91影视

Assume \(A = \left( {\begin{aligned}{{}}a&b\\b&d\end{aligned}} \right)\).

To show that \(Q\)is positive definite we need to prove that all the eigenvalues of the matrix\(A\)are positive.

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,

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

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

Since, determinant of A is positive,\({\lambda _1}{\lambda _2} > 0\), hence\({\lambda _1}\)and\({\lambda _2}\)has the same sign. Also\({\lambda _1} + {\lambda _2} = a + d > 0\)which proves that both eigenvalues are positive. Therefore,\(Q\)is positive definite

It is verified that \(Q\) is positive definite if \(\det A > 0\) and \(a > 0\).

As the determinant \(A\) is positive, \({\lambda _1}{\lambda _2} > 0\)hence\({\lambda _1}\)and\({\lambda _2}\)have the same sign. Also\({\lambda _1} + {\lambda _2} = a + d < 0\)which proves that both eigenvalues are negative. Therefore,\(Q\)is a negative definite.

It is verified that \(Q\) is negative definite if \(\det A > 0\) and \(a < 0\).

As the determinant \(A\) is negative, \({\lambda _1}{\lambda _2} < 0\)hence\({\lambda _1}\)and\({\lambda _2}\)have the opposite sign. Therefore,\(Q\)is indefinite.

It is verified that \(Q\) is indefinite if \(\det A < 0\).