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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!

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2 Answers 2

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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.

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  • $\begingroup$ 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 $A\pm A^T$. $\endgroup$ Sep 29, 2022 at 18:16
  • $\begingroup$ Just wondering: Where is it used that the order of the matrix is even? $\endgroup$
    – Martin R
    Sep 29, 2022 at 18:53
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    $\begingroup$ @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 $\endgroup$ Sep 29, 2022 at 19:39
  • $\begingroup$ @user8675309: Yes, of course. Thanks. $\endgroup$
    – Martin R
    Sep 29, 2022 at 19:53
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    $\begingroup$ @user8675309 -- great and intuitive answer! Thank you! $\endgroup$ Sep 29, 2022 at 19:57
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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 $1$.

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  • $\begingroup$ 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? $\endgroup$
    – Martin R
    Sep 29, 2022 at 18:57
  • $\begingroup$ @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. $\endgroup$ Sep 29, 2022 at 19:01
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    $\begingroup$ $A$ is given, but what are $B, C, D$? And how does this lead to $\det(A-A^T)$ and $\det(A+A^T)$? $\endgroup$
    – Martin R
    Sep 29, 2022 at 19:12

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