# A normal matrix with real eigenvalues is Hermitian

$A$ 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?

• I thought that was the definition of normal matrix. ?? – amcalde Aug 29 '14 at 13:58
• @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. – David H Aug 29 '14 at 14:02

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.