Question: Show that if (lambda) is an eigenvalue of the problem (mathbf{A x}=lambda mathbf{x}), it is also an eigenvalue of the adjoint problem (mathbf{A}^{mathrm{T}} mathbf{y}=lambda mathbf{y}).

Show that if $\lambda$ is an eigenvalue of the problem $\mathbf{A x}=\lambda \mathbf{x}$, it is also an eigenvalue of the "adjoint" problem $\mathbf{A}^{\mathrm{T}} \mathbf{y}=\lambda \mathbf{y}$.

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