# Prove that matrix $A = U^{-1}HU$ is Hermitian

This is one of my Mathematica Assignment problems. Assuming that the matrix $H$ is Hermitian and the matrix $U$ is unitary, prove that the matrix $A = U^{-1}HU$ is Hermitian. Can someone help? I'm unable to do it!

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I believe you meant to post this on Mathematics. This site is for the software program Mathematica, not mathematics. –  rcollyer Aug 5 '13 at 11:55
The definitions of Unitary and Hermitian matrices are $U^{-1} = U^*$ and $H=H^*$. You just need to check that $A^* = A$. The identity $(MN)^*=N^*M^*$ will be helpful. –  Anthony Carapetis Aug 5 '13 at 13:24

## migrated from mathematica.stackexchange.comAug 5 '13 at 13:20

This question came from our site for users of Mathematica.

The matrix $H$ satisfies ${}^t \overline{H} = H$ and $U$ satisfies ${}^t \overline{U} U = I_{n}$, which means that $\left( {}^t \overline{U} \right)^{-1} = U$. Consider ${}^t \overline{A}$, you have
$${}^t \overline{A} = {}^t \overline{U} {}^t \overline{H} {}^t \overline{U^{-1}}$$
$${}^t \overline{A} = U^{-1} H U = A$$
So $A$ is hermitian.