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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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

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