The square matrix \(\mathbf { A }\) has \(\lambda\) as an eigenvalue with \(\mathbf { e }\) as a corresponding eigenvector. Show that if \(\mathbf { A }\) is non-singular then
- \(\lambda \neq 0\),
- the matrix \(\mathbf { A } ^ { - 1 }\) has \(\lambda ^ { - 1 }\) as an eigenvalue with \(\mathbf { e }\) as a corresponding eigenvector.
The \(3 \times 3\) matrices \(\mathbf { A }\) and \(\mathbf { B }\) are given by
$$\mathbf { A } = \left( \begin{array} { r r r }
- 2 & 2 & - 4 \\
0 & - 1 & 5 \\
0 & 0 & 3
\end{array} \right) \quad \text { and } \quad \mathbf { B } = ( \mathbf { A } + 3 \mathbf { I } ) ^ { - 1 }$$
where \(\mathbf { I }\) is the \(3 \times 3\) identity matrix. Find a non-singular matrix \(\mathbf { P }\), and a diagonal matrix \(\mathbf { D }\), such that \(\mathbf { B } = \mathbf { P D P } \mathbf { P } ^ { - 1 }\).