Meaning of the identity $\det(A+B)+\text{tr}(AB) = \det(A)+\det(B) + \text{tr}(A)\text{tr}(B)$ (in dimension $2$)

Question

Throughout, $A$ and $B$ denote $n \times n$ matrices over $\mathbb{C}$. Everyone knows that the determinant is multiplicative, and the trace is additive (actually linear). \begin{align*} \det(AB) = \det(A)\det(B) && \mathrm{tr}(A+B)= \mathrm{tr}(A) + \mathrm{tr}(B). \end{align*} On the other hand, the opposite equations \begin{align} \det(A+B) = \det(A)+\det(B) && \mathrm{tr}(AB)= \mathrm{tr}(A)\mathrm{tr}(B) \tag{1}. \end{align} don't hold for all $A,B$ unless $n=1$. For instance, taking $A=B=I$ in (1), we get \begin{align*} \det(A+B) = 2^n && \det(A)+\det(B) = 2 && \mathrm{tr}(AB)= n && \mathrm{tr}(A)\mathrm{tr}(B) = n^2. \end{align*} When $n=2$ a very curious thing happens, which is that, even though the equations (1) are typically false, their sum is actually valid. That is, \begin{align} \det(A+B) + \mathrm{tr}(AB)= \det(A)+\det(B)+ \mathrm{tr}(A)\mathrm{tr}(B) \tag{2} \end{align} for all $A$ and $B$ when $n=2$. However, (2) does not hold when $n>2$. Indeed, specializing to $B=I$ in (2), we get \begin{align} \det(A+I) + \mathrm{tr}(A)= \det(A)+1+ n \cdot \mathrm{tr}(A) \end{align} or, equivalently, \begin{align} \det(A+I) = \det(A)+(n-1) \cdot \mathrm{tr}(A) +1 \tag{3}. \end{align} If $n \geq 2$ and $A$ is projection onto the first coordinate, we have \begin{align} \det(A+I) = 2 && \det(A)+(n-1)\cdot \mathrm{tr}(A) +1 = n, \end{align} so (3) is only an identity for $n=2$ (or, trivially, when $n=1$).

Question: Is there any special significance to the equation \begin{align} \det(A+B) + \mathrm{tr}(AB)= \det(A)+\det(B)+ \mathrm{tr}(A)\mathrm{tr}(B) , \end{align} which is valid for all $2 \times 2$ matrices? It seems very strange to me that the sum of two obviously false equations should turn out true. Are there any nice applications of this identity?

Answer 1

This is the case $n=2$ of a 1980 theorem of Amitsur. It is described (in more abstract notation) here and this question gives the reference to Amitsur's paper.

Answer 2

In a quite pedestrian way, this is just saying that $\det$ is a quadratic form with the trace as its polar form. Namely, for any matrix $A$, let $A^\star = \mathrm{Tr} A - A$ be its conjugate (as in “the conjugate root of the characteristic polynomial”). Then your formula is equivalent to $$ \det (A+B) - \det(A) - \det(B) = \mathrm{Tr}(A^\star \cdot B).$$ The left-hand side is the polar form associated to the determinant form.

In a slightly more abstract way, for any field $k$, the matrix algebra $k^{2\times 2}$ is a (split) quaternion algebra. The formula above is just the polar relation for the norm form in this algebra (which is the determinant of a matrix).