trace equals determinant
- To: mathgroup at smc.vnet.net
- Subject: [mg105463] trace equals determinant
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Fri, 4 Dec 2009 04:30:28 -0500 (EST)
- Reply-to: rlbagulatftn at yahoo.com
This idea as a general form came from some special matrices that I ran into:
These three matrices in 4x4 matrices share
the property that their determinant is equal to their trace:
a = {{0, 1, 1, 1},
{1, -1, 0, 0},
{1, 0, -1, 0},
{1, 0, 0, -1}}
atrace = Sum[a[[n, n]], {n, 1, Length[a]}]
-3
adet = Det[a]
-3
a2 = a.a
{{3, -1, -1, -1},
{-1, 2, 1, 1},
{-1, 1, 2, 1},
{-1, 1, 1, 2}}
a2trace = Sum[a2[[n, n]], {n, 1, Length[a]}]
9
adet = Det[a2]
9
aa = a2 + 2*a
{{3, 1, 1, 1},
{1, 0, 1, 1},
{1, 1, 0, 1},
{1, 1, 1, 0}}
Det[a2 + 2*a]
3
Sum[aa[[n, n]], {n, 1, Length[a]}]
3
A little analysis applied the idea of the determinant equaling the
trace in the 2x2 matrix
gave:
{{a,1},
{-1,1}}
Using that in a matrix Markov recursion
I got two ways to look at it:
1) varying "a" as an integer in the rows
2) looking at the polynomial in "a" generated as coefficients
One of them gave a sequence already in OEIS by another method of recursion.
The general 3x3 problem of the determinant equaling the trace isn't this
simple.
%I A167925
%S A167925 0,1,1,1,2,3,0,2,6,12,1,0,9,32,75,1,4,9,80,275,684,0,8,0,192,1000,
%T A167925 3240,8232,1,8,27,448,3625,15336,47677,122368,1,0,81,1024,13125,
%U A167925 72576,276115,835584,2158569,0,16,162,2304,47500,343440,1599066
%V A167925
0,1,1,1,2,3,0,2,6,12,-1,0,9,32,75,-1,-4,9,80,275,684,0,-8,0,192,1000,
%W A167925
3240,8232,1,-8,-27,448,3625,15336,47677,122368,1,0,-81,1024,13125,
%X A167925 72576,276115,835584,2158569,0,16,-162,2304,47500,343440,1599066
%N A167925 A triangular sequence of the Matrix Markov type based on the
2x2 matrix: m={{a,1},{-1,1}}; which has determinant equal to trace.
%C A167925 Row sums are:
%C A167925 {0, 2, 6, 20, 115, 1043, 12656, 189420, 3356913, 68661516,
1591360540,...}
%C A167925 Each row is a specific Markov sequence with a different
limiting ratio.
%e A167925 {0},
%e A167925 {1, 1},
%e A167925 {1, 2, 3},
%e A167925 {0, 2, 6, 12},
%e A167925 {-1, 0, 9, 32, 75},
%e A167925 {-1, -4, 9, 80, 275, 684},
%e A167925 {0, -8, 0, 192, 1000, 3240, 8232},
%e A167925 {1, -8, -27, 448, 3625, 15336, 47677, 122368},
%e A167925 {1, 0, -81, 1024, 13125, 72576, 276115, 835584, 2158569},
%e A167925 {0, 16, -162, 2304, 47500, 343440, 1599066, 5705728,
16953624, 44010000},
%e A167925 {-1, 32, -243, 5120, 171875, 1625184, 9260657, 38961152,
133155495, 390500000, 1017681269}
%t A167925 Clear[m, a, n, v];
%t A167925 m = {{a, 1}, {-1, 1}};
%t A167925 v[0] := {0, 1};
%t A167925 v[n_] := v[n] = m.v[n - 1];
%t A167925 Table[v[n][[1]], {n, 0, 10}, {a, 0, n}];
%t A167925 Flatten[%]
%K A167925 sign
%O A167925 0,5
%A A167925 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 15 2009
%S A129267 1,1,1,0,1,1,1,1,1,1,1,3,2,1,1,0,2,5,3,1,1,1,2,2,7,4,1,1,1,5,
%T A129267 7,1,9,5,1,1,0,3,12,15,1,11,6,1,1,1,3,3,21,26,4,13,7,1,1
%V A129267
1,1,1,0,1,1,-1,-1,1,1,-1,-3,-2,1,1,0,-2,-5,-3,1,1,1,2,-2,-7,-4,1,1,1,5,
%W A129267 7,-1,-9,-5,1,1,0,3,12,15,1,-11,-6,1,1,-1,-3,3,21,26,4,-13,-7,1,1
%E A129267 New method of calculation: The triangular sequence is a set
of matrix
Markov coefficients for the determinant equals trace 2x2 matrix: m {{a, 1},
{-1, 1}} which is an entirely different and new method of obtaining this
sequence. Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 15 2009
%C A129267 Contribution from Roger L. Bagula
(rlbagulatftn(AT)yahoo.com), Nov 15
2009: (Start)
%C A129267 Row sums are ( for the first row as zero):
%C A129267 {0, 1, 2, 2, 0, -4, -8, -8, 0, 16, 32,...}
%C A129267 This sequence is closely related to A167925. (End)
%e A129267 Contribution from Roger L. Bagula
(rlbagulatftn(AT)yahoo.com), Nov 15
2009: (Start)
%e A129267 {0},
%e A129267 {1},
%e A129267 {1, 1},
%e A129267 {0, 1, 1},
%e A129267 {-1, -1, 1, 1},
%e A129267 {-1, -3, -2, 1, 1},
%e A129267 {0, -2, -5, -3, 1, 1},
%e A129267 {1, 2, -2, -7, -4, 1, 1},
%e A129267 {1, 5, 7, -1, -9, -5, 1, 1},
%e A129267 {0, 3, 12, 15, 1, -11, -6, 1, 1},
%e A129267 {-1, -3, 3, 21, 26, 4, -13, -7, 1, 1} (End)
%t A129267 Contribution from Roger L. Bagula
(rlbagulatftn(AT)yahoo.com), Nov 15
2009: (Start)
%t A129267 Clear[m, a, n, v];
%t A129267 m = {{a, 1}, {-1, 1}};
%t A129267 v[0] := {0, 1};
%t A129267 v[n_] := v[n] = m.v[n - 1];
%t A129267 Table[CoefficientList[v[n][[1]], a], {n, 0, 10}];
%t A129267 Flatten[%] (End)
%Y A129267 A167925 [From Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 15
2009]
Some messing around with 4x4 matrices prompted by
posts in true number theory by Kermit Rose gave rise to
some new thoughts on the problem:
Looking at a two diagonal ( infinite) matrix:
the middle diagonal is:
1/n^s
The second diagonal is 1/(1-1/Prime[n]^s)
with an {0,Infinity} element of one.
Since the
Limit[1/n^s,n->Infinity]=0
Limit[1/(1-1/Prime[n]^s),n->Infinity]=1
The result is:
Trace=Zeta[s]
Determinant=Zeta[s]
CharacteristicPolynomial[M,x]=Product[1/n^s-x,{n,1,Infinity}]+Zeta[s]
That is a very strange infinite polynomial with roots
maybe impossible to find?
Don't even know if it works really.
I invented this two column matrix set to test the
Zeta[s] infinite form that I invented:
%I A168676
%S A168676 1,0,1,1,1,1,2,1,2,1,3,1,3,3,1,4,1,4,6,4,1,5,1,5,10,10,5,1,
%T A168676 6,1,6,15,20,15,6,1,7,1,7,21,35,35,21,7,1,8,1,8,28,56,70,56,
%U A168676 28,8,1,9,1,9,36,84,126,126,84,36,9,1
%V A168676 -1,0,-1,1,-1,1,2,-1,2,-1,3,-1,3,-3,1,4,-1,4,-6,4,-1,5,-1,5,-10,10,-5,1,
%W A168676 6,-1,6,-15,20,-15,6,-1,7,-1,7,-21,35,-35,21,-7,1,8,-1,8,-28,56,-70,56,
%X A168676 -28,8,-1,9,-1,9,-36,84,-126,126,-84,36,-9,1
%N A168676 Coefficients of characteristic polynomials for a two diagonal Matrix type that has determinant equal to trace:M(n)=Table[If[ k ===
m && m < n, 1, If[k == m + 1, 1, If[k == 1 && m == n, (-1)^(n + 1)*(n - 1), 0]]], {k, n}, {m, n}]
%C A168676 This set of matrices was constructed so that the determinant would be equal to the trace.
%C A168676 This system is interesting because the result contains a signed Pascal's triangle.
%C A168676 The initial term is adjusted to {-1} to fit the first column.
%C A168676 Row sums are:{-1, -1, 1, 2, 3, 4, 5, 6, 7, 8, 9,...}.
%e A168676 {-1},
%e A168676 {0, -1},
%e A168676 {1, -1, 1},
%e A168676 {2, -1, 2, -1},
%e A168676 {3, -1, 3, -3, 1},
%e A168676 {4, -1, 4, -6, 4, -1},
%e A168676 {5, -1, 5, -10, 10, -5, 1},
%e A168676 {6, -1, 6, -15, 20, -15, 6, -1},
%e A168676 {7, -1, 7, -21, 35, -35, 21, -7, 1},
%e A168676 {8, -1, 8, -28, 56, -70, 56, -28, 8, -1},
%e A168676 {9, -1, 9, -36, 84, -126, 126, -84, 36, -9, 1}
%t A168676 Clear[M, n, m, k]
%t A168676 M[n_] := Table[If[ k == m && m < n, 1, If[k == m + 1, 1, If[k == 1 && m == n, (-1)^(n + 1)*(n - 1), 0]]], {k, n}, {m,n}]
%t A168676 TableForm[Table[M[n], {n, 1, 10}]]
%t A168676 Table[Det[M[n]], {n, 1, 10}]
%t A168676 Table[Sum[M[n][[k, k]], {k, 1, n}], {n, 1, 10}]
%t A168676 a = Join[{{-1}}, Table[CoefficientList[CharacteristicPolynomial[M[n], x], x], {n, 1, 10}]]
%t A168676 Flatten[a]
%K A168676 nonn
%O A168676 0,7
%A A168676 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 02 2009
Getting this to work was a little hard:
%I A169593
%S A169593 1,1,1,1,1,1,2,1,2,1,6,4,9,6,1,24,121,264,166,24,1,120,44616,
%T A169593 93340,52950,4345,120,1,720,296321796,605003244,321204409,12686988,
%U A169593 164746,720,1,5040,49349521382400,99624831647040,51206316902496
%V A169593 -1,1,-1,1,-1,1,2,-1,2,-1,6,-4,9,-6,1,24,-121,264,-166,24,-1,120,-44616,
%W A169593 93340,-52950,4345,-120,1,720,-296321796,605003244,-321204409,12686988,
%X A169593 -164746,720,-1,5040,-49349521382400,99624831647040,-51206316902496
%N A169593 Coefficients of characteristic polynomials of determinant equals trace matrices using Eulerian trace and factorial determinant.
%C A169593 Row sums are:
%C A169593 {-1, 0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880,...}.
%C A169593 Traces are:
%C A169593 Table[Sum[M[n][[k, k]], {k, 1, n}], {n, 1, 10}]
%C A169593 {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880,...}
%C A169593 Determinants are:
%C A169593 Table[Det[M[n]], {n, 1, 10}]
%C A169593 {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880,...}
%e A169593{-1}, %e A169593 {1, -1}, %e A169593 {1, -1, 1}, %e A169593 {2, -1, 2, - 1},
%e A169593 {6, -4, 9, -6, 1}, %e A169593 {24, -121, 264, -166, 24, -1}, %e A169593 {120, -44616, 93340, -52950, 4345, -120, 1}, %e A169593 {720, -296321796, 605003244, -321204409, 12686988, -164746, 720, -1}, %e A169593 {5040, -49349521382400, 99624831647040, -51206316902496, 936232732785, -5234439280, 8349390, -5040, 1}, %e A169593 {40320, -274297679317746753201, 550979304410032093440, -279071418382631643820, 2395643740989790080, -5853386029582998, 2935936463360, -550407180, 40320, -1}, %e A169593 {362880, -65390418299618584017607843840, 131052209744019041903924775936, -65933467896655077725833452000, 271979805136698554372590800, -303398293776695489224080, 45414063262861346088, -2365327872234750, 45644404725, -362880, 1} %t A169593 Clear[M, n, m, k, a0, b]
%t A169593 (*Eulerian number expansion*)
%t A169593 p[t_] = (1 - x)/(1 - x*Exp[t*(1 - x)])
%t A169593 b = Table[ CoefficientList[ FullSimplify[ ExpandAll[(n!/ x)*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]]], x], {n, 1, 10}]
%t A169593 (* using the Eulerian numbers down the main diagonal since their sum is a factorial*)
%t A169593 M[n_] := Table[If[ k == m && m < n, b[[n - 1]][[ k]], If[k == m + 1, 1, If[k == 1 && m == n, (-1)^(n + 1)*( n - 1)!, 0]]], {k, n}, {m, n}]
%t A169593 TableForm[Table[M[n], {n, 1, 10}]]
%t A169593 Table[Det[M[n]], {n, 1, 10}]
%t A169593 Table[Sum[M[n][[k, k]], {k, 1, n}], {n, 1, 10}]
%t A169593 a = Join[{{-1}}, Table[CoefficientList[CharacteristicPolynomial[M[n], x], x], {n, 1, 10}]]
%t A169593 Flatten[a] %K A169593 sign
%O A169593 0,7
%A A169593 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 02 2009
So I haven't yet any general form
for the determinant equals trace problem,
I have developed a new matrix type that is useful.
--
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
alternative email: roger.bagula at gmail.com