MATH 38061/MATH48061/MATH68061: MULTIVARIATE STATISTICS Solutios to Poblems o Matix Algeba 1 Let A be a squae diagoal matix takig the fom a 11 0 0 0 a 22 0 A 0 0 a pp The ad So, log det A t log A t log A t p log det A log a ii log a ii log a ii log a 11 0 0 0 log a 22 0 0 0 log a pp 2 Let A be ay squae o-sigula matix By the spectal decompositio, we ca wite A λ i e i e T i So, ad p log det A log λ i log λ i t log A t log λ i e i e T i log λ i t e i e T i log λ i t e T i e i log λ i 1
Hece, log det A t log A 3 Suppose a ad x ae p 1 vectos By defiitio, a T x/ x 1 a T a T x/ x 2 x/ x a T x/ x p Sice a T x a k x k, k1 we have a T p x / x i k1 a k x k / x i a i fo i 1, 2,, p So, a T x/ x a 1, a 2,, a p T a 4 Suppose x is a p 1 vecto ad A a p p symmetic matix By defiitio, x T Ax/ x 1 x T x T Ax/ x 2 Ax/ x x T Ax/ x p Sice x T Ax k1 l1 a kl x k x l a kk x 2 k + k1 k,l1,k l a kl x k x l, we have p x T Ax / x i a kl x k x l / x i k1 l1 a kk x 2 k + a kl x k x l / x i k1 k,l1,k l 2a ii x i + a il x l + a li x l l1,l i l1,l i 2a ii x i + 2 a il x l l1,l i 2 a il x l l1 2
fo i 1, 2,, p So, x T Ax/ x 2 a 1l x l l1 2 a 2l x l l1 2 a pl x l l1 2 a 1l x l l1 a 2l x l l1 a pl x l l1 2Ax 5 Let be the covaiace matix about x a The i we have Sa 1 x i a x i a T Sa 1 x i x + x a x i x + x a T 1 x i x x i x T + x i x x a T + x a x i x T + x a x a T 1 x i x x i x T + x i x x a T + x a x i x T + x a x a T 1 1S + 0 + 0 + x a x a T 1 S + x a x at ii usig equatio 56 i the otes, we have Sa 1 S + x a x at { } 1S/ 1 + x a T / 1S 1 x a iii we have mi Sa a 1S/ mi 1S/ a { with the miimum attaied whe a x } 1 + x a T / 1S 1 x a 3
iv we have so tacesa tace 1 S + tace x a x at 1 taces + x at x a, with the miimum attaied whe a x 6 Assume A 1 ad B 1 exist 7 Let mi tace a Sa 1/tS i The A T 1 A 1 T sice A T A T 1 A T A 1 T A 1 A T I T I ii The AB 1 B 1 A 1 sice ABAB 1 ABB 1 A 1 AIA 1 AA 1 I ad The D 2 s x x s T S 1 x x s, g s x x T S 1 x s x q s x T S 1 x s i we have D 2 s x T x T s S 1 x x s x T S 1 x + x T s S 1 x s x T S 1 x s x T s S 1 x x T S 1 x + x T s S 1 x s 2x T S 1 x s q + q ss 2q s ii we have g s x T x T S 1 x s x x T S 1 x s + x T S 1 x x T S 1 x x T s S 1 x q s + x T S 1 x x T S 1 x x T s S 1 x iii we have D 2 s x x + x x s T S 1 x x + x x s x x x s x T S 1 x x x s x x x T x s x T S 1 x x x s x x x T S 1 x x + x s x T S 1 x s x x x T S 1 x s x x s x T S 1 x x x x T S 1 x x + x s x T S 1 x s x 2 x x T S 1 x s x g + g ss 2g s 4
iv we have v we have g s x x T S 1 x s x T x x S 1 x s x 0 g x x T S 1 x x tace x x T S 1 x x tace S 1 x x x x T tace S 1 x x x x T tace S 1 1S 1taceI 1p vi we have Ds 2 g + g ss 2g s g + g ss 2 g s 1p + g ss + 0 1p + g ss vii we have Ds 2 s 1p + g ss s 1p + 1p 2 1p 8 Let X have the covaiace matix Σ 4 0 0 0 9 0 0 0 1 The Σ 1 1/4 0 0 0 1/9 0 0 0 1 5
The eigevalues of Σ ae 4, 9, 1 The coespodig eigevectos ae 1, 0, 0, 0, 1, 0, 0, 0, 1 The eigevalues of Σ 1 ae 1/4, 1/9, 1 The coespodig eigevectos ae the same as those fo Σ 9 A quadatic fom x T Ax is said to be positive-defiite if the matix A is positive-defiite Note that we ca wite 3x 2 1 + 3x 2 2 2x 1 x 2 x 1, x 2 A x 1, x 2 T, whee A 3 1 1 3 The eigevalues of A ae 2, 4 So, A is positive defiite ad so is 3x 2 1 + 3x2 2 2x 1x 2 10 Fo a abitay p matix A we have A T A o-egative defiite sice x T A T Ax Ax T Ax y T y 0 fo all x, whee y Ax 11 Fist fid the eigevalues of A Note that A λi 13 λ 4 2 4 13 λ 2 2 2 10 λ λ 9 2 λ 18 So, the eigevalues of A ae 9, 18 Hece, the maximum ad miimum values of the atio x T Ax/x T x ae 18 ad 9, espectively 12 Usig equatio 25 i the otes, we have A PΛP 1 P 1 PΛ IΛ Λ λ 1 λ 2 λ p 13 A matix Q is othogoal if Q T Q I So, 1 I Q T Q Q Q Q 2, implyig that Q ±1 14 Suppose λ is a eigevalue of a k k positive defiite matix A The Ae λe fo some e So, 0 < e T Ae λe T e, implyig that λ must be geate tha zeo 15 To show that Σ 0 we fid its eigevalues Usig equatio 56 i the otes, we have Σ λi 1 α λi + α11 T 1 α λ p α I + 1 α λ 1 α 1 α λ 1T } 1 α λ {1 p α α + 1 α λ 1T 1 1 α λ } 1 α λ {1 p αp + 1 α λ 1 α λ p 1 1 α λ + αp The eigevalues λ 1 α, 1 α+αp These ae o-egative if ad oly if 1 α 1/p 1 To show that Σ 1 1 α 1 I α {1 + p 1α} 1 11 T 6
ote that ΣΣ 1 1 αi + α11 T 1 α 1 I α {1 + p 1α} 1 11 T 1 α 1 α1 α 1 αi 1 + p 1α 11T + α11 T α 2 1 + p 1α 11T 11 T 1 α 1 α1 α 1 αi 1 + p 1α 11T + α11 T α 2 p 1 + p 1α 11T { 1 α 1 1 α1 α αi + 1 + p 1α + α α 2 } p 11 T 1 + p 1α 1 α 1 1 αi I 16 Let H I 1/J, whee J 11 T i The H T H ad H 2 H sice I 1/J T I T 1/J T I 1/ 11 T T I 1/11 T H ad I 1/J I 1/J I T 1/J 1/JI + 1/ 2 JJ I T 2/J + 1/ 2 11 T 11 T I T 2/J + 1/ 11 T I T 2/J + 1/ J I T 1/J ii The H1 0 ad HJ JH 0 sice H1 I 1/J 1 I1 1/J1 1 1/11 T 1 1 1 0 ad JH J T H T HJ T 0 iii The Hx x x1 sice Hx I 1/J x Ix 1/Jx x 1/11 T x x x1 7
iv The x T Hx x i x 2 sice 17 Coside the covaiace matix: x T Hx x T I 1/J x Σ x T Ix 1/x T Jx x T Ix 1/x T 11 T x x 2 2 i 1/ x i x i x 2 σ 2 σ 2 ρ 0 σ 2 ρ σ 2 σ 2 ρ 0 σ 2 ρ σ 2 Its tace is σ 2 + σ 2 + σ 2 3σ 2 The detemiat is σ 2 σ 2 ρ 0 Σ σ 2 ρ σ 2 σ 2 ρ 0 σ 2 ρ σ 2 σ 6 1 ρ 2 σ 6 ρ 2 σ 6 1 2ρ 2 The ivese is Σ 1 1 σ 6 1 2ρ 2 1 σ 2 1 2ρ 2 σ 4 1 ρ 2 σ 4 ρ σ 4 ρ 2 σ 4 ρ σ 4 σ 4 ρ σ 4 ρ 2 σ 4 ρ σ 4 1 ρ 2 1 ρ 2 ρ ρ 2 ρ 1 ρ ρ 2 ρ 1 ρ 2 The eigevalues ae obtaied by solvig: σ 2 λ σ 2 ρ 0 σ 2 ρ σ 2 λ σ 2 ρ 0 0 σ 2 ρ σ 2 λ 2 σ 2 λ σ 2 λ 2σ 4 ρ 2 0 σ 2 λ σ 2 λ + 2σ 2 ρ σ 2 λ 2σ 2 ρ 0, so the eigevalues ae λ 1 σ 2, λ 2 1 + 2ρσ 2, λ 3 1 2ρσ 2 The eigevecto, say e T 1 x, y, z, coespodig to λ 1 σ 2 is obtaied by solvig: 0 σ 2 ρ 0 x 0 σ 2 ρ 0 σ 2 ρ y 0 0 σ 2 ρ 0 z 0 8
We have y 0 ad x + z 0, so e T 1 1, 0, 1 o et 1 1/ 2, 0, 1/ 2 i stadadized fom The eigevecto, say e T 2 x, y, z, coespodig to λ 2 1+ 2ρσ 2 is obtaied by solvig: 2σ 2 ρ σ 2 ρ 0 σ 2 ρ 2σ 2 ρ σ 2 ρ 0 σ 2 ρ 2σ 2 ρ x y z We have 2x + y 0, y 2z 0 ad x 2y + z 0, so e T 2 e T 2 1/2, 1/ 2, 1/2 i stadadized fom 0 0 0 1/ 2, 1, 1/ 2 o The eigevecto, say e T 3 x, y, z, coespodig to λ 3 1 2ρσ 2 is obtaied by solvig: 2σ 2 ρ σ 2 ρ 0 x 0 σ 2 ρ 2σ 2 ρ σ 2 ρ y 0 0 σ 2 ρ 2σ 2 ρ z 0 We have 2x + y 0, y + 2z 0 ad x + 2y + z 0, so e T 3 e T 3 1/2, 1/ 2, 1/2 i stadadized fom 1/ 2, 1, 1/ 2 o 9