Advanced book on Mathematics Olympiad

Algebra 425

∑n

k= 1

cos(m±j)kα=

(− 1 )m+j+^1

2

−

1

2

.

It follows that form =j, the(m, j )entry of the matrixA^2 is zero. HenceA^2 is a diagonal
matrix with nonzero diagonal entries. This shows thatA^2 is invertible, and so isA.

Remark.This matrix appears in topological quantum field theory. A matrix of this type is
used in the discrete Fourier transform, which has found applications to the JPEG encoding
of digital photography.

226.IfA+iBis invertible, then so isA†−iB†. Let us multiply these two matrices:

(A†−iB†)(A+iB)=A†A+B†B+i(A†B−B†A).

We have

〈(A†A+B†B+i(A†B−B†A))v, v〉

=〈A†Av, v〉+〈B†Bv,v〉+〈i(A†B−B†A)v, v〉

=||Av||^2 +||Bv||^2 +〈i(A†B−B†A)v, v〉,

which is strictly greater than zero for any vectorv =0. This shows that the product
(A†−iB†)(A+iB)is apositive definitematrix (i.e.,〈(A†−iB†)(A+iB)v, v〉> 0
for allv =0). The linear transformation that it defines is therefore injective, hence an
isomorphism. This implies that(A†−iB†)(A+iB)is invertible, and so(A+iB)itself
is invertible.

227.First solution: The fact thatA−Inis invertible follows from the spectral mapping
theorem. To find its inverse, we recall the identity

1 +x+x^2 +···+xk=

xk+^1 − 1

x− 1

,

which by differentiation gives

1 + 2 x+···+kxk−^1 =

kxk+^1 −(k+ 1 )xk+ 1

(x− 1 )^2

.

SubstitutingAforx, we obtain

(A−In)^2 (In+ 2 A+···+kAk−^1 )=kAk+^1 −(k+ 1 )Ak+In=In.

Hence

(A−In)−^1 =(A−In)(In+ 2 A+···+kAk−^1 ).