

A174997


Integer part of the greatest eigenvalues of the matrix n X n whose elements are the Fibonacci numbers F(n) (A000045) such that n X n = ((F(0),F(1),...,F(n1)),(F(n),F(n+1),...,F(2n1)),...,(F(n(n1)),F(n(n1)+1),...,F(n^21))), for n=1,2,...


1



0, 2, 24, 670, 49104, 9556612, 4912086816, 6644887923672, 23608681537374780, 220028639470801004558, 5375052124451092722363120, 344018604775369204515020274376, 57670543415219994487318191998268528
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OFFSET

0,2


COMMENTS

We consider an important property of the matrix n X n whose elements are the Fibonacci numbers, because this matrix has only two eigenvalues of the form lambda1 = a + b*sqrt(c) > 0, lambda2 = a  b*sqrt(c) < 0 with the interesting property: lambda1 + lambda2 = floor(lambda1). The other eigenvalues are = 0 with the multiplicity = n2. Consequently, floor(lambda1) = trace (n X n) if n >=3, (trace (n X n) is defined by the sum of the elements on the main diagonal). This property is checked with a computer for a large values of n using two methods: calculation the trace of n X n and the eigenvalues with the maple instruction "eigenvals".
Property: det(n X n) = 0 if n > = 3. Proof: we consider every 2 X 2 matrix included in the n X n matrix, and we show that det((F(n),F(n+1)), F(n+k),F(n+k+1)) = F(k) if k is even, and  F(k) if k is odd. Examples: With n=3, k=8, det((F(3),F(4)), F(11),F(12)) = F(8) = 21 = det((2,3), (89,144)); with n=6, k=11, det((F(6),F(7)), F(17),F(18)) = F(11) = 89 = det((8,13), (1597,2584)); replacing F(n) = (phi1^n  phi2^n) /sqrt(5) (with phi1 = (1+sqrt(5))/2 and phi2 = (1sqrt(5))/2) in the preceding expression is sufficient to prove the assertion. Now, if we consider the 3 X 3 matrix = ((F(0),F(1),F(2)), (F(3),F(4),F(5)),(F(6),F(7),F(8)), we show immediately that det(3 X 3) = 0. We have k=3, and det(3 X 3)= F(1)*F(3) + F(2)*F(3) = 0. Step by step we can prove that det(n X n) = 0.
Fisher proves that the determinant is zero for n>2.


REFERENCES

G. H. Golub and C. F. van Loan, Matrix Computations, Johns Hopkins, 1989, p. 336.
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.


LINKS



FORMULA

a(n) = trace(n X n) for n > = 3.


EXAMPLE

(with notation Maple): n = 1, eigenvals(matrix(1,1, [[0]])) = 0 and a(1) = 0 ; n = 2, eigenvals(matrix(2,2, [[0,1],[1,2]]) = {1+sqrt(2), 1sqrt(2)} and floor(1+sqrt(2)) = 2 and a(2) = 2 ; n = 3, eigenvals(matrix(3,3, [[0,1,1],[2,3,5],[8,13,21]]))= {0, 12+2*sqrt(39), 122*sqrt(39)} = {0, 24.48999599679679679641,  0.48999599679679679641} and floor(lambda1) = lambda1 + lambda2 = a(3) = 24 ; n = 4, the eigenvalues are 0,0, 670.4340471687, 0.43404716873926 => a(4); n = 5, the eigenvalues are 0,0,0, 49104.4667599815,  0. 46675998157216 => a(5); n = 6, the eigenvalues are 0,0,0,0, 9556612.4358632337,  0.43586323375096 => a(6); n = 7, the eigenvalues are 0,0,0,0,0, 4912086816.45572053826,  0.45572053826 => a(7).


MAPLE

T:=array(0..20000):u0:=0:u1:=1: T[0]:=0:T[1]:=1:for p from 2 to 10000 do:s:=u0+u1:u0:=u1:u1:=s: T[p]:=s:od:for n from 2 to 20 do:som:=0:for k from 1 by n+1 to n^2 do:som:=som + T[k1]:od:print(som):od:


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



