Signature of quadratic form $Q(p)=p(1)p(2)+p(3)p(4)$ I was asked to find the signature of the quadtratic form $Q(p)=p(1)p(2)+p(3)p(4)$ where $p$ is a polynomial in $\mathbb R_n[x]$
I tried doing it via finding the symmetric matrix that $Q$ corresponds to, and I did that by trying to find out what is the image of a general polynomial, $$p(x)=\sum_{i=0}^n \alpha_ix^i$$
$$Q(p)=\sum_{i=0}^n \alpha_i \sum_{j=0}^n \alpha_i2^i + \sum_{k=0}^n \alpha_i3^i\sum_{r=0}^n \alpha_i4^i$$
How do we go on from here?
We could also try finding the maximal subspace for which $Q$ is definite positive / definite negative, but that doesn't seem easier.
 A: When $\mathbf{n=0}$, the quadratic form has signature $\mathbf{(1,0)}$ since $Q(1)=2>0$.

When $\mathbf{n=1}$, the quadratic form has matrix
$$\mathrm{Mat}(\beta,(2-X,X-1))=
\begin{pmatrix}
0&\frac12(1+0+6+3)\\
\frac12(0+1+6+3)&0
\end{pmatrix}=\begin{pmatrix}
0&5\\
5&0
\end{pmatrix}$$
and thus signature $\mathbf{(1,1)}$.

When $\mathbf{n=2}$, the  matrix of $Q$ w.r.t. the basis $\left(\frac{(X-2)(X-3)}{(1-2)(1-3)},\frac{(X-1)(X-3)}{(2-1)(2-3)},\frac{(X-1)(X-2)}{(3-1)(3-2)}\right)$ is
$$\begin{pmatrix}
0&\frac12&\frac12\\
\frac12&0&-\frac32\\
\frac12&-\frac32&1
\end{pmatrix}$$
Since the subspace $\Bbb R$ is positive for $Q$, and the determinant of the above matrix equals $-1$, $Q$ necessarily has signature $\mathbf{(2,1)}$.

When $\mathbf{n\geq 3}$, consider the basis $(\phi_1,\phi_2,\phi_3,\phi_4,\dots,\phi_n,\phi_{n+1})$ of $(\Bbb R_n[X])^*$ where $\phi_i:\Bbb R_n[X]\to\Bbb R$ is the linear form that maps $P$ to $P(i)$. By definition, $Q(P)=\phi_1(P)\phi_2(P)+\phi_3(P)\phi_4(P)$. Its antedual basis is the basis $(L_i)_{i=1,\dots,n+1}$ of $\Bbb R_n[X]$ where
$$L_i=\prod_{j\neq i}\,\frac{X-j}{i-j}$$
Let $\beta$ be the polar form of $Q$, that is, for all $A,B\in\Bbb R_n[X]$, $$\beta(A,B)=\frac12\Big(\phi_1(A)\phi_2(B)+\phi_2(A)\phi_1(B)+\phi_3(A)\phi_4(B)+\phi_4(A)\phi_3(B)\Big)$$ the matrix of $\beta$ w.r.t. this basis is
$$\mathrm{Mat}(\beta,(L_i))=
\begin{pmatrix}
0&\frac12\\
\frac12&0\\
&&0&\frac12\\
&&\frac12&0\\
&&&&0\\
&&&&&\ddots\\
&&&&&&0
\end{pmatrix}$$
Thus the signature of $Q$ is the same as that of the quadratic form defined by the symmetric matrix
$$\begin{pmatrix}
0&1\\
1&0\\
&&0&1\\
&&1&0\\
\end{pmatrix}$$
that is, $\mathbf{(2,2)}$.
To prove this, just notice that the matrix of $\beta$ in the basis $\mathcal B=(L_1+L_2,L_1-L_2,L_3+L_4,L_3-L_4,L_5,\dots,L_{n+1})$ is
$$\mathrm{Mat}(\beta,\mathcal B)=
\begin{pmatrix}
1\\
&-1\\
&&1\\
&&&-1\\
&&&&0\\
&&&&&\ddots\\
&&&&&&0
\end{pmatrix}$$
