= 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 $\langle\alpha e_j,e_j\rangle=\lambda_j>0$. 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 \b[the positive-definite Hermitian square root is unique], with eigenvalues $\sqrt{\lambda_j}$.
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