91Ó°ÊÓ

If \(q\left( {\overrightarrow x } \right)\) is a positive definite quadratic form, then so is \(kq\left( {\overrightarrow x } \right)\), for any scalar \(k\).

A matrix is said to be positive definite matrix, when it is symmetric matrix and all its eigenvalues are positive.

Consider\(q\left( {\overrightarrow x } \right) = {\overrightarrow x ^T}\left[ {\begin{array}{*{20}{l}}1&2\\3&4\end{array}} \right]\overrightarrow x \),which is a positive definite matrix.

The quadratic form is written as,

\(\begin{array}{c}q\left( x \right) = {x^T}\left[ {\begin{array}{*{20}{l}}1&2\\3&4\end{array}} \right]x\\ = \left[ {\begin{array}{*{20}{l}}{{x_1}}&{{x_2}}\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1&2\\3&4\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}}{{x_1}}&{{x_2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1} + 2{x_2}}\\{3{x_1} + 4{x_2}}\end{array}} \right]\\ = x_1^2 + 2{x_1}{x_2} + 3{x_1}{x_2} + 4x_2^2\\ = x_1^2 + 5{x_1}{x_2} + 4x_2^2\end{array}\)

If \(q\left( x \right)\) is positive definite, then \( - q\left( x \right)\) is a negative definite. That is, the result is not true for \(k = - 1\).

The given statement is FALSE.