Advanced book on Mathematics Olympiad

Algebra 415

we see that we could almost use the equality from the statement, but the factors in two
terms come in the wrong order. Another property of the trace comes to the rescue, namely,
tr(XY )=tr(Y X). We thus have

tr(AAt+BtB)−tr(AB+BtAt)=tr(AAt)+tr(BtB)−tr(AB)−tr(BtAt)

=tr(AAt)+tr(BBt)−tr(AB)−tr(AtBt)= 0.

It follows that tr[(A−Bt)(A−Bt)t]=0, which impliesA−Bt=On, as desired.

Remark.The Hilbert–Schmidt norm plays an important role in the study of linear trans-
formations of infinite-dimensional spaces. It was first considered by E. Schmidt in his
study of integral equations of the form

f(x)−

∫b

a

K(x, y)f (y)dy=g(x).

Here the linear transformation (which is a kind of infinite-dimensional matrix) is

f(x)→

∫b

a

K(x, y)f (y)dy,

and its Hilbert–Schmidt norm is
(∫b

a

∫b

a

|K(x, y)|^2 dxdy

) 1 / 2

.

For a (finite- or infinite-dimensional) diagonal matrixD, whose diagonal elements are
d 1 ,d 2 ,··· ∈C, the Hilbert–Schmidt norm is
√
trDDt=(|d 1 |^2 +|d 2 |^2 +···)^1 /^2.

206.The elegant solution is based on the equality of matrices
⎛
⎝

(x^2 + 1 )^2 (xy+ 1 )^2 (xz+ 1 )^2

(xy+ 1 )^2 (y^2 + 1 )^2 (yz+ 1 )^2

(xz+ 1 )^2 (yz+ 1 )^2 (z^2 + 1 )^2

⎞

⎠=

⎛

⎝

1 xx^2

1 yy^2

1 zz^2

⎞

⎠

⎛

⎝

111

2 x 2 y 2 z

x^2 y^2 z^2

⎞

⎠.

Passing to determinants and factoring a 2, we obtain a product of two Vandermonde
determinants, hence the formula from the statement.
(C. Co ̧sni ̧t ̆a, F. Turtoiu,Probleme de Algebr ̆a(Problems in Algebra), Editura Tehnic ̆a,
Bucharest, 1972)

207.Consider the matrix

M=

(

11

10

)

,