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A8_soln

# A8_soln - Math 235 Assignment 8 Solutions 1 For each...

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Unformatted text preview: Math 235 Assignment 8 Solutions 1. For each quadratic form QUE), determine the corresponding symmetric matrix A. By diagonalizing A, express 62(5) in diagonal form and give an orthogonal matrix that diago— nalizes A. Classify each quadratic form. a) Q(:r, y) = 7m2 +123cy + 123/2. . . __ 7 6 Solution. We have A 2— [6 12 The characteristic equation is 0 2 det(A — AI) 2 A2 —19/\ + 48 2 (A — 3)(/\ ~16). The roots are 3 and 167 so these are the eigenvectors of A. For /\ = 3 we get 46 23 c,_—3 A_a=k9}{042m-[4. —9 6 ——3 2 2 __ 2 Mpg .4240 42.2243}. Normalizing the vectors, we get that the orthogonal matrix which diagonalizes A is P : _3/\/ﬁ 2/13 __ 2 2 m_ T27 [ﬁx/E 3/m’aner3ivi—t15yuwhere M 2P y. Since all the eigenvalues of A are positive, it follows that 62(33, 3/) is positive deﬁnite. _ 7—)\ 6 1soA—AI—[6 12_)\]. For /\ 2 16 we get b) CM, y) = 932 + 69621 — 72/2 Solution: We have A =2 [g 37 The characteristic equation is 0=det(A—/\I)=/\2+6/\—16:(/\—2)(/\+8). The roots are 2 and —8, so these are the eigenvectors of A. For /\ 2: 2 we get 213 1—3 u_3 A_m-[3_42l00]2a2bl 9 3 3 1 _. _ ——1 A+a2klyl0J2a_[J. Normalizing the vectors, we get that the orthogonal matrix which diagonalizes A is P =2 3/\/T6—1/10 -2 2 :17__ :c [—Uxﬁa 3/\/1—5 ,andQ22x1—8y1,where [y:]2PT [y] Since one of the eigenvalues of A are positive and the other is negative, it follows that Q(m, y) is indeﬁnite. JW 3 l- 3 —7—/\ For A : —8 we get 2 c) Q(:z:, y, z) 2 2:172 + 23/2 — 322 — 4583/ + 6% + 6yz. 2 ——2 3 2 — A ——2 3 Solution: We have A 2 ——2 2 3 so A — AI 2 ——2 2 —— A 3 . 3 3 ~3 3 3 —3 — A The characteristic equation is 0 2 det(A —— AI) 2 ~(A — 3)(A — 4)(A + 6). The roots are 3, 4 and —6, so these are the eigenvectors of A. For A 2 3 we get —1 —2 3 1 0 —1 A—3I2 —2 ~13 ~ 01—1 3 3 —6 0 0 0 1 Thus, 1 is a corresponding eigenvector. For A 2 4 we get 1 —2 —2 3 1 1 0 A — 41 2 ——2 ——2 3 N O 0 1 3 3 ——7 0 0 0 ——1 Thus, 1 is a corresponding eigenvector. For A 2 —6 we get 0 1 Thus, 1 is a corresponding eigenvector. —2 Normalizing the vectors7 we get that the orthogonal matrix which diagonalizes A is P 2 1/\/§ —1/\/3 1/\/é , and Q = 2x? + 43;? —— 62%, where yl 2 PT y _ 1N3 0 —2/\/6 21 2 Since A has both positive and negative eigenvalues it follows that Q(\$, y, z) is indeﬁnite. 3 2. Sketch the graph of each of the following equations showing both the original and new axes For any hyperbola, ﬁnd the equation of the asymptotes. a) 7:172 +12xy +12y2 2 48. K‘ a. Solution: The corresponding symmetric matrix is [7 6 ] The eigenvalues are /\1 2 16 and /\2 2 3 with .1 ’1 ‘ El ; 612 corresponding eigenvectors 271 2 [g], 272 2 [—23] Thus we have the ellipse L 165:2 + 33f = 48, with principal axis {71, and 272. This is the equation of 7x1+12zy+123f2‘. an ellipse with ﬁ-intercepts (ix/3,0) and Q—intercepts (0, 3:4). Graphing this gives: b) 3:2 + 6xy — 73/2 2 32. Solution: The corresponding symmetric matrix is [:13 E7] The eigenvalues are A1 : 2 and /\2 = 28 with corresponding eigenvectors 271 2 [if], 212 2 ——1 3 ]. Thus we have the hyperbloa 25:2 — 8g? 2 32, with principal axis 171, and 272. Since this is a hy— perbola we graph the asymptotes. They are when 0 = 2932 — 83;? = (ﬁr: + x/éyJXx/ii — VS?) We have (ﬁnding the inverse of the change of basis matrix by taking the transpose) ill = a [—31 él [El = 7—6 lix—Sil- So, the asymptotes are 0: Vim x/égr— x/i(3x+y)i\/§(:c 23y) = (Nita/2395+ Mimi/ﬂy, (3\/§:l:2\/§) 2\/2::6\/2 hence y 2 an 2> y 2 95, y 2 »—%:c. Graphing this gives the diagram to the right. 4 3. Let Q(:c,y) : :ETAf with A = [a b] and detA gé 0. b c a) Prove that Q is positive deﬁnite if det A > 0 and a > 0. Solution: Let /\1, A2 be the eigenvalues of A. If det A > 0 then ac —— b2 > 0 so a and c must both have the same sign. Thus, c > 0. We know that detA : /\1/\2 and /\1 + /\2 = a + c and so /\1 and /\2 must have the same sign since det A > 0 and we have /\1 + /\2 : a + c > 0 so we must have /\1 and A2 both positive so Q is positive deﬁnite. b) Prove that Q is negative deﬁnite if detA > O and a < 0. Solution: Let /\1, /\2 be the eigenvalues of A. If det A > 0 then ac — b2 > 0 so a and c must both have the same sign. Thus, c < 0. We know that detA = /\1/\2 and /\1 + /\2 = a + c and so /\1 and /\2 must have the same sign since det A > 0 and we have /\1 + /\2 = a + c < 0 so we must have A1 and A2 both negative so Q is negative deﬁnite. c) Prove that Q is indeﬁnite if det A < 0. Solution: Let /\1, /\2 be the eigenvalues of A. If det A < 0 then det A : /\1/\2 < 0 so /\1 and /\2 must have different signs thus Q is indeﬁnite. 4. Let A be an n X n matrix and let sage R“. Deﬁne < f,§>: fTAg’. Prove that < ,> is an inner product on R" if and only if A is a positive deﬁnite, symmetric matrix. Solution: If A is positive deﬁnite, then for all f E R" we have < 55,2? >= 3111455 > 0 and fFAf : 0 if and only if f : 6 < i,g>=ﬂAg=5-Ag=Agz—g=gtAsz=gTA5=< 15> < 21?, aﬁ+b2j >= fAMiZ—l—bﬁ’) = f~(aA21’+bA;J) 2 aﬁAQ‘H—bf-Ag’: a < 55,1? > H? < ivy > Hence, < , > is an inner product. If < ,> is an inner product, then observe that the ij—th entry of A is given by (Alij :< 51759” > where {51, . . . ,En} is the standard basis for R". Then, (Ale =< (53¢?) >:< 52351 >= (Am so A is symmetric. Also, we have fTAa? = (if, if) > O for all f # 6 and hence A is positive deﬁnite. MATH 235 Assignment 8 MATLAB Solutions Quadratic Forms (8) A = [3 1 0 2 —2; 1 0 —3 4 0; A x 3 1 0 2 —2 1 0 ~3 4 0 0 -3 -4 1 «1 2 4 1 7 0 —2 0 «1 0 5 ('0) {RD} = eiglA) R = ~0.0991 -0.1405 0.6759 —0.4342 0.1900 -0.5684 —0.3697 0.8473 0.2592 0.8145 0.4717 —0.0852 ~0.0383 0.0616 0.3816 D = —9.8297 0 0 0 —4.1886 0 0 0 0.7781 0 0 0 0 0 0 (C) .742 O -3 -4 1 ~1; .4315 .6560 .2759 .2989 .4668 ONOOO 241—70; @0000 .498 .5721 .1481 .0459 .1321 .7945 00000 -2 O —1 O 5] Thus, Q(X) = —9.8297)E12 —4.1886fc22 + 0.77812,2 +3.7422ic,2 +6.4980252. (0') Since the eigenvalues are neither all positive, nor all negative, Therefore, the quadratic form is indeﬁnite. ...
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A8_soln - Math 235 Assignment 8 Solutions 1 For each...

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