6
$\begingroup$

The Pfaffian $\text{pf}$ is defined for a skew-symmetric matrix which is also a polynomial of matrix coefficients. One property for Pfaffian is that $\operatorname {pf} (A)^{2}=\det(A)$ holds for every skew-symmetric matrix A .

As for determinants we have Cayley-Hamilton theorem, here is my question: \begin{align} w &= \begin{pmatrix} -A \cdot a & -I \\ I & B \cdot b \end{pmatrix} \\ \operatorname{Pf}(w) &= \sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor} P_i \cdot a^i b^i \\ A \cdot B &= C \end{align}

Where $A,B \in M_n(\Bbb R)$ are skew-symmetric, and $a,b \in \mathbb R$. $\text{Pf}(w)$ is expressed as polynomial of $ab$ ($P_i \in \mathbb R$ are coefficients).

Then can we show the following vanishing property? $$ \sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor} P_i \cdot C^{\left\lceil n/2\right\rceil - i} = 0 $$

My idea is that we can show the square of $P_iC^i$ is zero by Cayley-Hamilton theorem, but how to proceed?

Last but not least, can we proof something more general about analogs of Cayley-Hamilton theorem for Pfaffian?

$\endgroup$
  • $\begingroup$ What do you mean by "determinant of $\operatorname{Pf}(w)$"? The Pfaffian $\operatorname{Pf}(w)$ is just a number. $\endgroup$ – darij grinberg Aug 21 '17 at 11:34
  • $\begingroup$ Oh,I will edit this mistake. $\endgroup$ – sawdada Aug 21 '17 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.