Short proof for the determinant of a $4$ by $4$ matrix

Question

Prove that $\det \begin{bmatrix}x&y&z&t\\-y&x&-t&z\\-z&t&x&-y\\-t&-z&y&x\end{bmatrix} = (x^2+y^2+z^2+t^2)^2$

I'm looking for an elegant proof that doesn't involve bruteforce.

Since the answer is given, I'm thinking we can argue that the determinant here is a homogeneous polynomial $P(x,y,z,t)$ with degree $4$, that is invariant under $x\to -x$ and permutations of $x,y,z,t$.

As a result, $P(x,y,z,t) = \lambda (x^4+y^4+z^4+t^4) + \delta (x^2y^2+x^2z^2 + x^2t^2+y^2z^2 + y^2t^2 + z^2t^2)$

$\lambda$ and $\delta$ can be found by computing $P(0,0,0,1)$ or some such.

The problem is, it doesn't look easy to prove that $P$ doesn't change under permutation of $x,y,z,t$, neither that it's invariant when the variables are negated.

Can you suggest another short proof, or prove the two claims above ?

Answer 1

Hint:

$$ \begin{pmatrix} x & y & z & t \\ -y & x & -t & z \\ -z & t & x & -y \\ -t & -z & y & x \end{pmatrix} \begin{pmatrix} x & -y & -z & -t \\ y & x & t & -z \\ z & -t & x & y \\ t & z & -y & x \end{pmatrix} = (x^2+y^2+z^2+t^2) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

Answer 2

The matrix has the two-by-two block form $$M=\begin{pmatrix} xI+yJ & zI+t J \\ -zI+t J & xI-yJ \end{pmatrix}$$ where $I$ is the 2-by-2 identity matrix and $J=\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}$. Since these four matrices commute, the determinant of $M$ by treating the elements as scalars and then taking the determinant of the result: \begin{align}\det{M}&=\det[(xI+y J)(xI-yJ)-(-zI+tJ)(-zI-tJ)]\\&=\det[(x^2+y^2+z^2+t^2)I]\\&=(x^2+y^2+z^2+t^2)^2.\end{align}