Question

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.)

Short answer

\(\left( {\frac{1}{\mu } + \alpha } \right)\) is an eigenvalue of \(A\) in terms of \(\mu \) and \(\alpha \).

Step-by-step solution

Write the definition eigenvector and eigenvalue

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 \(\lambda \).

Find the eigenvalue of \(A\) in terms of \(\mu \) and \(\alpha \)

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.

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

Solve the obtained result for \(A{\bf{x}}\)

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 \({\bf{x}}\).