A determinant from analytic geometry?

Question

I have a question regarding the following determinant:

$\begin{vmatrix} +ax - by - cz & bx+ay & cx+az \\ bx+ay& -ax+by-cz & bz+cy \\ cx+az & bz+cy & -ax-by+cz \end{vmatrix} = (a^2 + b^2 + c^2)(x^2 + y^2 +z^2)(ax+by+cz).$

I can prove the above equality by performing row operations and column operations. However the above equation has a lot of geometrical terms and it seems to me that this equation could have other interpretations that I am missing.

So I have two questions:

1) Can we write the given determinant as product of two determinants (or three even)?

2) Is there a conceptual proof of the above equality through linear algebra or analytic geometry?

Answer 1

I do not know the answer to the first question. But the problem can be solved by using linear algebra.

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Proof: Define $u = (a,b,c)$, $v = (x,y,z)$ and $P = uv^T + vu^T$[Thanks erfink].The given problem asks us to prove the following: $$\det(P - (u^Tv)I) = ||u||^2 ||v||^2 u^Tv.$$

We note that $P$ is a $3 \times 3$ matrix and we make the following observations:

1. $w = u \times v$ is in the null space of $P$ since $v^Tw = u^Tw=0$. So zero is an eigenvalue.
2. We can assume that $a^2 + b^2 + c^2 = x^2 + y^2 + z^2 = 1$ because of homogeneity.
3. Since we may assume $||u|| = ||v|| = 1$, it follows that $(uv^T + vu^T)(u+v) = (u^Tv + 1)(u+v)$, $(uv^T+vu^T)(u-v) = (u^Tv -1)(u-v).$ Thus $u^Tv + 1$, $u^Tv-1$ are eigenvalues of $P$ as well.

From these observation it follows that the characteristic polynomial is $$t(t+1-u^Tv)(t-1-u^Tv) = \det(tI - P).$$

Substituting $t = u^Tv$, we get $$u^Tv = \det(P - (u^Tv)I).\blacksquare$$