 Solution

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.