91Ó°ÊÓ

• Let us consider q(x,y,z) to be an indefinite quadratic form defined by a symmetric matrix B with det (B) < 0.
• As per the theorem 8.2.2 , we know that the quadratic form q is diagonalizable with respect to the orthonormal eigenbasis of B
• Let us consider the diagonalized form of with the diagonal matrix.

$\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{λ}}_1& 0& 0\\ 0& {\mathrm{λ}}_2& 0\\ 0& 0& {\mathrm{λ}}_3\end{array}\right]$

• With respect to the theorem 8.2.4, we are aware that a matrix is a positive definite if and only if all its eigenvalues are positive.
• Similarly, a symmetric matric X is negative definite if and only if X is positive definite.
• Hence, we have all eigenvalues of a negative definite matrix as negative or non-positive.
• As A is indefinite, some eigenvalues must be positive and some must be negative too.
• As $\det \mathrm{A}={\mathrm{λ}}_1{\mathrm{λ}}_2{\mathrm{λ}}_3<0$, one of the eigenvalues must be positive, whereas all the other eigenvalues must be negative.
• Without the loss of generality, let ${\mathrm{λ}}_1>0,{\mathrm{λ}}_2>0,{\mathrm{λ}}_3<0$, and so the quadratic form is.

$q\left(x\right)=x^{T}Ax=\left[\begin{array}{ccc}x& y& z\end{array}\right]\left[\begin{array}{ccc}{\mathrm{λ}}_1& 0& 0\\ 0& {\mathrm{λ}}_2& 0\\ 0& 0& {\mathrm{λ}}_3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]={\mathrm{λ}}_1{x}^2+{\mathrm{λ}}_2{y}^2+{\mathrm{λ}}_3{z}^2$

• The quadratic form $\mathrm{q}\left(\mathrm{x}\right)=1$is indefinite and level curve represents a one-sheeted hyperboloid.
• The quadratic form $\mathrm{q}(\mathrm{x},\mathrm{y},\mathrm{z})={\mathrm{x}}^2+2{\mathrm{y}}^2-{\mathrm{z}}^2$is indefinite with det(A) < 0 and the level curve q(x,y,z) = 1.
• The level curve is sketched below:

Result:

The level curve represents one–sheeted hyperboloid.