# Reason: non-integer powers of Hermitian matrix not Hermitian?

Let $$\alpha = 2.1$$. If $$A$$ is symmetric real, $$A^\alpha$$ remains symmetric (although it could be complex-valued). If $$B$$ is Hermitian, $$B^\alpha$$ isn't Hermitian (losses the symmetry).

However, for integer powers e.g. when $$\alpha=3$$, $$B^\alpha$$ remains Hermitian.

May I know why the non-integer powers of Hermitian matrices behave this way?

PS: I computed the powers with the eigen decomposition approach.

A Hermitian matrix $$B$$ has eigendecomposition $$B=U\Lambda U^{*}$$ where $$U$$ is unitary and $$\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$$ has only real-valued entries. By eigendecomposition approach, I assume you mean that you are defining $$B^{\alpha}\equiv U\Lambda^{\alpha}U^{*}.$$
Lemma. Let $$\alpha$$ be a real number and $$B$$ be a Hermitian matrix. Then, $$B^{\alpha}$$ is not Hermitian if and only if one of the eigenvalues of $$B$$ is negative and $$\alpha$$ is not an integer.
Proof. $$B^{\alpha}$$ being Hermitian is equivalent to $$\Lambda^{\alpha}$$ being real since $$(B^{\alpha})^{*} =U\Lambda^{\alpha}U^{*} =U(\Lambda^{\alpha})^{*}U^{*}.$$ The result follows from noting that for a real number $$\lambda$$, $$\lambda^{\alpha}$$ is not real if and only if $$\lambda$$ is negative and $$\alpha$$ is not an integer.