Are these square matrices always diagonalisable?

Solution 1:

All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $\mathbb C$, by the spectral theorem.

Solution 2:

The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $\delta = 0$ and off-diagonal entries $\tau = 1$ and $\sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$\lambda_k = 2i\cos\left(\dfrac{k\pi}{n+1}\right),$$ for $k = 1,\ldots,n$, and the corresponding eigenvectors $v_1,\ldots,v_n$ have entries $$v_k[m] = i^m\sin\left(\dfrac{mk\pi}{n+1}\right).$$

Solution 3:

Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.