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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.

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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.

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