(a) Is it possible for a quadratic form to be positive, q(𝐱+)>0, at only one point 𝐱+ ∈ℝ^n ? (b)

step 1 So we could consider here a quadratic form, Q of X, Y, being equal to X, Y. This is a valid quad

(a) Is it possible for a quadratic form to be positive,...

(a) Is it possible for a quadratic form to be positive, $q\left(\mathbf{x}_{+}\right)>0$, at only one point $\mathbf{x}_{+} \in \mathbb{R}^n$ ? (b) Under what conditions is $q\left(\mathbf{x}_0\right)=0$ at only one point?

Key Concepts

Homogeneity of Quadratic Forms

A quadratic form is a homogeneous polynomial of degree two, meaning that if you scale its argument by any real number t, its value scales by t². This property implies that if the quadratic form takes a positive value at some nonzero point, then by scaling that point you get an entire ray (and in fact an open cone) of points where the quadratic form is positive. This homogeneity is key in understanding why a quadratic form cannot be positive at only one isolated point.

Positive (or Negative) Definiteness

A quadratic form associated with a symmetric matrix is said to be positive definite if it gives a positive value for every nonzero vector; similarly, it is negative definite if it gives a negative value for every nonzero vector. For a positive (or negative) definite form, the value zero is attained only at the zero vector. Thus, a quadratic form can vanish at only one point (namely the origin) if and only if it is nondegenerate and definiteness holds.

Kernel and Degeneracy of Quadratic Forms

The kernel (or null space) of a quadratic form is the set of all vectors that are mapped to zero. If the associated symmetric matrix has a nontrivial kernel (i.e. is degenerate), then there exist infinitely many nonzero vectors (typically forming a subspace) for which the quadratic form is zero. Therefore, the condition that the quadratic form equals zero at only one point (the zero vector) is equivalent to the statement that the quadratic form (or its matrix) is nondegenerate.

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