Show that if (lambda) is an eigenvalue of the problem (mathbf{A x}=lambda mathbf{x}), it is also an
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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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