Determinant of the symmetric part of a matrix.

Question

Define the symmetric part of a matrix $A$ as: $$ A^+ := \frac{A+A^t}{2}. $$

Is there a formula relating the determinants of $A$ and $A^+$?

Thanks!

Answer 1

Let $n\geq1$ be the number of columns of the $n\times n$-matrix $A$. Let $C_{i}$ and $L_{i}$ be the columns and the lines of $A$, respectively. Then, as the lines of $A$ are the columns of $A^{t}$ and conversely, one has $$\det\left(\frac{A+A^{t}}{2}\right)=2^{-n}\det\left(A+A^{t}\right)=2^{-n}\det\left(\left(C_{1}+L_{1},\ldots,C_{n}+L_{n}\right)\right)$$ $$=2^{-n}\det\left(\left(C_{1},\ldots,C_{n}\right)\right)+2^{n}\det\left(\left(L_{1},\ldots,L_{n}\right)\right)+2^{-n}\sum_{I\subset\left\{ 1,\ldots,n\right\} ,1\leq\#I\leq n-1}\det\left(\left(C_{I},L_{I^{\complement}}\right)\right)_{<} $$ $$=2^{-n}\det\left(A\right)+2^{-n}\det\left(A^{t}\right)+2^{-n}\sum_{I\subset\left\{ 1,\ldots,n\right\} ,1\leq\#I\leq n-1}\det\left(\left(C_{I},L_{I^{\complement}}\right)\right)_{<} $$ $$=2^{-n+1}\det\left(A\right)+2^{-n}\sum_{I\subset\left\{ 1,\ldots,n\right\} ,1\leq\#I\leq n-1}\det\left(\left(C_{I},L_{I^{\complement}}\right)\right)_{<} $$ where the symbol $\sum_{I\subset\left\{ 1,\ldots,n\right\} ,1\leq\#I\leq n-1}\det\left(\left(C_{I},L_{I^{\complement}}\right)\right)_{<}$ means that the sum is taken on all the possible ordered indexes for the columns and the lines, with at least one line and one column. For example, if $I=\left\{ 2,4,n\right\} $, then $$\det\left(\left(C_{I},L_{I^{\complement}}\right)\right)_{<}=\det\left(\left(L_{1},C_{2},L_{3},C_{4},L_{5},L_{6},\ldots,L_{n-2},L_{n-1},C_{n}\right)\right). $$