Exercises 23 and 24 show how to classify a quadratic form \(Q\left( {\bf{x}} \right) = {{\bf{x}}^T}A{\bf{x}}\) when \(A = \left( {\begin{aligned}{{}}a&b\\b&d\end{aligned}} \right)\)and\(detA \ne {\bf{0}}\), without finding the eigenvalues of \(A\).

23. If \({\lambda _{\bf{1}}}\) and \({\lambda _{\bf{2}}}\) are the eigenvalues of \(A\), then the characteristic polynomial of \(A\) can be written in two ways: \(det\left( {A - \lambda I} \right)\) and \(\left( {\lambda - {\lambda _{\bf{1}}}} \right)\left( {\lambda - {\lambda _{\bf{2}}}} \right)\). Use this fact to show that \({\lambda _{\bf{1}}} + {\lambda _{\bf{2}}} = a + d\) (the diagonal entries of \(A\)) and \({\lambda _{\bf{1}}}{\lambda _{\bf{2}}} = detA\).

As it is given that if \({\lambda _1}\) and \({\lambda _2}\) are the eigenvalues of \(A\), then the characteristic polynomial of \(A\) can be written in two ways: \(\det \left( {A - \lambda I} \right)\) and \(\left( {\lambda - {\lambda _1}} \right)\left( {\lambda - {\lambda _2}} \right)\).

Find the characteristic polynomial.

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= \det \left( {\begin{aligned}{{}}{a - \lambda }&b\\b&{d - \lambda }\end{aligned}} \right)\\ &= \left( {a - \lambda } \right)\left( {d - \lambda } \right) - bd\\ &= {\lambda ^2} - \left( {a + d} \right)\lambda + ad - {b^2}........\left( 1 \right)\end{aligned}\)

Also, the expansion can be written as:

\(\left( {\lambda - {\lambda _1}} \right)\left( {\lambda - {\lambda _2}} \right) = {\lambda ^2} - \left( {{\lambda _1} + {\lambda _2}} \right)\lambda + {\lambda _1} \cdot {\lambda _2}......\left( 2 \right)\)

On equating both equations we get,

\({\lambda _1} + {\lambda _2} = a + d\)

\(\begin{aligned}{}{\lambda _1}{\lambda _2} &= ad - {b^2}\\ &= \det A\end{aligned}\)

Hence it is proved that \({\lambda _1} + {\lambda _2} = a + d\) and \({\lambda _1}{\lambda _2} = \det A\).