A curious determinantal identity A real $n\times n$ matrix $A$ can be uniquely decomposed into the sum of a symmetric matrix $B$ and a skew-symmetric matrix $C$ by setting $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$.
When $n=2$ there is an interesting result for the determinants of these matrices: if $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, then $B=\begin{bmatrix}a & \frac{b+c}{2} \\ \frac{b+c}{2} & d\end{bmatrix}$ and $C=\begin{bmatrix}0&\frac{b-c}{2} \\ -\frac{b-c}{2} & 0\end{bmatrix}$, hence
$$ \det(B)+\det(C)=ad-\frac{(b+c)^2}{4}+\frac{(b-c)^2}{4}=ad-bc=\det(A) $$
By testing a few examples, I've come to the conclusion that this identity does not hold for $n=3$ or $n=4$. My question, therefore, is whether the equality $\det(A)=\det(B)+\det(C)$ for $2\times 2$ matrices is simply a coincidence, or whether there is something interesting going on.
 A: Once you distill away all the multilinear algebra and restrict to linear transformations $\mathbb{R}^2 \to \mathbb{R}^2$, one can define the adjugate or classical adjoint of a matrix $A \in \mathbb{R}^{2 \times 2}$ to be the unique matrix $\operatorname{adj}(A) \in \mathbb{R}^{2 \times 2}$ such that 
$$
 \forall v_1,v_2 \in \mathbb{R}^2, \quad \det(A v_1 \vert v_2) = \det(v_1 \vert \operatorname{adj}(A)v_2),
$$
where for $v_1,v_2 \in \mathbb{R}^2$, $(v_1 \vert v_2)$ denotes the $2 \times 2$ matrix whose first column is $v_1$ and whose second column is $v_2$; a direct computation then shows that
$$
 \operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{pmatrix}d&-b\\-c&a\end{pmatrix}.
$$ 
The defining property of $\operatorname{adj}(A)$ immediately implies the following facts:


*

*for every $A \in \mathbb{R}^{2 \times 2}$, $\operatorname{adj}(A)A = \det(A)I_2$ (the well-known general property of adjugates);

*the assignment $A \mapsto \operatorname{adj}(A)$ defines a linear transformation $\operatorname{adj} : \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2}$ (a property specific to the $2 \times 2$ case);


which allow us to directly compute
$$
\begin{align}
 &\left(\det\left(\tfrac{1}{2}(A+A^t)\right)+\det\left(\tfrac{1}{2}(A-A^t)\right)\right)I_2\\ 
=\, &\operatorname{adj}\left(\tfrac{1}{2}(A+A^t)\right)\left(\tfrac{1}{2}(A+A^t)\right) + \operatorname{adj}\left(\tfrac{1}{2}(A-A^t)\right)\left(\tfrac{1}{2}(A-A^t)\right)\\
=\, &\frac{1}{4}\left(\operatorname{adj}(A)A+\operatorname{adj}(A)A^t+\operatorname{adj}(A^t)A + \operatorname{adj}(A^t)A^t+\operatorname{adj}(A)A-\operatorname{adj}(A)A^t-\operatorname{adj}(A^t)A + \operatorname{adj}(A^t)A^t\right)\\
=\, &\frac{1}{2}\left(\operatorname{adj}(A)A+ \operatorname{adj}(A^t)A^t\right)\\
=\, &\frac{1}{2}\left(\det(A)+\det(A^t)\right)I_2\\
=\, &\det(A)I_2.
\end{align}
$$
So, at the end of the day, as loup blanc's answer points out, what you're really seeing here is the special fact that the operator of taking the adjugate of a square matrix is actually linear in the case of $2 \times 2$ matrices; if you're comfortable with multilinear algebra, then this really boils down to the numerological coincidence that $\wedge^{n-1}\mathbb{R}^n = \mathbb{R}^n$ when $n=2$ (and not just $\wedge^{n-1}\mathbb{R}^n \cong \mathbb{R}^n$, which holds in general).
A: The relation $\det(A)=\det(B)+\det(C)$ is true when $n=2$ because the application $A\in M_2\rightarrow adj(A)\in M_2$ is linear ($adj(A)$ is the classical adjoint of $A$).
Proof. Let $f:A\in M_2\rightarrow \det(A)-\det(B)-\det(C)$; note that $f=0$ iff, for every $A,H\in M_2$,  $Df_A(H)=0$. One has 
$Df_A(H)=tr(Hadj(A)-1/4Hadj(A+A^T)-1/4H^Tadj(A+A^T)-1/4Hadj(A-A^T)$
$+1/4H^Tadj(A-A^T))=tr(H(adj(A)-1/2adj(A+A^T)-1/2adj(A-A^T)))=0$.
That is equivalent to $2adj(A)-adj(A+A^T)-adj(A-A^T)=0$. We conclude thanks to the linearity of $adj(.)$.
