Let
$$ (D_n)_{ij}=\mathbf 1_{i\mid j} $$
be the divisibility matrix, and let \(A_n\) be the Redheffer matrix:
$$ (A_n)_{ij} = \mathbf 1_{i\mid j\ \mathrm{or}\ j=1}. $$
The famous identity
$$ \det A_n=M(n) $$
puts the Mertens function inside a very concrete \(0\)-\(1\) matrix.
But the useful matrix is not quite \(A_n\). Normalize away the ordinary divisibility geometry first:
$$ \boxed{ B_n=D_n^{-1}A_n. } $$
Then every singular value of \(B_n\) except two is exactly \(1\).
I did not expect the entire Redheffer anomaly to collapse that hard.
Möbius inversion turns the extra column into a Mertens staircase#
The Redheffer matrix differs from \(D_n\) only in its first column. If
$$ u=(0,1,\ldots,1)^T, $$
then
$$ A_n=D_n+ue_1^T = D_n(I+v_ne_1^T), $$
where
$$ v_n=D_n^{-1}u. $$
The inverse divisibility matrix is literal Möbius inversion:
$$ (D_n^{-1})_{ij} = \begin{cases} \mu(j/i),&i\mid j,\\ 0,&i\nmid j. \end{cases} $$
Therefore
$$ \boxed{ v_n = \left( M(n)-1, M(\lfloor n/2\rfloor), \ldots, M(1) \right)^T. } $$
So
$$ \boxed{ B_n=I+v_ne_1^T. } $$
This is a rank-one perturbation of the identity. On the orthogonal complement of the plane spanned by \(e_1\) and the part of \(v_n\) perpendicular to \(e_1\), it acts as the identity exactly.
The entire nontrivial singular spectrum is two-dimensional.
The exceptional pair has a closed formula#
Put
$$ r_n^2 = \sum_{i=2}^nM(\lfloor n/i\rfloor)^2, $$
and
$$ T_n=1+M(n)^2+r_n^2. $$
The two exceptional squared singular values are
$$ \boxed{ \sigma_\pm(B_n)^2 = \frac{ T_n\pm\sqrt{T_n^2-4M(n)^2} }{2}. } $$
In particular,
$$ \boxed{ \sigma_+(B_n)\sigma_-(B_n)=|M(n)|. } $$
Everything else is \(1\).
The verifier constructs \(D_n\), \(D_n^{-1}\), and \(A_n\) over the integers, checks the factorization exactly, and then compares the numerical singular pair with the closed formula. At \(n=24\), the pair error is below \(9\times10^{-16}\), while the other \(n-2\) singular values differ from \(1\) by less than \(7\times10^{-16}\).
The formula concerns \(D_n^{-1}A_n\), not \(A_n\) itself. That distinction is load-bearing: divisibility normalization is what reveals the rank-one geometry.
The remaining energy is positive and exactly RH-sized#
The staircase energy can be regrouped by the values of \(\lfloor n/i\rfloor\):
$$ \boxed{ r_n^2 = \sum_{m\le n/2} \left( \left\lfloor\frac nm\right\rfloor - \left\lfloor\frac n{m+1}\right\rfloor \right) M(m)^2. } $$
Every coefficient is nonnegative. This is a very tempting place to start believing that matrix positivity has bought something.
It has not.
The term \(i=2\) in \(r_{2m}^2\) is literally \(M(m)^2\). Thus
$$ \boxed{ \mathrm{RH} \Longleftrightarrow r_n^2=O_\epsilon(n^{1+\epsilon}) \quad\text{for every }\epsilon>0. } $$
The forward implication is the Littlewood bound applied to the staircase. The reverse implication is already embedded in that one \(i=2\) summand.
Positivity has repackaged the Mertens problem without weakening it.
This is still useful. It says exactly why generic matrix inequalities, Schur complements, or norm estimates cannot close the route: the positive energy already contains the pointwise Mertens obstruction at half scale.
The staircase jumps on proper divisors#
Incrementing \(n\) changes \(\lfloor n/i\rfloor\) exactly when \(i\mid n+1\). That gives the signed recursion
$$ \boxed{ r_{n+1}^2-r_n^2 = 1+ \sum_{\substack{d\mid n+1\\1<d<n+1}} \left( M(d)^2-M(d-1)^2 \right). } $$
Since
$$ M(d)^2-M(d-1)^2 = 2\mu(d)M(d-1)+\mu(d)^2, $$
the positive global energy has sharply signed local feedback. A bounded scout through \(n=2{,}000{,}000\) found increments from \(-3615\) to \(3791\). There is no hidden contraction waiting in the divisor update.
So the exact picture is:
$$ \text{Redheffer} \xrightarrow{\ D_n^{-1}\ } \text{identity plus rank one} \xrightarrow{\ \mathrm{SVD}\ } \text{two exceptional values} \xrightarrow{\ } \text{one positive Mertens staircase}. $$
That compression is beautiful. It is also mercilessly honest about the boundary. The matrix problem does not become easier merely because its nontrivial singular geometry is only two-dimensional; all the arithmetic difficulty has been concentrated into \(M(n)\) and \(r_n\).
Any successor has to prove something genuinely new about the signed proper-divisor feedback. Another generic matrix norm bound will only walk around the same two-dimensional room.
Notebook references: K-0162