Let's say we are given hermitian matrix $H$.
How to prove that the matrix $M$, formed from eigenvectors of $H$ is unitary?
Thanks
2026-04-03 06:03:57.1775196237
How to prove that the corresponding matrix is unitary
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This is false. The identity matrix $I_n$ is Hermitian. Any basis is a basis of eigenvectors for $I_n$. This yields any $M$ invertible. But not every invertible matrix is unitary. Worse, just take one eigenvector, $x$. Then the set of eigenvectors $\{x,\ldots,x\}$ yields a rank $1$, hence non invertible for $n\geq 2$, matrix $M$.
What is true is that if $H$ is Hermitian, there exists an orthonormal basis of eigenvectors. In which case, indeed, the corresponding matrix $M$ is unitary. In other words, there exists $M$ unitary such that $MHM^*$ is (real) diagonal.