Largest eigenvalue of a symmetric matrix close to the previous prime number

Consider the infinite matrix starting:

\displaystyle T(n,k) = - \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \end{bmatrix}

Notice the minus sign in front of the matrix. I was too lazy to change the sign in front of every element in the matrix. List the largest eigenvalues from each matrix:

{-1., 1.41421, 2.65544, 3.43931, 4.77106, 5.24392, 6.84437, 7.15539, \
7.47476, 7.57341, 10.9223, 11.096, 12.9021, 13.0453, 13.259, 13.4055, \
16.9724, 17.0824, 18.9443, 19.0552, 19.2282, 19.307, 22.9972, \
23.0759, 23.1576, 23.2173, 23.2976, 23.3972, 29.0103, 29.0407, \
30.963, 31.0104, 31.1008, 31.1505, 31.268, 31.34, 37.0284, 37.0658, \
37.1289, 37.174, 41.029, 41.0503, 42.9921, 43.0326, 43.0807, 43.1149, \
46.996, 47.0293, 47.0619, 47.1025, 47.1582, 47.2011, 53.0192, \
53.0497, 53.1076, 53.1419, 53.1893, 53.2117, 59.0477, 59.0681, \
61.0248, 61.0474, 61.0812, 61.1071}

As rounded to nearest integer the values are:
{-1, 1, 3, 3, 5, 5, 7, 7, 7, 8, 11, 11, 13, 13, 13, 13, 17, 17, 19, \
19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, \
37, 37, 37, 37, 41, 41, 43, 43, 43, 43, 47, 47, 47, 47, 47, 47, 53, \
53, 53, 53, 53, 53, 59, 59, 61, 61, 61, 61}

Which is close to the previous prime number.

Plotting them we get:

A useful lower bound for the largest eigenvalue would say something about the distribution of prime numbers.

Or as a ListPlot:

Mathematica 8 code for plot of largest eigenvalues:

Clear[a, t, n, k, i, j]
t[n_, 1] = -1;
t[1, k_] = -1;
t[n_, k_] :=
t[n, k] =
If[n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], -Sum[
t[k - i, n], {i, 1, n - 1}]];
nn = 64;
a = Range[1, nn]*0;
b = Range[1, nn]*0;
Do[m = Table[Table[t[n, k], {k, 1, j}], {n, 1, j}];
a[[j]] = Eigenvalues[m], {j, 1, nn}]

Round[Table[Max[N[a[[i]]]], {i, 1, nn}], 0.00001]
ListLinePlot[Table[Max[N[a[[i]]]], {i, 1, nn}]]
ListLinePlot[
Flatten[{0, Differences[Table[Max[N[a[[i]]]], {i, 1, nn}]]}],
Ticks -> {Range[nn]}]

Excel cell formulas for the matrix:

European:
=if(or(row()=1; column()=1); 1; if(row()>=column(); -sum(indirect(address(row()-column()+1; column(); 4)&":"&address(row()-1; column(); 4); 4)); -sum(indirect(address(column()-row()+1; row(); 4)&":"&address(column()-1; row(); 4); 4))))

American:
=if(or(row()=1, column()=1), 1, if(row()>=column(), -sum(indirect(address(row()-column()+1, column(), 4)&":"&address(row()-1, column(), 4), 4)), -sum(indirect(address(column()-row()+1, row(), 4)&":"&address(column()-1, row(), 4), 4))))

Edit 20.3.2012: Added a picture of the eigenvalues as radicals:

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