Said more specifically, suppose $A,B\in M_n(K)$, $K$ a commutative ring. An $r$-th principal minor of a square matrix is the determinant $$\det\begin{bmatrix}a_{k_1k_1} & \cdots & a_{k_1k_r}\\ \vdots & \ddots & \vdots \\ a_{k_rk_1} & \cdots & a_{k_rk_r}\end{bmatrix}$$ where $1\le k_1<\cdots<k_r\le n$, $A$ is an $n\times n, n>r$ square matrix. Prove that the sum of all of $AB$'s $r$-th principal minors is equal to that of those of $BA$'s.

Hint: use Cauchy-Binet formula.

I simply have no idea what use to be made of the hint. I just can't represent the terms in the way Cauchy-Binet formula does, anyway for me it seems impossible to construct any of $P_r(AB)$ from sub-blocks of $A,B$.

Any help?

Edit: I'm sorry I made a mistake. It should be principal minor, no leading.

  • $\begingroup$ The definition of principal minor is not the one you gave. What you gave is a leading principle minor. Which one do you mean? There is precisely one leading principle minor of size $k\times k$ for each $k$ (the corner), but there are $\binom{n}{k}$ principle minors (which do not have to be corners). It's likely that you're stuck because you're using the wrong definition. $\endgroup$ – EuYu Sep 6 '15 at 16:36
  • 1
    $\begingroup$ Then the statement doesn't seem to be true. Consider $$A=\begin{pmatrix}1&2\\3&4\end{pmatrix},B=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$ The sum of the leading principle minors for $AB$ is $4$ and the sum of the leading principle minors for $BA$ is $5$. $\endgroup$ – EuYu Sep 6 '15 at 16:46
  • $\begingroup$ @EuYu I'm sorry. You're right. It should be principle minors, no leading. $\endgroup$ – Vim Sep 6 '15 at 23:20

You may begin with $P_r(AB)=\det(A_{[r],[n]}B_{[n],[r]})$ and apply Cauchy-Binet formula to decompose it into a sum of products of determinants.

  • $\begingroup$ That works. Thank you very much. $\endgroup$ – Vim Sep 7 '15 at 2:11

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.