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?
1 Answer
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2 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 $\det(B_0+C_0)=0\ne1=\det(B_1+C_1)$.
- $\begingroup$ From what you say, is it correct to deduce that, even for odd n, there is no function f such that det(A) = f(det(B))? $\endgroup$mfnx– mfnx2020-11-09 13:28:38 +00:00Commented Nov 9, 2020 at 13:28
- $\begingroup$ @mfnx That is correct. $\det(A)$ is not a function of $\det(B)$ and $\det(C)$. $\endgroup$user1551– user15512020-11-09 13:39:48 +00:00Commented Nov 9, 2020 at 13:39