13.9. AN APPLICATION TO STATISTICS 355

and so B commutes with every matrix which commutes with A. To see this, suppose CA =AC, then

BC = p(A)C =Cp(A) = B

This shows that if B is such a square root, then it commutes with every matrix C whichcommutes with A. It also shows, by a repeat of the argument 13.15 - 13.16 that B2 = A.

Could there be another such Hermitian square root which has all nonnegative eigen-values? It was just shown that any such square root commutes with every matrix whichcommutes with A. Suppose B1 is another square root which is self adjoint, and has nonneg-ative eignevalues. Since both B,B1 are nonnegative,

(B(B−B1)x,(B−B1)x)≥ 0,

(B1 (B−B1)x,(B−B1)x)≥ 0 (13.18)

Now, adding these together, and using the fact that the two commute because they bothcommute with every matrix which commutes with A,

((B+B1)(B−B1)x,(B−B1)x)≥ 0((B2−B2

1)x,(B−B1)x

)= ((A−A)x,(B−B1)x) = 0.

It follows that both inner products in 13.18 equal 0. Next use the existence part shownabove to take the square root of B and B1 which is denoted by

√B,√

B1 respectively. Then

0 =(√

B(B−B1)x,√

B(B−B1)x)

0 =(√

B1 (B−B1)x,√

B1 (B−B1)x)

which implies√

B(B−B1)x=√

B1 (B−B1)x= 0. Thus also,

B(B−B1)x= B1 (B−B1)x= 0

Hence0 = (B(B−B1)x−B1 (B−B1)x,x) = ((B−B1)x,(B−B1)x)

and so, since x is arbitrary, B1 = B. ■

Corollary 13.8.2 The U in Theorem 13.7.5 is unique.

13.9 An Application To StatisticsA random vector is a function X : Ω→Rp where Ω is a probability space. This means thatthere exists a σ algebra of measurable sets F and a probability measure P : F → [0,1].In practice, people often don’t worry too much about the underlying probability space andinstead pay more attention to the distribution measure of the random variable. For E asuitable subset of Rp, this measure gives the probability that X has values in E. Thereare often excellent reasons for believing that a random vector is normally distributed. Thismeans that the probability that X has values in a set E is given by∫

E

1

(2π)p/2 det(Σ)1/2 exp(−1

2(x−m)∗Σ

−1 (x−m)

)dx

13.9. AN APPLICATION TO STATISTICS 355and so B commutes with every matrix which commutes with A. To see this, suppose CA =AC, thenBC = p(A)C=Cp(A)=BThis shows that if B is such a square root, then it commutes with every matrix C whichcommutes with A. It also shows, by a repeat of the argument 13.15 - 13.16 that B* = A.Could there be another such Hermitian square root which has all nonnegative eigen-values? It was just shown that any such square root commutes with every matrix whichcommutes with A. Suppose B, is another square root which is self adjoint, and has nonneg-ative eignevalues. Since both B, B; are nonnegative,(B(B—B,)x,(B—B))x) > 0,(B| (B—B,)x,(B—B,)x) >0 (13.18)Now, adding these together, and using the fact that the two commute because they bothcommute with every matrix which commutes with A,((B+B1)(B—Bi)x,(B—B)a) > 0((B? — By) x, (B—B))x) = ((A—A) a, (B—B)) x) =0.It follows that both inner products in 13.18 equal 0. Next use the existence part shownabove to take the square root of B and B; which is denoted by WB, \/B| respectively. Then0 = (VB(B—B))x, VB(B-B,)z)0 = (VBi(B-B;)«, VB) (B—B))x)which implies VB (B—B,) x = \/B, (B—B,)x =0. Thus also,B(B—B,)« = B,(B—B,)a=0HenceO= (B(B—B,)x—B, (B—B,)x,2) = ((B—B,)x,(B—B,)x)and so, since x is arbitrary, Bj = B. @Corollary 13.8.2 The U in Theorem 13.7.5 is unique.13.9 An Application To StatisticsA random vector is a function X :Q— R? where Q is a probability space. This means thatthere exists a o algebra of measurable sets Y and a probability measure P : ¥ — (0, 1].In practice, people often don’t worry too much about the underlying probability space andinstead pay more attention to the distribution measure of the random variable. For E asuitable subset of IR’, this measure gives the probability that X has values in E. Thereare often excellent reasons for believing that a random vector is normally distributed. Thismeans that the probability that X has values in a set E is given bytg _! x—m)*x!(a—m) |) dxhomtaaw P( 2 yen i)a