822 CHAPTER 39. STATISTICAL TESTS

(A2−A

)vk = 0−0 = 0. Thus A2−A = 0 because this matrix sends every vector in a

basis to 0.Conversely, suppose A2 = A. Why are all eigenvalues 1 or 0? Say Av = λv and say

λ ̸= 0. Then for each v an eigenvector, A2v = Av = λAv and so A(1−λ )v = 0. If λ ̸= 1,then Av = 0 which is assumed not to be so. Hence λ = 1. Thus all eigenvalues are either0 or 1. ■

This proves the following interesting theorem.

Theorem 39.4.5 Let X1, · · · ,Xn be independent and n(0,σ2

). Let A be a real symmetric

matrix. Then for X =(

X1 · · · Xn

)T, X

T AXσ2 is X 2 (r) for some r ≤ n if and only if

A2 = A if and only if the eigenvalues of A are 0 or 1. In fact, r is the rank of A.

At this point, it might be good to recall that the distribution of

nS2/σ2 ≡

n

∑k=1

(Xk− X̄)2

σ2

is X 2 (n−1) where

X̄ =1n

n

∑k=1

Xk

In showing this, first there was some algebra.

X 2(n)︷ ︸︸ ︷n

∑k=1

(Xk−µ)2

σ2 =n

∑k=1

((Xk− X̄)+(X̄−µ))2

σ2

After some simple manipulations,

=n

∑k=1

(Xk− X̄)2

σ2 +n

∑k=1

(X̄−µ)2

σ2 =nS2

σ2 +n

∑k=1

(1n ∑

nj=1 (Xk−µ)

)2

σ2

=nS2

σ2 +n

∑k=1

(1n

n

∑j=1

(Xk−µ)

σ

)2

=

X 2(1)︷ ︸︸ ︷

n

∑j=1

(Xk−µ)√nσ

2

+nS2

σ2

Then it was proved that the two random variables at the end are independent. This wasdone by using the special form of the normal distribution. Then from this, we obtained onlooking at the moment generating functions,

(1

1−2t

)n/2

= E(

exp(

tnS2

σ2

))E

exp

t

(n

∑j=1

(Xk−µ)√nσ

)2

= E(

exp(

tnS2

σ2

))1

(1−2t)1/2