Question

Question: Let \(\lambda \) be an eigenvalue of an invertible matrix A. Show that \({\lambda ^{ - {\bf{1}}}}\) is an eigenvalue of \({A^{ - {\bf{1}}}}\). (Hint: Suppose a nonzero x satisfies \(A{\bf{x}} = \lambda {\bf{x}}\))

Short answer

\({\lambda ^{ - 1}}\) is an eigenvalue of \({A^{ - 1}}\).

Step-by-step solution

Write the given information

For an invertible matrix A, \(\lambda \) is an eigenvalue.

Check for the eigenvalue of \({A^{ - {\bf{1}}}}\)

If \(\lambda \) is an eigenvalue of A, then there exists a nonzero vector \(x\) such as that \(A{\bf{x}} = \lambda {\bf{x}}\).

Since A is invertible, then \({A^{ - 1}}A{\bf{x}} = {A^{ - 1}}\left( {\lambda {\bf{x}}} \right)\).

As \(x \ne 0\), so \(\lambda \) cannot be equal to zero. Therefore, \({\lambda ^{ - 1}}{\bf{x}} = {A^{ - 1}}{\bf{x}}\).

The above equation shows that, \({\lambda ^{ - 1}}\) is an eigenvalue of \({A^{ - 1}}\).