eigenvalues of a Hermitian matrix are real

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Suppose $\lambda$ is an eigenvalue of the self-adjoint matrix $A$ with non-zero eigenvector $v$ . Then $Av=\lambda v$ .

\lambda^{{\ast}}v^{{H}}v=\left(\lambda v\right)^{{H}}v=\left(Av\right)^{{H}}v=% v^{{H}}A^{{H}}v=v^{{H}}Av=v^{{H}}\lambda v=\lambda v^{{H}}v

Since $v$ is non-zero by assumption, $v^{H}v$ is non-zero as well and so $\lambda^{{*}}=\lambda$ , meaning that $\lambda$ is real. ∎

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