Number Theory: An Introduction to Mathematics

4 Some Matrix Theory 239

It should be noted that, if U is any orthogonal matrix such thatUtHU =
diag[λ 1 ,...,λn] then, sinceUUt=I,thecolumnsx 1 ,...,xnofUsatisfy

Hxj=λjxj( 1 ≤j≤n).

That is,λjis aneigenvalueofHandxja correspondingeigenvector( 1 ≤j≤n).
A real symmetric matrixAispositive definiteifxtAx>0 for every real vector
x =0(andpositive semi-definiteifxtAx≥0 for every real vectorxwith equal-
ity for somex =0). It follows from Theorem 10 that two real symmetric matrices
can be simultaneously diagonalized, if one of them is positive definite, although the
transforming matrix may not be orthogonal:

Proposition 11If A and B are n×n real symmetric matrices, with A positive definite,
then there exists an n×n nonsingular real matrix T such that TtAT and TtBT are
both diagonal matrices.

Proof By Theorem 10, there exists a real orthogonal matrixUsuch thatUtAUis a
diagonal matrix:

UtAU=diag[λ 1 ,...,λn].

Moreover,λj> 0 ( 1 ≤j≤n),sinceAis positive definite. Hence there existsδj> 0
such thatδ^2 j= 1 /λj.IfD=diag[δ 1 ,...,δn], thenDtUtAU D=I. By Theorem 10
again, there exists a real orthogonal matrixVsuch that

Vt(DtUtBU D)V=diag[μ 1 ,...,μn]

is a diagonal matrix. Hence we can takeT=UDV.

Proposition 11 will now be used to obtain an inequality due to Fischer (1908):

Proposition 12If G is a positive definite real symmetric matrix, and if

G=

[

G 1 G 2

Gt 2 G 3

]

is any partition of G, then

detG≤detG 1 ·detG 3 ,

with equality if and only if G 2 = 0.

Proof SinceG 3 is also positive definite, we can writeG=QtHQ,where

Q=

[

I 0

G− 31 Gt 2 I

]

, H=

[

H 1 0

0 G 3

]

,

andH 1 =G 1 −G 2 G− 31 Gt 2 .SincedetG=detH 1 ·detG 3 , we need only show that
detH 1 ≤detG 1 , with equality only ifG 2 =0.
SinceG 1 andH 1 are both positive definite, they can be simultaneously diagonal-
ized.Thus,ifG 1 andH 1 arep×pmatrices, there exists a nonsingular real matrixT