I am currently working through a proof given in Longuet-Higgins (1957) but am unsure about how a line given on page 352 follows. To summarize what is confusing me, I give the below (note I have tried to make the notation simpler/clearer than in the paper so it may not match exactly what is given in the reference).
Let $M$ be a symmetric positive definite matrix defined as:
$$M = \begin{bmatrix} m_{40} & m_{31} & m_{22} \\ m_{31} & m_{22} & m_{13} \\ m_{22} & m_{13} & m_{04} \\ \end{bmatrix}$$
Where the elements $m_{i,j}$ are real constants (apologies the subscript indexing is strange, this carries over from a previous part of the proof). Let $T$ be the matrix:
$$T = \begin{bmatrix} 0 & 0 & \frac{1}{2} \\ 0 & -1 & 0 \\ \frac{1}{2} &0 & 0 \\ \end{bmatrix}$$
Let $D$ represent the diagonal matrix of eigenvalues of the matrix $MT$ (i.e. $MT=PDP^{-1}$ for some orthogonal matrix of eigenvectors $P$) and denote the diagonal elements of $D$ (eigenvalues of $MT$) like so:
$$D = \begin{bmatrix} d_{1} & 0 & 0 \\ 0 & d_{2} & 0 \\ 0 &0 & d_{3} \\ \end{bmatrix}$$
The paper states that, as $d_i$ (for all $i \in \{1,2,3\}$) is an eigenvalue for $MT$, the following holds:
$$\det(MT-d_iI)$$
Expanding this determinant we obtain:
$$4d_i^3 - 3Hd_i - \det(M) = 0$$
Where $3H=m_{40}m_{04}-4m_{31}m_{13}+3m_{22}^2$.
So far this makes sense to me. However, in the next line, without proof, they state that the above implies:
$$(1) \hspace{2cm} d_1+d_2+d_3=0$$
And:
$$(2) \hspace{1.6cm} d_1d_2d_3=\frac{1}{4}\det(M)$$
I cannot seem to work out how $(1)$ and $(2)$ follow from what I have stated. Is anyone able to prove $(1)$ and $(2)$ from what I have stated in this question?
Let me represent the characteristic polynomial in the following way:
$$(x-d_1)(x-d_2)(x-d_3)=x^3-(d_1+d_2+d_3)x^2+(d_1d_2+d_1d_3+d_2d_3)x-d_1d_2d_3$$
Upon expanding it, we notice that.
Compare this with the given expression divided by $4$ to make the leading coefficient $1$,
$$x^3-0\cdot x^2-\frac34Hx-\frac14\det(M)$$
We can conclude that $\sum_{i=1}^3d_i=0, \prod_{i=1}^3d_i=\frac{\det(M)}{4}$