TITLE: Is there a name for the factorization $\frac12(a+c)(b-d)+\frac12(a-c)(b+d)=ab-cd$?
QUESTION [7 upvotes]: I came across this factorization:
$$
\begin{align}
\frac12 (a + c)(b - d) + \frac12 (a - c)(b + d) &= \frac12(ab-ad+cb-cd+ab+ad-cb-cd)
\\[4pt]
&= \frac12(2ab - 2cd) \\[4pt]
&= ab - cd
\end{align}
$$
Is there a technical mathematical name for this? I've seen it briefly being referred to as "twisted factorization" but it doesn't seem prevalent.

REPLY [2 votes]: Not sure about the name, but this follows from the factorization of matrices: $$\left[\begin{array}{cc}\frac{a+c}2&\frac{b+d}2\\-(a-c)&b-d\end{array}\right]=\left[\begin{array}{cc}\frac 12&\frac 12\\ -1&1\end{array}\right]\left[\begin{array}{cc}a&d\\ c&b\end{array}\right].$$ Geometrically, this can be explained as follows: Let $v_1=<a,d>,v_2=<c,b>$ be two vectors. Then $$v_1\wedge v_2=\frac 12(v_1+v_2)\wedge (v_2-v_1),$$ i.e. the area of the parallelogram spanned by $v_1,v_2$ is half of the area of the parallelogram spanned by the vectors representing the diagonals of the previous parallelogram.