In a work about the Wave Equation, I encountered the following symmetric matrix of size $1+n$, whose entries are quadratic forms. The arguments are a scalar $a$ and a vector $X\in k^n$. $$S(a,X)=\begin{pmatrix} \frac12(a^2+|X|^2) & -aX^T \\ -aX & X\otimes X+\frac12(a^2-|X|^2)I_n \end{pmatrix}.$$ It turns out that the determinant factorizes at a high degree: $$\det S=\left(\frac{a^2-|X|^2}2\right)^{1+n}.$$

Is there any conceptual explanation ? Is this an example of some theory, or of a more general family ?

For the sake of completeness, the solutions of the WE $\partial_t^2u=c^2\Delta u$ satisfy the additional conservation laws ${\rm Div}\,S(\partial_t u,c\nabla u)=0$ (read the divergence row-wise). The first line of this vector-valued equation is the conservation of energy. The determinant is a power of $(\partial_tu)^2-c^2|\nabla u|^2$, which is a null-form, in Klainerman's terminology.

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    $\begingroup$ You mean other than the fact that $S(a,X)$ is orthgonally equivalent to the sum of the $n=1$ case and a linear transformation on a space of dimension $(n{-}1)$ that is $\tfrac12(a^2-|X|^2)$ times the identity? (The $n=1$ case is simply the Pythagorean triple identity.) $\endgroup$ Dec 27, 2017 at 10:29
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    $\begingroup$ $X \otimes X$ means $XX^T$ ? $\endgroup$ Dec 27, 2017 at 11:00
  • $\begingroup$ @darijgrinberg. Yes I do. $\endgroup$ Dec 27, 2017 at 11:33

1 Answer 1


Let's put $x:=(0,X)\in k^{n+1}$ and $y:=-ae_1+x\in k^{n+1}$. Then $S$ writes as a symmetric rank-$2$ perturbation of a multiple of the identity, $S=S_0+ \lambda I_{n+1},$ with $$S_0:= -\big(a^2-|x|^2\big)\ e_1\!\otimes e_1+y \otimes y,$$ and $$\lambda:={1\over2}\big(a^2-|x|^2\big).$$ This gives $\lambda$ as eigenvalue of $S$ of multiplicity $n-1$, with eigenspace $\ker S_0 =e_1^{\perp}\cap u^{\perp}$. On the orthogonal of the latter, $(\ker S_0)^\perp=\mathrm{span}(e_1,u)$, writing the operator $S$ in the orthogonal basis $(e_1,x/|x|)$ one finds the eigenvalues $|x|^2+\lambda\pm a|x|$, and $$\det(S_0+\lambda I_{n+1})=\lambda^{n-1}\big((\lambda+|x|^2)^2-a^2|x|^2 \big)$$ This of course holds for any value of $\lambda$, but $\lambda:={1\over2}\big(a^2-|x|^2\big)$ is special in that it makes the determinant equal to $\lambda^{n+1}$.

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    $\begingroup$ (this is more or less an expansion of Robert Bryant's comment) $\endgroup$ Dec 27, 2017 at 17:42

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