Question

Theorem. Consider the quadratic form Q(x) = Ar where A is anxn symmetric matrix and A, and denote the largest and smallest eigenvalues of A, respectively. Then max Q(x) = 2 = max Q() = 1 and Q0.) = 1, where is any unit eige vector corre sponding to ii) in (r) and QU.) where is any unit eigen vector corresponding to do 1. - Find max Q(x) and min Q(x). 1) Q(1) = 3x + 43273 +673 ii) Q(z) = 4x + 127432 - 1) Q(1) = 2x - 3x3 iv) Q(z) = 2x - 3:3 - 533 v) Q(x) = 3x - 4x3 - 2x + 1* 1. Find max Q(x) andmin Q(x). S = {1: |-|| = 1) 1) (+) == + 47122+4=; ii) Q() -1 + 48172 + 4x2 S - {x: 4x} + 25x - 100) i) Q(x) = *} +21,22 s={: 10)

Answer 1

1. Ano: ( P(x) = 3 x ² + 4x, 72 + 6 x ² Hace A - 1 & The characteristic polynomial of A is given ley- 14-15120 1828310 x² – 9x + 14 =0 (1-7) (7-2) 20 x 27,2 Hence the eigenvalues of A are & andz. max p(x) = max Q(x) = 7 Xxli21 11x11 = 1 and min 1x1 z 1 Q(x) = 2. ( 9 (X) = Но А. 4x² + 12x, 22 – x2 Г. , The characteristic polynomial of A x² – 37 – 40=0 is given ley- (-8) (x+5) 20. 728.-5. Or the eigenvalues & A wee 8 and-s

Hence max 9(x) = 8 and ming(x) 2-5 HX121 Cilis 9(x) = 2x, 2 – 3 x 2 Here Ar 12 of 0 -3 A are 2 and 3. Thus the Hence eigen values of max 9(x) = 2 and minglu) 2-3 11x1121 (N) 8(x) = 2x2 – 3x – 5022 Here A. 1 2 0 07 To -3 o 0 0 g eigenvalues of A is the are 2-3 and 5 Hence max & (x) = I! X1121 7 and min g(x) 2-5. 1x1 z 1 (1) G(x) = 34,² - 4x² – 273 + x Here A = 3 o oo 1o – 4 o o o o 2o L 0 0 0 1 1 This the eigenvalues of A are 3-4,-2, 1 Hence max 9(x) = 3 118111 2 and min 4(x) = - 4 and 11% 11 21 - END .
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