Signs, Bounds, and Support of Quadratic Forms

Why the Eigenvalues Matter

A quadratic form can look complicated in its original coordinates. For a Gaussian vector, however, an orthogonal change of coordinates turns it into a weighted sum of squares. The signs of those weights—the eigenvalues of the matrix—then reveal whether the form can be negative or positive, give exact bounds when the vector’s length is fixed, and identify the support of its distribution.

Throughout, let \(U\sim N_p(\mu,I_p)\), where \(I_p\) is the \(p\times p\) identity matrix, and let \(A\) be a real symmetric \(p\times p\) matrix. We study \(Q=U^\mathsf{T}AU\). The identity covariance ensures that an orthogonal change of coordinates preserves the standard-normal structure; the mean may be nonzero.

Diagonal Coordinates and Eigenvalue Bounds

By the spectral theorem, there is an orthogonal matrix \(H\) and a diagonal matrix \(\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_p)\) such that \(A=H\Lambda H^\mathsf{T}\). Set \(V=H^\mathsf{T}U\) and \(\nu=H^\mathsf{T}\mu\). Then \(V\sim N_p(\nu,I_p)\), and

Thus the eigenvalues are the weights on the squared coordinates. Since the squared coordinates are nonnegative, nonnegative eigenvalues give a nonnegative form; nonpositive eigenvalues give a nonpositive form. Eigenvalues of both signs permit both signs of the form.

Proof. Write \(x=Hy\), which preserves length because \(H\) is orthogonal. Then \(\lVert x\rVert^2=\sum_i y_i^2\) and \(x^\mathsf{T}Ax=\sum_i\lambda_i y_i^2\). For every \(i\), \(\lambda_{\min}\leq\lambda_i\leq\lambda_{\max}\); multiplying by \(y_i^2\geq 0\) and summing gives the two inequalities. If \(y\) is a unit coordinate vector for an eigenvalue \(\lambda_{\min}\) or \(\lambda_{\max}\), the corresponding bound is an equality. Scaling that eigenvector to length \(r\) proves the sphere statement. \(\square\)

Sign and Support for a Gaussian Vector

The matrix determines the sign for every possible input vector. For a Gaussian vector with identity covariance, it also determines which signs occur with positive probability. A Gaussian density with positive-definite covariance is positive everywhere, so every nonempty open ball has positive probability. In particular, if the form is negative at even one vector, continuity makes it negative in a neighborhood of that vector, and the Gaussian vector enters that neighborhood with positive probability.

Proof. First, a positive semidefinite \(A\) makes \(u^\mathsf{T}Au\geq 0\) for every \(u\), so \(Q\geq 0\) almost surely. Conversely, if \(A\) is not positive semidefinite, there is a vector \(x\) with \(x^\mathsf{T}Ax<0\). Continuity gives an open ball around \(x\) on which the form remains negative. Since \(U\) has a density positive everywhere, that ball has positive probability. Thus \(Q\geq 0\) almost surely is impossible. Applying the same argument to \(-A\) proves the negative-semidefinite assertion and shows that an indefinite form has both signs with positive probability.

To identify support, let \(g(x)=x^\mathsf{T}Ax\). Continuity and full support of \(U\) imply that the support of \(g(U)\) is the closure of \(g(\mathbb{R}^p)\): if \(t=g(x)\), every neighborhood of \(t\) contains values of \(g\) on an open neighborhood of \(x\), which has positive Gaussian probability. Conversely, if \(t\) lies outside the closure of the image, some neighborhood of \(t\) cannot be reached by \(g(U)\).

If \(A\) is nonzero positive semidefinite, its image under \(g\) is exactly \([0,\infty)\): zero is attained at \(x=0\), and for a unit eigenvector \(v\) with eigenvalue \(\lambda>0\), \(g(cv)=c^2\lambda\) ranges over all nonnegative values as \(c\) varies. The negative-semidefinite case gives \((-\infty,0]\). If \(A\) is indefinite, choose vectors with positive and negative form values. Scaling those vectors produces arbitrarily large values of both signs, and the continuous path between them takes every intermediate value. Its image is therefore \(\mathbb{R}\). Finally, for nonzero positive semidefinite \(A\), \(g(U)=0\) requires \(U\) to lie in the proper subspace \(\ker(A)\), which has probability zero under a density; the negative-semidefinite case follows by considering \(-A\). If \(A=0\), the form is identically zero. \(\square\)

Common Pitfall and Practical Use

A common mistake is to say that a positive-definite form has support \((0,\infty)\) because its value is strictly positive almost surely. That reasoning confuses values attained with positive probability and points in the support. A neighborhood of zero still has positive probability, however small it may be; therefore zero belongs to the support. The same distinction applies at the upper or lower endpoints of supports more generally.

The eigenvalue check is useful before attempting a more detailed distributional calculation. It establishes whether negative values are possible, supplies sharp bounds when the vector length is fixed, and reveals whether a proposed support is plausible. It also shows why the mean vector does not change these sign and support conclusions in this setting: a nondegenerate Gaussian vector still has positive density throughout \(\mathbb{R}^p\). The mean changes probabilities, but not the range available to the quadratic map.