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$A$ is a normal matrix (i.e. $AA^*=A^*A$, where * denotes the hermitian conjugate). If all its eigenvalues are real, prove that it is Hermitian (i.e. $A^*=A$).

I have tried many things but could not complete a proof. Could anybody please provide some help?

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  • $\begingroup$ I thought that was the definition of normal matrix. ?? $\endgroup$ – amcalde Aug 29 '14 at 13:58
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    $\begingroup$ @amcalde The definition of a normal matrix is that it commutes with its conjugate transpose: $AA^*=A^*A$. This doesn't necessarily mean that a normal matrix equals its conjugate transpose. $\endgroup$ – David H Aug 29 '14 at 14:02
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A matrix $A$ is normal if and only it is diagonalized by some unitary matrix, i.e., there exists a unitary matrix $U$ ($UU^*=U^*U=I$), such that $$ A=U^*DU, $$ with $D$ diagonal, containing the eigenvalues of $A$ in the diagonal. (See here.)

In our case the eigenvalues of $A$ are real. Then $$ A^*=(U^*DU)^*=U^*D^*U=U^*DU=A, $$ as $D^*=D$, since the eigenvalues are real.

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