Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 11E (Linear Algebra)

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

The adjoint reverses products, so for the two Hermitian operatorsThus $\alpha \beta$ is Hermitian if and only if $\alpha \beta =\beta \alpha$. No positivity hypothesis is needed for this equivalence.

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 $\sqrt{{\lambda }_j}$.