Relationship determinant symmetric matrix and skew-symmetric counterpart

Question

Say we have a square matrix A and we write it as the sum of its symmetric and skew-symmetric counterpart. Is there any formula which relates the determinants of A, its symmetric and skew-symmetric parts?

Answer 1

Let $A=B+C$, where $B$ is symmetric and $C$ is (hollow and) skew-symmetric.

When $n\le2$, one may directly verify that $\det(A)=\det(B)+\det(C)$.

When $n\ge3$, $\det(A)$ is not a function of $\det(B)$ and $\det(C)$. It suffices to construct two symmetric matrices $B_0,B_1$ and two skew-symmetric matrices $C_0,C_1$ such that $\det(B_0)=\det(B_1)$ and $\det(C_0)=\det(C_1)$ but $\det(B_0+C_0)\ne\det(B_1+C_1)$.

Here are our constructions (that work over all fields, including those of characteristic is $2$). When $n\ge3$ is odd, let $$ B_k=\pmatrix{1&1&0\\ 1&1&0\\ 0&0&k}\oplus I_{n-3}, \quad C_0=C_1=\pmatrix{0&-1&0\\ 1&0&0\\ 0&0&0}\oplus0. $$ Then $\det(B_0)=\det(B_1)=\det(C_0)=\det(C_1)=0$ but $\det(B_0+C_0)=0\ne1=\det(B_1+C_1)$.

When $n\ge4$ is even, let \begin{aligned} B_0&=I_2\oplus\pmatrix{1&1\\ 1&1}\oplus I_{n-4},\\ B_1&=I_2\oplus\pmatrix{0&0\\ 0&0}\oplus I_{n-4},\\ C_0&=\pmatrix{0&-1\\ 1&0}\oplus\pmatrix{0&0\\ 0&0}\oplus0,\\ C_1&=\pmatrix{0&0\\ 0&0}\oplus\pmatrix{0&-1\\ 1&0}\oplus0,\\ \end{aligned} Again, $\det(B_0)=\det(B_1)=\det(C_0)=\det(C_1)=0$ but $\det(B_0+C_0)=0\ne1=\det(B_1+C_1)$.