The Cross-Alignment Matrix

Write the six fractions $1/7$ through $6/7$. Compare every row with every other row, digit by digit, and put the proportion of matches in a square table.

Each row matches itself perfectly. No two distinct rows agree at any position. The table has six ones down its diagonal and zeros everywhere else. In the language of linear algebra, it is the identity matrix $I_6$.

At thirteen, something appears off the diagonal.

 1/13 = 0.|076923|
11/13 = 0.|846153|

The third and sixth digits agree. The square where these two fractions meet therefore contains $1/3$. So does the square on the other side of the diagonal, where their order is reversed. Five other pairs do the same.

The Coherence Decomposition averages these agreements into a background score. Here I want to keep their addresses. Which rows agree with which? Do the agreements gather into groups, or run between them? At thirteen, the six pairs will give us twelve independent directions, divided between two eigenvalue levels. None of that arrangement survives in the pair average $1/33$.

Three fields, three shapes

Call the table $\mathbf A_b(n)$, the cross-alignment matrix for denominator $n$ in base $b$. The row and column labels are the numerators $1$ through $n-1$. Entry $(j,k)$ records the agreement between $j/n$ and $k/n$.

As before, the digits stay on a common clock. Move every row past the same clearing depth, then compare over a common period. Do not rotate one row to make it agree better with another. Two terminating rows score one; a terminating row paired with a repeating row scores zero.

Seven gives the identity. Thirteen gives an identity with six additional matching pairs. Twelve gives solid blocks. After two decimal places, its eleven rows fall into the terminating group of size three, the repeating threes of size four, and the repeating sixes of size four. Put each group together and the matrix becomes

where $J_r$ is an $r\times r$ square of ones. Every comparison within a group succeeds. Every comparison between groups fails.

Reordering rows and columns together changes where the squares sit, but not which fractions agree. This modest freedom will make the arithmetic much easier to see.

The three averages, recovered from the table

For $n\geq3$, the mean above the diagonal is $\sigma_b(n)$. The symmetric mean below it gives the same number. Self-matches belong to neither average.

Recovering $\alpha_b(n)$ also requires the marked reference row and the termination convention. Write $n=qs$, where $q$ is coprime to the base and every prime factor of $s$ divides the base. If $q>1$, the reference row of the matrix gives zero to each of the $s-1$ terminating rows. The reference score gives them full credit. Thus

$$\alpha_b(n)=\frac{\sum_{k=1}^{n-1}A_b(n)_{1,k}+s-1}{n-1}.$$

Then $F_b(n)=\alpha_b(n)-\sigma_b(n)$. If $q=1$, all rows terminate and both averages are one.

At twelve, the reference row sums to four. Add the three terminating rows and divide by eleven to recover $7/11$. The off-diagonal mean is $3/11$, leaving $F=4/11$. Those are the scalar counts of the preceding article. Here they serve as checks on the table, rather than the destination.

Ten slots per position

There is a reason these tables have well-behaved eigenvalues. Each is a table of inner products, a Gram matrix.

To build the vectors, give each decimal position ten slots, one for each possible digit. A six-digit word gets sixty slots. Occupy one slot at each position, according to the digit there, and give every occupied slot weight $1/\sqrt6$. The resulting vector has length one.

Two such vectors overlap only where both the position and the digit agree. For $1/13$ and $11/13$, exactly two slots are shared. Their inner product is $2/6$.

In base $b$ with common period $L$, use $bL$ slots and weight $1/\sqrt L$. All terminating rows receive the same separate unit vector, perpendicular to the digit slots. This implements the termination rule exactly. A terminated row is not mistaken for a repeating zero digit.

The construction gives

Every vector has length one, so the diagonal entries are one. More importantly, a Gram matrix is positive semidefinite. Its eigenvalues cannot be negative. We have reached a constraint on the whole spectrum without calculating a single eigenvalue.

For a prime $p$ not dividing $b$, the identity case is precisely the digit-partitioning case of Digit-Partitioning Primes and the Alignment Formula.

At or below that boundary, different nonzero remainders always emit different digits. Their vectors occupy disjoint slots. Above it, two remainders must emit the same digit somewhere, producing a positive entry off the diagonal. In decimal, eleven is the last prime with an identity matrix. Thirteen is the first departure.

One pair, two directions

An eigenvector is a direction that a matrix scales without turning. Its eigenvalue is the scale factor. The identity at seven leaves every direction unchanged, so its six eigenvalues are all one.

Thirteen is only a little harder. Its six matching pairs are disjoint. Place the two members of each pair next to one another, and the twelve-row matrix separates into six copies of

Give the two rows equal weights $(1,1)$. Multiplication by $B$ raises both to $4/3$. With opposite weights $(1,-1)$, the magnitudes shrink to $2/3$ and the signs stay as they were. The sum grows; the difference shrinks. Neither changes direction.

Each pair supplies one direction of each kind. Across six pairs, the eigenvalues are $4/3$ six times and $2/3$ six times. All twelve directions survive, but the single level at seven has split in two.

Twelve behaves differently. Within each solid block, only the sum of the row weights survives multiplication. Differences within the group disappear. The three blocks give eigenvalues $3,4,4$; the other eight eigenvalues are zero. Eleven rows, but only three surviving directions.

The block spectrum for a digit-partitioning core

Take $n=pm$, where $p$ is prime, $p\nmid b$, $p\leq b+1$, and every prime factor of $m$ divides $b$. Group the rows by their eventual tails. The matrix becomes

$$J_{m-1}\oplus\underbrace{J_m\oplus\cdots\oplus J_m}_{p-1\text{ blocks}}.$$

For $m\geq2$, its nonzero eigenvalues are $m-1$ once and $m$ repeated $p-1$ times. There are $pm-p-1$ zeros. Multiplying the denominator by a larger supported factor adds rows within these groups, while the rank stays at $p$.

At $m=1$, the terminating block is absent and the matrix is $I_{p-1}$. The rank there is $p-1$.

Slide the six digits

The six-pair calculation settles thirteen’s eigenvalues. To understand why the pairs occur, return to long division.

Multiplying a numerator by ten modulo the denominator advances its repeating word by one place. At seven, this visits all six nonzero numerators. The six rows are the six shifts of $142857$.

Slide that word against itself and count equal digits. At shift zero, all six agree. At every other shift, none do. This list of six counts is its digit-equality autocorrelation. Divide by six and the list is

Its discrete Fourier transform is six ones, the spectrum of $I_6$. Equality is the test throughout. Each digit has its own occupied-or-empty channel, just as in the slot construction.

Thirteen has two cycles, represented by $076923$ and $153846$. Each word contains six distinct digits. Compare either with its own shifts and the counts are exactly those at seven. Look only within either cycle and nothing has changed.

Now compare the cycles with each other. Shift $153846$ three places and it becomes $846153$, the word for $11/13$. Its third and sixth digits agree with $076923$. At every other shift there are no matches.

The change at thirteen lives entirely between the two cycles. Their individual autocorrelations cannot see it.

Two cycles, six small matrices

Order the numerators by those cycles.

 1, 10, 9, 12, 3, 4
 2,  7, 5, 11, 6, 8

The matrix now has four six-by-six blocks. The two diagonal blocks are $I_6$. The other two record the half-turn that brings the matching digits together. If $P$ shifts six coordinates by one place, these blocks are $P^3/3$.

The symbol $\sim$ here means that rows and columns have been reordered together. Each block is circulant, with successive rows obtained by cyclic shifts. Fourier coordinates diagonalize those shifts. In frequency $j$, a half-turn contributes $(-1)^j$, leaving the small matrix

The sign alternates, but each $M_j$ has the same two eigenvalues, $2/3$ and $4/3$. We recover the six-pair answer by a route that works beyond six isolated pairs.

For a general prime $p\nmid b$, let $L$ be the repeating period and $h=(p-1)/L$ the number of multiplication cycles. Arrange the matrix by those cycles. Fourier transformation reduces its $h^2$ circulant blocks to $L$ matrices of size $h\times h$. Together, their eigenvalues give the full spectrum.

No primitive-root assumption is needed. If there is only one cycle, each small matrix is just a number. With several cycles, the cross-correlations remain in the calculation. Nor do two cycles generally mean two eigenvalue levels. At thirteen, the six frequency matrices happen to share their eigenvalues; elsewhere the levels can vary with frequency.

The digit channels behind the Fourier calculation

For cycle $a$ and digit $d$, let $x_{a,d}(t)$ be one when position $t$ contains $d$, and zero otherwise. Use the unnormalized Fourier transform

$$X_{a,d}(j)=\sum_{t=0}^{L-1}x_{a,d}(t)e^{-2\pi ijt/L}.$$

With a consistent Fourier convention, the frequency matrix has entries

$$M_j(a,a')=\frac1L\sum_{d=0}^{b-1}\overline{X_{a,d}(j)}X_{a',d}(j).$$

This is another Gram matrix, now of complex Fourier coefficients. It is Hermitian and positive semidefinite. With one cycle, it reduces to the sum of the squared magnitudes of the digit-channel coefficients, divided by $L$.

Sixteen directions, nine levels

Seventeen supplies one cycle, long enough for repeated digits to enter the word.

1/17 = 0.|0588235294117647|

At shift zero, all sixteen positions agree. At shifts $\pm1$, $\pm3$, and $\pm5$ modulo sixteen, exactly two agree. Every other nonzero shift gives none. The normalized autocorrelation is therefore one at zero and $1/8$ at those six shifts.

Take its Fourier transform. Pairing each positive shift with its negative gives

These sixteen eigenvalues occupy nine distinct levels, from $1/4$ to $7/4$.

The spectrum counts multiplicities as well as heights. Seven has six directions at one level. Thirteen has twelve directions at two levels. Seventeen has sixteen directions at nine levels. A pair average can register an increase in agreement; it cannot recover this splitting of directions.

Thirty rows to a thousand, in two orderings

Each plate below shows the same exact matrix twice. On the left, numerators run in their ordinary order. On the right, multiplication cycles sit together, exposing the circulant blocks. The eigenvalues beneath them belong to both arrangements. Relabeling a matrix does not change its spectrum.

The color key is shared within each plate, but changes between denominators. Nonlinear contrast brings out small differences around the pair average. Blank cells score zero; the self-match diagonal is gray. The eigenvalue axis is linear below $0.2$ and logarithmic above it, so a large leading eigenvalue does not hide the rest.

These are computed examples, not additional general theorems. Select a plate to inspect either grid at full resolution.

Thirty to ninety-six rows

One hundred fifty to one thousand eight rows

Three matrices, shift counts, and exact spectra

These wider plates collect the small examples for side-by-side comparison.

The matches between the cycles

The ingredients have precedents. Lempel and Greenberger studied cyclic Hamming correlation; Kak and Chatterjee studied correlations in reciprocal digit sequences. Armstrong and Armstrong described the multiplication structure of repetends. Gram matrices and the Fourier diagonalization of circulants are standard linear algebra, treated by Horn and Johnson and Davis.

The construction here assembles the synchronized comparisons across the whole fractional field, including every multiplication cycle and the terminating rows. It gives the digit-partitioning boundary a matrix form, yields exact block spectra for the supported families, and keeps the cross-cycle terms when the base does not generate all nonzero residues.

At thirteen, either cycle considered alone still gives an identity matrix. Put the two together and the six matching pairs split the spectrum. The change was in the relation between the cycles. Keeping every comparison made it available to calculate.

The distance between long division and spectral analysis turns out to be one matrix.