Page 133 - Textos de Matemática Vol. 44

LARGEST EIGENVALUE 123
Assume now that A gives equality in (2.1). The first inequality in (2.3), to reduce to equality, states that
a2ijx2i =a2ijx2j
foralli,j=1,...,n.Henceaij =0orxi =xj.Lety1 >···>ys(≥0)bethe distinct values of x1,...,xn, and let yt appear for kt times. Denoting by y(k) that y repeats for k times, let P be a permutation matrix such that
Then
Px=(y(k1) ...y(ks))T. 1s
B1 O ... O
TO B2 ... O A=P  . . . . P,
.... O O ... Bs
where each Bi is a nonnegative symmetric ki × ki matrix with zero diagonal. Sincethevectory=(y(k1))T ̸=0satisfiesB y=λy,wehaveλ(B )=λ.But
111 λ=λ(B1)≤ 1 −1+ 1+4∥B1∥2
2
s
1
≤2 −1+ 1+4 ∥Bi∥2 i=1
= 1 −1+ 1+4∥A∥2 =λ, 2
and so B2, . . . , Bs are zero matrices. Hence, denoting B = B1, T B O T˜
(2.5) A=P OOP=PBP.
Next, consider the second inequality in (2.3). To get equality, we must have
a2ijxixj = aijxixj
for all i,j = 1,...,n. If aij > 0, then, by (2.5), xi = xj = y1(> 0), and so aij = 1. We have thus proved that, to attain equality throughout in (2.3), A is necessarily of the form (2.5), where B is a symmetric k×k matrix (2 ≤ k ≤ n) with entries zero or one and diagonal zero.
In studying additional conditions required to equality in (2.2), we can, in- steadofA,considerB˜.Thenx=(y0)T wherey=(1(k))T/√k.Ifi>k,then everything is zero in (2.2), and so equality holds. Next let i = 1. The first row vector of B˜ (considered as a column vector) and the vector x(1) must be linearly