Start with a strictly totally positive matrix \(B\). Split its one-based indices into odd and even sets,

$$ O=\{1,3,5,\ldots\}, \qquad E=\{2,4,6,\ldots\}, $$

and write

$$ B= \begin{pmatrix} B_{OO}&B_{OE}\\ B_{EO}&B_{EE} \end{pmatrix}. $$

Define the even-channel feedback

$$ H=B_{EE}-B_{EO}(I+B_{OO})^{-1}B_{OE}. $$

Then every principal minor of \(H\) is positive. In other words, \(H\) is a P-matrix.

This is not the same as saying \(H\) is entrywise positive. It need not be. It is a subtler positivity that survives exactly where the determinant expansion needs it.

Where the matrix comes from#

The original problem has two reciprocal spectral choices at each coordinate. If \(U\) is the all-low matrix and \(V\) the all-high matrix, replacing any subset of columns of \(U\) by columns of \(V\) produces \(2^r\) determinant sheets.

The transition

$$ B=DU^{-1}V, \qquad D=\operatorname{diag}(1,-1,1,-1,\ldots), $$

is strictly totally positive. Every positive sheet magnitude is

$$ \det(U)\det B_{S,S}, $$

and the complete folded determinant is

$$ \det(U+V)=\det(U)\det(I+DB). $$

So the exponential-size sheet sum is already the principal-minor expansion of one matrix.

Odd-even elimination makes the sign visible#

Block elimination gives

$$ \det(I+DB) =\det(I+B_{OO})\det(I-H). $$

The first factor is positive. The entire folded sign is therefore the sign of \(\det(I-H)\).

Now fix \(I\subseteq E\). The submatrix \(H_{I,I}\) is the Schur complement of \(I+B_{OO}\) inside

$$ \begin{pmatrix} I+B_{OO}&B_{O,I}\\ B_{I,O}&B_{I,I} \end{pmatrix}. $$

Expanding the identity entries in the odd block gives the exact formula

$$ \det H_{I,I} = \frac{\displaystyle\sum_{S\subseteq O} \det B_{S\cup I,S\cup I}} {\det(I+B_{OO})}. $$

Every determinant in the numerator is positive because \(B\) is strictly totally positive. The denominator is positive for the same reason. Hence every principal minor of \(H\) is positive.

No spectral approximation is happening. The P-matrix property is a direct quotient of positive principal-minor sums.

Two to the r becomes floor(r/2)+1#

Let

$$ a_k=\sum_{\substack{I\subseteq E\\|I|=k}}\det H_{I,I}>0. $$

Then

$$ \det(I-zH)=\sum_{k=0}^{|E|}(-z)^k a_k. $$

At \(z=1\), this is the original reciprocal folded sign. The \(2^r\) Boolean sheets have been grouped into only \(|E|+1\) positive layers, with alternation depending solely on how many even channels are active.

That is a very large compression.

The P-matrix edge#

A P-matrix can have negative entries and nonreal eigenvalues. Therefore

$$ \det(I-H)>0 $$

does not follow from the theorem. There are strictly totally positive matrices whose feedback has a nonreal conjugate eigenvalue pair, and physical reciprocal packets whose feedback has gauge-invariant negative directed cycles.

So the diagonal Perron-Frobenius route is closed. What survives is more precise and more useful: the sign problem is not an unstructured \(2^r\)-term cancellation. It is one alternating polynomial with explicit positive coefficients. Any successful continuation has to control that layer polynomial, the special arithmetic spectrum of \(H\), or a different invariant cone.

Notebook references: C-0194, C-0195, C-0199, O-0219, O-0220