424 CHAPTER 19. EIGENVALUES AND EIGENVECTORS
where A1 is an (p−1)× (p−1) matrix. Summarizing, there is an orthogonal matrix Usuch that
UT AU =
(λ aT
0 A1
)I claim that a= 0. To see this, take the transpose of both sides, using symmetry of A toobtain (
λ aT
0 A1
)=UT AU =
(UT AU
)T=
(λ 0T
a A1
)Thus a= 0 as claimed. Now by induction, there is an orthogonal matrix Û such that
ÛT A1Û = D
where D is a diagonal matrix. Now note that from the way we multiply matrices,(1 0T
0 Û
)T ( 1 0T
0 Û
)=
(1 0T
0 ÛT
)(1 0T
0 Û
)
=
(1 0T
0 ÛT
)(1 0T
0 Û
)=
(1 0T
0 ÛTÛ
)=
(1 0T
0 I
)= I
Thus (1 0T
0 Û
)≡ Ũ
is an orthogonal matrix. Now
ŨTUT AUŨ = ŨT(
λ 0T
0 A1
)Ũ
=
(1 0T
0 ÛT
)(λ 0T
0 A1
)(1 0T
0 Û
)
=
(λ 0T
0 ÛT A1Û
)=
(λ 0T
0 D
)which is a diagonal matrix. This shows the first part. Now if
U =(u1 u2 · · · up
)D =
λ 1 0. . .
0 λ p
, UT AU = D
then, multiplying on both sides by U,
AU =UD
and so, from the way we multiply matrices, this yields
AU =(
Au1 Au2 · · · Aup)=UD
=(
λ 1u1 λ 2u2 · · · λ pup)
which shows that Au j = λ ju j for each j. This shows the columns of U form an orthonor-mal set of eigenvectors and the diagonal entries of D are the eigenvalues of A.