If we have a given hermitian matrix

$$\left(\begin{matrix} \lambda & e^i \\ e^{-i} & \lambda \\ \end{matrix}\right)$$

wherein $\lambda$ is a real number, through the properties of a hermitian matrix its eigenvalues must be a real number, and since the only real numbers in this matrix is $\lambda$, is it safe to assume that the eigenvalues of this matrix is $\lambda$ without calculating for it?

  • 3
    $\begingroup$ There is no reason to think the eigenvalues of a matrix will be foound among the entries of the matrix. $\endgroup$ – Gerry Myerson Oct 8 '15 at 8:20

no it's not and you gave already the explanation.... just calculate the eigenvalues of your matrix... they are $\lambda \pm 1$ therefor not $\lambda$


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