Consider an indefinite quadratic form q on $\mathbb{R}^{3}$ with symmetric matrix A. If det A < 0, describe the level surface $q(\vec{x})=0.$

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Consider $q(x,y,z)$ to be an indefinite quadratic form defined by a symmetric matrix $B$ with det $(B)<0.$ By Theorem 8.2.2., we know that the quadratic form $q$ is diagonalizable with respect to the orthonormal eigenbasis of $B.$

Consider the diagonalized form of $q$ with the diagonal matrix

A=\begin{bmatrix} \lambda_1 & 0 & 0\\0 & \lambda_2 & 0\\0 & 0 & \lambda_3 \end{bmatrix}

. Since $A$ is a diagonal matrix, thus the eigenvalues of $A$ are the diagonal entries of $A$ , viz., $\lambda_1$ , $\lambda_2$ and $\lambda_3.$

From Theorem 8.2.4, we know that a matrix is positive definite (positive semi-definite) if and only if all its eigenvalues are positive (non-negative). Also, a symmetric matrix $X$ is negative definite (semi-definite) if and only if $-X$ is positive definite (positive semi-definite). Thus, we have all eigenvalues of a negative definite (semi-definite) matrix as negative (non-positive).

Recall that $A$ is indefinite so, by the previous discussion, some eigenvalues must be positive and some eigenvalues must be negative. Since det $A=\lambda_1 \cdot \lambda_2 \cdot \lambda_3<0$ , one of the eigenvalues must be negative, while the other eigenvalues must be positive. Without loss of generality let $\lambda_1 > 0$ , $\lambda_2 > 0$ , $\lambda_3 < 0$ .

Hence, the quadratic form

q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}=\begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & 0\\0 & \lambda_2 & 0\\0 & 0 & \lambda_3 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}=\lambda_1x^2+\lambda_2y^2+\lambda_3z^2

is indefinite and the level curve $q(\mathbf{x})=0$ represents an infinite elliptic cone.

The quadratic form $q(x,y,z)=x^2+2y^2-z^2$ is indefinite with det $(A)<0$ and the level curve $q(x,y,z)=0$ is an infinite elliptic cone, sketched below: