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Mathematics-Online course: Linear Algebra - Normal Forms - Diagonalisation

Diagonal From of Hermitian Matrices

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The eigenvalues $ \lambda_i$ of a hermitian matrix $ A \; (A=A^*)$ are real, and there exists an ONB consisting of eigenvectors $ v_i$. Therefore

$\displaystyle U^*AU=$diag$\displaystyle (\lambda_1,\ldots,\lambda_n)\,, $

where $ U=(v_1,\ldots,v_n)$ is an unitary matrix.

The special case of real symmetric matrices $ (A=A^{\operatorname t})$ provides real eigenvalues.

(Authors: Höllig/Reble/Höfert)
The unitary diagonalisation is a conclusion of the common result about normal matrices, which includes the hermitian case. It remains to show, that an eigenvalue $ \lambda$ is real. Is $ v$ the corresponding eigenvector, the proposition follows from
$\displaystyle \lambda v^* v$ $\displaystyle =$ $\displaystyle v^*(\lambda v) = v^*(Av) = (A^*v)^*v$  
  $\displaystyle =$ $\displaystyle (Av)^*v = (\lambda v)^* v = \overline{\lambda} v^* v \,,$  

for $ \lambda = \overline{\lambda} \Rightarrow \lambda \in \mathbb{R}$.
(Authors: Höllig/Reble/Höfert)
  automatically generated 4/21/2005