# Determinant of $A+A^T$ is an odd integer if $\text{det}(A-A^T)=1$.

Let $$A\in\text{Mat}(2n\times 2n;\mathbb{Z})$$ be an integer matrix such that $$\text{det}(A-A^T)=1$$. I want to show that $$\text{det}(A+A^T)$$ is an odd integer. Murasugi claims in his book "Knot Theory and its Applications" that this is trivial and it probably follows immeditiately from some determinant property of skew-symmetric resp. symmetric matrices. But I just can not proof it. Any help would be greatly appreciated!

The point is that, so long as we enforce zeros on the diagonal, then skew-symmetric and symmetric matrices coincide in characteristic 2.

More formally, consider the ring homomorphism $$\phi:\mathbb Z\longrightarrow R\cong \mathbb F_2$$
then define $$\Phi$$ to apply $$\phi$$ component-wise to the matrix $$B\in \mathbb Z^{m\times m}$$. Then

$$\phi\Big(\det\big(B\big)\Big)=\det\Big(\Phi\big(B\big)\Big)$$
and with $$B:=A+A^T$$ we have
$$\phi\Big(\det\big(A+A^T\big)\Big)=\det\Big(\Phi\big(A+A^T\big)\Big)=\det\Big(\Phi\big(A-A^T\big)\Big)=\phi\Big(\det\big(A-A^T\big)\Big)=\phi(1)=1$$
so $$\det\big(A+A^T\big)\%2 =1$$, i.e. the determinant is odd.

• The $\cdot ^T$ in the OP turns out to be a red herring, i. e., it works with general $A\pm B$ just as with $A\pm A^T$. Sep 29, 2022 at 18:16
• Just wondering: Where is it used that the order of the matrix is even? Sep 29, 2022 at 18:53
• @MartinR -- it is implied/required in the original problem statement when we are told $\text{det}(A-A^T)=1$ -- the determinant would have to be zero if it was odd order Sep 29, 2022 at 19:39
• @user8675309: Yes, of course. Thanks. Sep 29, 2022 at 19:53
• @user8675309 -- great and intuitive answer! Thank you! Sep 29, 2022 at 19:57

Given a matrix $$A_{2(n+1)\times 2(n+1)}\in \mathrm{Mat}(2(n+1)\times 2(n+1),\mathbb{Z})$$ there is matrices $$\begin{array}{c c c} A_{2n \times 2n}\in \mathrm{Mat}(2n\times 2n,\mathbb{Z}), & \quad & B_{2n \times 2}\in \mathrm{Mat}(2n\times 2,\mathbb{Z}), \\ && \\ C_{2 \times 2n}\in \mathrm{Mat}(2 \times 2n,\mathbb{Z}), & \quad & D_{2\times 2} \in \mathrm{Mat}(2 \times 2,\mathbb{Z}) \end{array}$$ such that $$A_{2(n+1)\times 2(n+1)} = \left\lgroup \begin{array}{cc} A_{2n \times 2n} & B_{2n\times 2} \\ C_{2 \times 2n } & D_{2 \times 2 } \end{array} \right \rgroup.$$ Now you just use induction and the formula $$\det\begin{pmatrix}A & B \\ C & D\end{pmatrix} = \det(A) \det\left(D - C A^{-1} B\right).$$ to proof that $$\det\big( A_{_{2(n+1)\times 2(n+1)}} + A_{_{2(n+1)\times 2(n+1)}}^T \big) \\ =\det\begin{pmatrix} A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T} & B_{_{2n\times 2}}^{}+C_{_{2\times 2n}}^{T} \\ C_{_{2\times 2n}}^{}+B_{_{2n\times 2}}^{T} & 2 D_{_{2\times 2}} \end{pmatrix} \\ =\det \begin{pmatrix} A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T} \end{pmatrix} \det\Big( 2D_{_{2\times 2}} - (C_{_{2\times 2n}}^{}+B_{_{2n\times 2}}^{T})(A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T})^{-1}(B_{_{2n\times 2}}^{}+C_{_{2\times 2n}}^{T}) \Big)$$ is equal to $$1$$.

• Please pardon my ignorance, this is not my area of expertise: Is $A_{2n \times 2n}$ the given matrix $A$? What are $B, C, D$? Is $A$ necessarily invertible? Sep 29, 2022 at 18:57
• @MartinR Yes, $A_{2n\times 2n}$ is a matrix of order $2n\times 2n$. Yes, $B$, $C$ and $D$ are matrices. Yes, $A$ is necessarily invertible. Sep 29, 2022 at 19:01
• $A$ is given, but what are $B, C, D$? And how does this lead to $\det(A-A^T)$ and $\det(A+A^T)$? Sep 29, 2022 at 19:12