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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?

Rodrigo de Azevedo
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Myshkin
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    @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

1 Answers1

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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 $$ A=U^*DU, $$ where $D$ is a diagonal matrix, containing the eigenvalues of $A$ in the diagonal. (See here.)

In particular, since the eigenvalues of $A$ are real, then $D^*=D$, and hence $$ A^*=(U^*DU)^*=U^*D^*U=U^*DU=A. $$

Yiorgos S. Smyrlis
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