91Ó°ÊÓ

Consider \(A\) and \(B\) as \(n \times n\) matrices.

1. \(A\)isinvertiblesuch that \(\det A \ne 0\).
2. \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)
3. \(\det {A^T} = \det A\)
4. When \(A\) is triangular, the product of the entries on the main diagonal of \(A\) is \(\det A\).

Use the property of determinant to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

\[\begin{array}\det \left( {{A^T} - \lambda I} \right) = \det \left( {{A^T} - \lambda {I^T}} \right)\\ = \det {\left( {A - \lambda I} \right)^T}{\rm{ }}\left( {{\mathop{\rm by}\nolimits} \,\,{\mathop{\rm transpose}\nolimits} \,\,{\mathop{\rm property}\nolimits} } \right)\\ = \det \left( {A - \lambda I} \right)\end{array}\]

Thus, it is proved that \(A\) and \({A^T}\) have the same characteristic polynomial.