I am not a Math student and I am having trouble finding some small proof for the Muir's identity.

Even a slightly lengthy but easy to understand proof would be helpful.

Muir's Identity

$$\det(A)= (\operatorname{pf}(A))^2;$$

the identity is given in the first paragraph of the following link


I am expecting a proof which uses minimal advanced mathematics.Any reference to a textbook or link would do.

I would be very grateful, if any of you could point me in that direction.

P.s- i have done all the googling required and i wasnt satisfied with their results,so dont post any results from google's 1st page

Thanks in Advance

  • 3
    $\begingroup$ Could you include (a clear reference to) Muir's identity? $\endgroup$ – WimC May 4 '13 at 15:37

This answer does not show the explicit form of $\textrm{pf}(A)$ but it proves that such a form must exist as a polynomial in the entries of $A$.

Let $A$ be a generic skew-symmetric $n \times n$ matrix with indeterminate entries $A_{i j}$ on row $i$ column $j$ for $0 \leq i < j \leq n$. I will prove by induction in $n$ that $\det(A)$ is the square of a polynomial in the indeterminates $A_{i j}$. If $n = 2$ then

$$ \det \begin{pmatrix} 0& A_{1 2} \\ -A_{1 2}& 0 \end{pmatrix} = A_{1 2}^2 $$ is a square polynomial.

Let $$ B = \begin{pmatrix}1 \\ & 1 \\ & -A_{1 3} & A_{1 2} \\ & -A_{1 4} & & A_{1 2} \\ & \vdots & & & \ddots \\ & -A_{1 n} & & & & A_{1 2} \end{pmatrix} $$

where all unlisted entries are equal to zero. Then the product

$$ C = B\,A\,B^{T} $$

is skew-symmetric and takes the form $$ C = \begin{pmatrix} 0& A_{1 2} & 0 & \dotsc & 0\\ -A_{1 2} & 0 & \ast & \dotsc & \ast\\ 0 & \ast & \ddots & \ddots & \vdots\\ \vdots & \vdots & \ddots & & \ast \\ 0 & \ast & \dotsc & \ast & 0 \end{pmatrix} $$

where each asterisk denotes some polynomial in the indeterminates. Let $A'$ be the bottom right $(n-2) \times (n-2)$ skew-symmetric sub-matrix of $C$. By induction $\det(A')$ is the square of a polynomial. From the explicit form of $C$ it follows that $$ A_{1 2}^{2n-4} \det(A) = \det(B \, A \, B^T) = \det(C) = A_{1 2}^2 \det(A') $$ or $$\det(A) = A_{1 2}^{6-2n} \det(A').$$ Now the right hand side must be a polynomial (because $\det(A)$ is) and since it is also a square we are done.

  • $\begingroup$ @WimC-- I thank you for taking the pain to write this lengthy and useful proof, yes it showed that the det(A) is actually a square of a polynomial... Only if I could show that this polynomial is the pfaffian, it would make my day.... But still, Thank you :) $\endgroup$ – Vk1 May 5 '13 at 0:45

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