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The following theorem is stated in a textbook of mine:

Suppose $A \in Mat_{n,n}(\mathbb C)$ is Hermitian. Then $A$ is Unitary Diagonalizable: $D = U^HAU$, where $U$ is Unitary and $D$ is diagonal with entries in $\mathbb R$.

It has made me think, whether there exist a Unitary matrix $U$ such that $D$ has at least one non-real entry ? Or does every Unitary Diagonalization of a Hermitian matrix imply that $D$ has real entries only ?

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Suppose that there there are two matrices $D$ and $U$ such that $U$ is unitary, and $D$ is diagonal, and $D=U^*AU$. Your question is, Does this imply that $D$ is real?

The answer is yes. Indeed, $$ \overline{D}=D^*=(U^*AU)^*= U^*A^*U=U^*AU=D $$ So $D$ is real.$\qquad\square$

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  • $\begingroup$ How do you show $(AB)^*) = A^*B^*$ ? Here $A^*$ denote the matrix obtained by complex conjugation of every entry in $A$ right ? $\endgroup$ – Shuzheng Apr 29 '14 at 21:17
  • $\begingroup$ No, $A^*$ is the conjugate transpose of $A$, this is a more standard notation than $A^H$ that you use. Generally $(AB)*=^*=B^*A^*$. $\endgroup$ – Omran Kouba Apr 30 '14 at 6:13

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