 Solution

Since $A$ is Skew-Hermitian, $H=iA$ is a Hermitian matrix. The spectral theorem gives a unitary matrix $Q$ with $Q^{†}HQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ and real eigenvalues ${\lambda }_j$. Hence $Q^{†}AQ=\mathrm{diag}(-i{\lambda }_j)$ andIn particular the determinant is nonzero, so $I+A$ is invertible. Applying the same argument to $-A$ proves that $I-A$ is also invertible. Equality holds precisely when all ${\lambda }_j=0$, equivalently $A=0$.