How do you show that eigenvectors of a Hermitian matrix form a complete set of basis?
2026-03-27 21:35:28.1774647328
Completeness of eigenvectors of Hermitian Matrix.
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In the same way how you do that for normal matrices.
Use or prove the Schur decomposition: for any square matrix $A\in\mathbb{C}^{n\times n}$, there is a unitary $Q$ and triangular $T$ such that $A=QTQ^*$. A simple but very strong result which can be shown quite simply by induction.
Show that $A$ is Hermitian iff $T$ is Hermitian. A Hermitian triangular matrix is necessarily diagonal.
The eigenvectors can be then found stuffed in the columns of $Q$.