \omega^{j(n+1)}}{\omega^j - 1}, $$ which is non-zero because $\omega^{j(n+1)} \neq 1$. On the other hand, it is easy to see that for $j = 0$, we compute $\lambda_j = 0$.
Thus, the matrix $M_n$ has $0$ as an eigenvalue with algebraic multiplicity $1$. It follows that the dimension of its null space is $1$, which means that its rank is $(2n + 1) - 1 = 2n$.
