Discussion
The eigenvalues of a Hermitian matrix are all
Answer: A
The eigenvalues of a Hermitian matrix are all real.
If \(H = H^{\dagger}\) and \(Hv = \lambda v\), then
\[ \begin{aligned} v^{\dagger}Hv &= \lambda\, v^{\dagger}v \\ (v^{\dagger}Hv)^{*} &= v^{\dagger}H^{\dagger}v = v^{\dagger}Hv \end{aligned} \]
so \(v^{\dagger}Hv\) is real, and since \(v^{\dagger}v \gt 0\), \(\lambda\) is real. This is why observables in quantum mechanics are Hermitian operators.
Purely imaginary eigenvalues belong to anti-Hermitian matrices, and eigenvalues of modulus one to unitary matrices. Real and positive is too strong: \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) is Hermitian with eigenvalue \(-1\). Positivity needs a positive-definite matrix.
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