Problem 11008
(American Mathematical Monthly, Vol.110, April 2003) Proposed by J. L. D´ıaz-Barrero and J. J. Egozcue (Spain).
Let A(z) =Pn
k=0akzk be a monic polynomial with complex coefficients and with zeros z1, . . . , zn. Prove that
1 n
n
X
k=1
|zk|2<1 + max
1≤k≤n|an−k|2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will prove that
n
X
k=1
|zk|2≤ n − 1 +
n
X
k=1
|an−k|2,
then the required inequality follows easily:
1 n
n
X
k=1
|zk|2≤ 1 − 1 n+ 1
n
n
X
k=1
|an−k|2<1 + 1 n
n
X
k=1
|an−k|2≤ 1 + max
1≤k≤n|an−k|2. Let M be the n × n companion matrix for the monic polynomial A(z)
M =
0 0 · · · 0 −a0 1 0 · · · 0 −a1 0 1 · · · 0 −a2 ... ... . .. ... ... 0 0 · · · 1 −an−1
.
By Schur’s theorem, there exist an upper triangular matrix T and a unitary matrix U such that M = U T U∗. Recall that the Frobenius norm of a n × n complex matrix C = [cij] is defined as kCkF =q
Pn
i,j=1|cij|2). Since this norm is invariant with respect to the multiplication by a unitary matrix we have that
kM kF = kU T U∗kF = kU T kF = kT kF.
Note that the main diagonal of T consists of the eigenvalues of M which are the zeroes z1, . . . , zn of its characteristic polynomial A(z). Therefore, computing the norms,
n
X
k=1
|zk|2≤ kT k2F = kM k2F = n − 1 +
n
X
k=1
|an−k|2.
