Suppose \(\mu \) is an eigenvalue of the \(B\) in Exercise \({\bf{15}}\), and that \({\rm{x}}\) is a corresponding eigenvector, so that \({\left( {A - \alpha I} \right)^{{\bf{ - 1}}}}{\bf{x}} = \mu {\bf{x}}\). Use this equation to find an eigenvalue of \(A\) in terms of \(\mu \) and \(\alpha \). (Note: \(\mu \ne {\bf{0}}\) because \(B\) is invertible.)

Eigenvector and Eigenvalue: An eigenvector of \(n \times n\) matrix \(A\) is a nonzero vector \({\bf{x}}\) such that \(A{\bf{x}} = \lambda {\bf{x}}\) for some scalar \(\lambda \) where scalar \(\lambda \) is called an eigenvalue of \(A\). If there is a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\) then \({\bf{x}}\) is called an eigenvector corresponding to \(\lambda \).

Suppose \(\mu \) is an eigenvalue of the \(B\) and that \({\bf{x}}\) is a corresponding eigenvector, so that \({\left( {A - \alpha I} \right)^{ - 1}}{\bf{x}} = \mu {\bf{x}}\).

Write the standard Matrix equation for eigenvalue and eigenvector.

\(\begin{aligned}{c}{\left( {A - \alpha I} \right)^{ - 1}}{\bf{x}} = \mu {\bf{x}}\\\left( {A - \alpha I} \right){\left( {A - \alpha I} \right)^{ - 1}} = \left( {A - \alpha I} \right)\mu {\bf{x}}\\I{\bf{x}} = \left( {A - \alpha I} \right)\mu {\bf{x}}\\{\bf{x}} = \left( {A - \alpha I} \right)\left( {\mu {\bf{x}}} \right)\\ = A\left( {\mu {\bf{x}}} \right) - \left( {\alpha I} \right)\left( {\mu {\bf{x}}} \right)\\ = \mu \left( {A{\bf{x}}} \right) - \alpha \mu {\bf{x}}\end{aligned}\)

Simplify the obtained equation.

\(\begin{aligned}{c}{\bf{x}} + \alpha \mu {\bf{x}} = \mu \left( {A{\bf{x}}} \right)\\\frac{1}{\mu }\left( {{\bf{x}} + \alpha \mu {\bf{x}}} \right) = A{\bf{x}}\\A{\bf{x}} = \frac{1}{\mu }\left( {1 + \alpha \mu } \right){\bf{x}}\\A{\bf{x}} = \left( {\frac{1}{\mu } + \alpha } \right){\bf{x}}\end{aligned}\)

Thus, \(\left( {\frac{1}{\mu } + \alpha } \right)\) is an eigenvalue of \(A\) corresponding to the eigenvector \({\bf{x}}\).