$A$ 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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$\begingroup$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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