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

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 ?

• The matrix is the sum of $xI$ and an anti-symmetric matrix $A$; i.e., $A$ equals the negative of its transpose. I wonder if that can't be used to compute $$\det(x I + A).$$ – avs Aug 24 '16 at 17:07
• This question can probably be reframed in terms of the real matrix representations of the quaternions (shown here.) It presumably reflects that $|x+i y+zj+t k|^2=x^2+y^2+z^2+t^2$. – Semiclassical Aug 24 '16 at 17:48
• en.wikipedia.org/wiki/Pfaffian – Jack D'Aurizio Aug 25 '16 at 1:12
• @JackD'Aurizio still, you gotta compute this pfaffian – Gabriel Romon Aug 25 '16 at 7:44
• If you limit yourself to the case of $x=0$, then the formula for the Pfaffian of a generic 4-by-4 skew-symmetric matrix (found in @JackD'Aurizio's link) immediately gives the determinant as $(y^2+z^2+t^2)^2$. (I imagine there's some simple way to restore the $x$-dependence.) Of course, there's' still the matter of how one obtains said formula in the first place... – Semiclassical Aug 25 '16 at 12:53

Hint:

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

• what's the rationale behind this ? Did you find it by luck while scrapping things ? – Gabriel Romon Aug 24 '16 at 17:34
• $$x+y i \sim \begin{pmatrix} x & y \\ -y & x \end{pmatrix}$$ Now $$\det \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix}= a\bar{a}+b\bar{b}$$ $$x+yi=a$$ $$z+ti=b$$ – Ng Chung Tak Aug 24 '16 at 17:37
• That's similar to the logic I had (see my answer). The impetus in my case was that it reminded me of $|x+i y|^2+|z+i t|^2=x^2+y^2+z^2+t^2$. – Semiclassical Aug 24 '16 at 17:41
• The same trick can be applied to any antisymmetric matrix $A$. In particular: $$(xI + A)(xI + A)^T = (xI + A)(xI - A) = x^2I - A^2$$ In this case, $A^2$ is a multiple of the identity. – Omnomnomnom Aug 24 '16 at 17:47

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}

• See this link for a discussion of the block-determinant result I'm using: ee.iisc.ac.in/people/faculty/prasantg/downloads/blocks.pdf – Semiclassical Aug 24 '16 at 17:42
• Nice, at least it's not too out of the blue – Gabriel Romon Aug 24 '16 at 17:48
• In terms of Kronecker products, we have $$M = \pmatrix{x&z\\-z&x} \otimes I + \pmatrix{y&z\\-z&-y} \otimes J$$ – Omnomnomnom Aug 24 '16 at 17:48
• It's also matrix representation of quaternion $q=x+y\mathbf{i}+z\mathbf{j}+t\mathbf{k}$ – Ng Chung Tak Aug 24 '16 at 17:51
• @NgChungTak Yep. See my comment to the question itself :). Your answer, then, corresponds to the fact that $|q|^2=q\overline{q}=x^2+y^2+z^2+t^2$. – Semiclassical Aug 24 '16 at 17:56