 Solution

By the spectral theorem, a Hermitian matrix has real eigenvalues and an orthonormal eigenbasis. Choose column vectors $u_j$ with $H{u}_j={\lambda }_j{u}_j$ and $u_i^{†}{u}_j={\delta }_{ij}$, and form $U=(u_1\cdots u_n)$. Then $U^{†}U=I$, so $U$ is unitary, and $HU=U\Lambda$ givesThe entries of $\Lambda$ are the real eigenvalues, repeated according to their multiplicities. Within a degenerate eigenspace, any orthonormal basis may be used. Equivalently $H=U\Lambda U^{†}$.