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Problem 10958

(American Mathematical Monthly, Vol.109, August-September 2002) Proposed by R. Chapman (UK).

Let An, be the n-by-n0, 1-matrix with 1s in exactly those positions (j, k) such that n ≤ j +k ≤ n+1.

Find the eigenvalues of An.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let Pn be the characteristic polynomial of An

Pn(x) = det (xIn−An)

Expanding the determinant with respect to the last row of the matrix, it is simple to verify that the sequence of polynomials Pn satisfy the linear recurrence equation:

Pn(x) = x · Pn1(x) − Pn2(x) with P0(x) = 1, P1(x) = x − 1.

By linearity, Tn:= Pn+ Pn1is another solution of the same recurrence equation but with different initial conditions:

Tn(x) = x · Tn1(x) − Tn2(x) with T0(x) = 2, T1(x) = x.

Note that Tn is just the Tchebycheff polynomial of order n and each Pn can be written in the following form

Pn(x) = Tn(x) − Pn1(x) = · · · = (−1)n·

" n

X

k=1

(−1)kTk(x) + P0(x)

# .

Since Tn(2 cos θ) = 2 cos(nθ), then

Pn(2 cos θ) = 2 · (−1)n·

" n

X

k=1

(−1)kcos(kθ) +1 2

# .

By the definition of the Dirichlet kernel δn(θ)

δn(θ) := sin(n +12)θ 2π sin12θ = 1

π ·

" n

X

k=1

cos(kθ) +1 2

#

hence it follows that

" n

X

k=1

(−1)kcos(kθ) + 1 2

#

= π · δn(θ + π) = sin(n +12)(θ + π)

2 sin12(θ + π) = (−1)n· cos(n +12)θ 2 cos12θ . Therefore

Pn(2 cos θ) = cos(n +12)θ cos12θ . Now it is easy to find the zeros of Pn (i. e. the eigenvalues of An):

xk= 2 cos θk= 2 cos2k + 1

2n + 1π for k = 0 . . . n − 1.



Riferimenti

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