Question

Suppose \(Ax = \lambda x\) with \(x \ne 0\). Let \({\bf{\alpha }}\) be a scalar different from the eigenvalues of \(A\), and let \(B\) and let \(B = {\left( {A - \alpha I} \right)^{{\bf{ - 1}}}}\) . Subtract \(\alpha x\) from both sides of the equation \(A{\rm{x}} = \lambda {\rm{x}}\), and use algebra to show that \(\frac{1}{{\lambda - \alpha }}\) is an eigenvalue of \(B\), with \({\rm{x}}\) a corresponding eigenvector.

Short answer

It is proved that \(\frac{1}{{\lambda - \alpha }}\) is an eigenvalue of the matrix \(B = {\left( {A - \alpha I} \right)^{ - 1}}\) with \({\rm{x}}\) a corresponding eigenvector.

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 \(\lambda \).

Show that \(\frac{{\bf{1}}}{{\lambda  - \alpha }}\)is an eigenvalue of \(B\), with \(x\) a corresponding eigenvector

As it is given that\(A{\bf{x}} = \lambda {\bf{x}}\)where\({\bf{x}} \ne 0\), and\(\alpha \left( { \ne \lambda } \right)\)is a scalar. Now we have to prove that\(\frac{1}{{\lambda - \alpha }}\)is an eigenvalue of\(B\)with the corresponding eigenvector\({\bf{x}}\)where\(B = {\left( {A - \alpha I} \right)^{ - 1}}\).

Write the standard Matrix equation for eigenvalue and eigenvector.

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

Furthermore,

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

Thus,\(\frac{1}{{\lambda - \alpha }}\)is an eigenvalue of the matrix\(B = {\left( {A - \alpha I} \right)^{ - 1}}\).

It is proved that\(\frac{1}{{\lambda - \alpha }}\)is an eigenvalue of the matrix\(B = {\left( {A - \alpha I} \right)^{ - 1}}\)with \({\rm{x}}\) a corresponding eigenvector.