Calculate the signature of $(p,q)=p(1)q(1)+p(-1)q(-1)-p(129)q(129)$ Calculate the signature of
$B:(p,q) \mapsto p(1)q(1)+p(-1)q(-1)-p(129)q(129)$
where $p,q \in Pol(\mathbb{R}^{3})$ and $B$ is the bilinearform
I could solve it the conventional way by assuming that $p=a x^3+b x^2+ cx+d$ and $q=sx^3+t x^2+ ux+v$ calculating the eigenvalues of the corresponding matrix $A$ with $B(p,q)=(a,b,c,d)A \begin{pmatrix}
s \\ t \\ u \\ v
\end{pmatrix}$. The problem is that I could do it but it gets really complicated, which leads me to the suspicion that there is a trick
 A: Here's a nice trick. First, consider the bilinear form $B_k(p,q) = p(k)q(k)$ for $k \in \Bbb R$. Picking up where you started, we have
$$
\begin{align}
(p,q)_k &= (ak^3 + bk^2 + ck + d)(sk^3 + tk^2 + uk + v) = \\
&\begin{array}{ccccc}
= && k^6 as & + & k^5at & + & k^4 au & + & k^3 av\\
&+&k^5bs & + & k^4bt & + & k^3 bu\\
&+&k^4cs  & + &\cdots
\end{array}
\\ &= \pmatrix{a & b & c & d} 
\underbrace{\pmatrix{k^6 & k^5 & \cdots \\
k^5 & k^4 & \cdots\\
\vdots & \vdots & \ddots&k^2&k\\&&&k&1}}_{A_k} \pmatrix{s\\t\\u\\v}. 
\end{align}
$$
Notably, the matrix $A_k$ above is rank-1 and positive semidefinite, with $A_k = v_kv_k^T$ where $v = (k^3,k^2,k,1)^T$.
Now, the matrix corresponding to $B = B_1 + B_{-1} + B_{129}$ is given by $A = A_1 + A_{-1} + A_{129}$. From the properties of Vandermonde matrices, we can deduce that the vectors $v_1,v_{-1},$ and $v_{129}$ are linearly independent. It follows that $A$ has rank $3$. Because $A$ is a sum of positive semidefinite matrices, $A$ is positive semidefinite.
Alternatively, we can write $A_k = MM^T$ with $M = [v_1 \ \ v_{-1}\ \ v_{129}]$, which gives us another way to show that $A_k$ is positive semidefinite with rank $3$.
Thus, we can deduce that the signature of $B$ is $n_+ = 3, n_- = 0, n_0 = 1$.
