Let $P$ be a $n$_degree polynomial over $\mathbb{R}$. Let the quadratic form on the vector space of $n$_degree polynomials:

$$H(P)=\int_0^\infty e^{-x^2}P(x)P(-x)dx$$

What is the signature of $H$?

In the usual basis of polynomials I could find the symmetric matrix representing this quadratic form $[M]_{ij}=(-1)^i\frac{(i+j-1)!!}{2^\frac{i+j}{2}}\frac{\sqrt{\pi}}{2}$ when ${i+j=0(\operatorname{mod}2)}$ and $[M]_{ij}=0$ otherwise. But in this form of $M$ it's likely that I can't diagonalized it to obtain the signature. What other approaches may be useful here?


1 Answer 1


One way of determining the signature is to decompose the vector space $V$ into $U\oplus W$, such that the quadratic form is positive definite on $U$ and negative definite on $W$, and such that $U$ and $W$ are orthogonal with respect to the quadratic form. Then the signature is the difference of the dimensions (this is easy to see by diagonalizing the form).

In this case, let $U$ be the space of even polynomials and $W$ the subspace of odd polynomials.

  • $\begingroup$ I don't think $U$ and $W$ are orthogonal with respect to the quadratic form... $\endgroup$ Commented Oct 27, 2014 at 1:07
  • $\begingroup$ Sorry - I didn't see this reply until now. $\endgroup$
    – JamesM
    Commented Nov 16, 2014 at 16:30
  • 1
    $\begingroup$ The orthogonality is w.r.t. the symmetric bilinear form $$G(P,Q) = \frac12 \int_0^\infty e^{-x^2} (P(x)Q(-x) + Q(x)P(-x))\,dx.$$ (Note that $G$ is symmetric, and satisfies $G(P,P)=H(P)$.) You'll then find if $P$ is even and $Q$ is odd this gives zero. $\endgroup$
    – JamesM
    Commented Nov 16, 2014 at 16:40
  • $\begingroup$ Yes I've found that out :) Thanks!! $\endgroup$ Commented Nov 17, 2014 at 0:49

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