A First Course in Linear Algebra, 2017a

412 Spectral Theory
where λ 1 ,λ 2 ,...,λ n are the (not necessarily distinct) eigenvalues of A. Let⃗x ∈ R n , ⃗x ≠⃗0, and define
⃗y = U T ⃗x. Then
⃗x T A⃗x =⃗x T (UDU T )⃗x =(⃗x T U)D(U T ⃗x)=⃗y T D⃗y.
Writing⃗y T = [ y 1 y 2 ··· y n
]
,
⃗x T A⃗x = [ ] y 1 y 2 ··· y n diag(λ1 ,λ 2 ,...,λ n ) ⎢
⎣
= λ 1 y 2 1 + λ 2y 2 2 + ···λ ny 2 n .
⎡
⎤
y 1
y 2
⎥
. ⎦
y n
(⇒) First we will assume that A is positive definite and prove that⃗x T A⃗x is positive.
Suppose A is positive definite, and⃗x ∈ R n ,⃗x ≠⃗0. Since U T is invertible,⃗y = U T ⃗x ≠⃗0, and thus y j ≠ 0
for some j, implying y 2 j > 0forsome j. Furthermore, since all eigenvalues of A are positive, λ iy 2 i ≥ 0for
all i and λ j y 2 j > 0. Therefore,⃗xT A⃗x > 0.
(⇐) Now we will assume⃗x T A⃗x is positive and show that A is positive definite.
If ⃗x T A⃗x > 0 whenever ⃗x ≠⃗0, choose ⃗x = U⃗e j ,where⃗e j is the j th column of I n .SinceU is invertible,
⃗x ≠⃗0, and thus
⃗y = U T ⃗x = U T (U⃗e j )=⃗e j .
Thus y j = 1andy i = 0wheni ≠ j, so
λ 1 y 2 1 + λ 2y 2 2 + ···λ ny 2 n = λ j,
i.e., λ j =⃗x T A⃗x > 0. Therefore, A is positive definite.
♠
There are some other very interesting consequences which result from a matrix being positive definite.
First one can note that the property of being positive definite is transferred to each of the principal
submatrices which we will now define.
Definition 7.75: The Submatrix A k
Let A be an n×n matrix. Denote by A k the k×k matrix obtained by deleting the k+1,···,n columns
and the k + 1,···,n rows from A. Thus A n = A and A k is the k × k submatrix of A which occupies
the upper left corner of A.
Lemma 7.76: Positive Definite and Submatrices
Let A be an n × n positive definite matrix. Then each submatrix A k is also positive definite.
Proof. This follows right away from the above definition. Let⃗x ∈ R k be nonzero. Then
⃗x T A k ⃗x = [ ⃗x T 0 ] [ ] ⃗x
A > 0
0