# 13th - EXERCISE SET 3.! 12. The characteristic polynomial...

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Unformatted text preview: EXERCISE SET 3.! 12. The characteristic polynomial of A is (a — up? — at + 2}; thus J. = 1 a the only real eigenvalue of —l 41 - G In ll of Li — A :is 2 and the geometric multiplicity of .l. = 1 is nullityflir — A] = 3 — 2 = 1. —l 2 — 1 ﬂ 2 A. The reduced row echelon form of the matrix 11' - A = [-2 T 41] is [o 1 4]; thus the rank 14. The characteristic polynomial of A is {Ji+2]l[li - I)“; thus the eigenvalues are it: —2 and J's =1 1 _ I L _ _ —2 l —] mth algebraic multiplicities l and 2 respectively. The matrix —2I — A = 1 —2 4] has rank 2 —1 —-1 —2 1 1 —l and the matrix 1! - A = l 1 1. 4] has rank 1. Thus J. = —2 has geometric multiplicity l, and —1 —l l J'- = 1 has geometric multiplicity 2 It follows that A has 1 + 2 — 31' ' - ' — me I [IE and this; is diagm-lulimbk_ ET I In pendent emmm 24. The characteristic polynomial of A is pili} = i}. + 2PM — 3]“; thus A has two eigenvalnw A. = —2 n 1 and It = 3, each of algebraic multiplicity 2. The vectors v1 = '1] and v: = 3 form a basis for the o I] s o eigenspaoe corresponding to A = —21 and the vectors ya =- i and “£4 = 1; form a basis for the o 1 eigenspace cormsponding to J. = 3. Thus A is diagonalisahle and the matrix P = [in y: V3 n] has the property that l] l -l l —2 I] [l- l] I] l [l [I —2 {I [l i} 1 [l i] U I] —2 5 —5 l I] l —l [i "-2 [II II} —1 P: = P A [I i] l [I ll i] 3 [I l] i] l I] [I I} 3 l} ll [I ll 1 [l l] i) 3 i] ll 0 l I] [l i} 3 D4. in} A. is a 6 x E matrix. [11} The eigenspece corresponding to A = 1 has dimension 1. The eigenspaoe corresponding to J. = 3 has dimension 1 or 2. The eigenspace corresponding to Jt = 4 has dimension 1. 2 or 3. (c) If A is diagonalisable, then the eigenspaces corresponding to l. = 1r .1 = 3, and i. = 4 have dimensions 1| 2, and .‘i rcspectiyely. {cl} These vectors must correspond to the eigenvalue A = 4. D5. {a} If in has geometric multiplicity 2 and Jr; has geometric multiplicity 3, then Jig must have multiplicity 1. Thus the sum of the geometric multiplicities is ti and so A. is diagonallsahle. {b} In this case the matrix is not disgonalisahle since the sum of the geometric multiplicities of the eigenvalues is less than 6. (c) The matrix may or may not he diagonalisahle. The geometric multiplicity of 2.3 must he 1 cr 2. II the geometric multiplicity of its is 2, then the matrix is diagcnclisahle. If the geometric multiplicity of Jig, is 1, then the matrix is not diagonaliaahle. EXERCISE SET 3.3 1a. The characteristic polynomial of A in pm} = J13 + 231.2 - 1125.1. — 375!) = it + 3:11:21 — item + on); thin the eigenvalues of A are 11 = —3, )1: = 25. and its = —5[l. Gmeaponciing eigenvoctors are [I —4 V1=1,V2= ﬂ,MdV3= 'D 3 I] P_ 1'1 V2 VEJ=E ti 3 a]. These vectors are mutually orthogonal, and the orthogonal 4 _ IlV1|| |l‘-'2|I ilvaJI has the property that U l D —2 n —nn 0 —% i —n n n PTAP= —§ n g n —2 n 1 n n = n 25 n =1) :1 n g —3n n —22 .3, g g n n —5n 13. The eigenvalues of A are J11 = —3, It: = 25. and A3 = Hm with corresponding normalized oigenvectora a. 3 ﬂ ‘3 E 111= H. u: = I] ,u; = n . Than a spectral decomposition :1le is 0 e 5 3 E —2n—2n n 4% i A: n —n n =—n1[n1n]+2n n [-g n g]—nn n {g n g] —nnn—22 n a 4 2 2 nnn ii'l-% hﬂh =—2n1n+2n nn n~ennnn "ii eel—i 23. {a} The characteristic polynomial of A in pH} = )1” — 6.312 + 12}. - 3. Computing succession powers S —5 4 it] -24 12 ofA,wehach2=|a —12 a andA3=[2«t —4o 2:1];th112 12 "24 [It 36 —T2 ea 2n —24 12 a —n a 2 -2 1 A“-ﬁA*+12A= 24 —4n 24 —n n —12 2 +12 2 —2 2 =2: an —72 44 12 —24 1e :1 -n n which shows that A satisﬁes its cliaxacteriatic equation, in. that pH} = D. [11} Since a3 = n14? — 12.4 + 3!, we have 21‘ = iii-13 — 1221i + tot = 24A“ — 5421+ 42:. {a} Sine-2113 — an“ +1214 — a! = n. we have ate: — n14 +122} = 2: and 21-1 = gist? — an +121}. D1. [a] True. The matrix MT is symmetric and hence in orthogonally diagonalizahle. {12] False. If A is diagonaliaahle hut not ammetﬁc [therefore not orthogonally diagonalizahle}, then there is a haaia for R“ {but not an orthogonal haeia} nona‘mting of eigenvectors of A. I. 1 (o) False. An orthogonal matrix need not he symmetric; for incarnate A = 3F . 35 E {:1} Tina. H A is an invertible orthogonally diagonalizahle matron then there is an orthogonal matrix P such that PTAP= D where D in a diagonal matrix with nonzero entries {the eigenvalues of A} on the main diagonal. It followa that PTA—1P = {Pram-1 = 3-1 and that 1'31"1 is a diagonal matrix with nonzero entries [the reciprocals of the eigenvalues} on the main diagonal. Thus!r the matrix A“ ie orthogonally diagonallzahla [a] True. If A is orthogonally diagonalicahle, then A is symmetric and thus has real eigmvaluea. P2. Suppe-se A = clulu? + cgugug' + - r - + enunul" where {111,112,” .. un} isanerthenermal hasisfer R". Since {uJ-uJTF' = ugTuf = ujuj' it fellows that AT = A; thus A is symmetric. Furthermerer since 113'“; = u..- - “j = 515, we have ﬂ. Auj = {eluluf + caning + - - - + managiuj = Zciut-uruj = Cju-f i=1 fereachj = 1.2,...m. Thus c1.egt...,cn areeigeneelum efA. EXERCISE SEI' 3.4 1s. {a} The eigenvalues of A = E _”] we .1 = 2 and l = ‘5; thus A a indeﬁnite. (11} negative deﬁnite {it} positive deﬁnite {d} negative semideﬁnite (e) peeitlee semideﬁnite 33. {a} Tier +3} = {x + ﬂTAlx +3r] = {JET +3rT]|AI[x + y} = xTAx +xTAy +yTAir + yTAy = He: + 2ft; + 31" A}; = The] + 2xTA1r + rte] lb} Ties} = tritium) = Eixfﬁxl = C’Tlx} 3 II. I"- 34. {a} Fereachi:1,....nwehave[ire-i}3=z§—2:r¢£+i2=.tE—2mﬁ215+ﬁg(z;rj) = n n 11 ﬂ _ _ I? _ a E “3:. + 14—1! + 2 E 2 min). Thus In the quadratic farm :1 3"] I- . . J J=1k=J+l 1 s3: . t—llizl—eFHe-t}2+~--+tmn~ei“} the meﬁﬁem of If 13 ﬁ [1_%+;1,n] = i. and the ceeﬁicient of Iiirj for tet-J' is —1-[—,, — g + gm] = —ﬂ. Imam that a: = Tex where 11—1 '1 1 _ _ 1 ; _'Iq. "—1 H Thin-1i 1 l _ 1 _aln—li 7-1 nits-1i A = _ 1 _ 1 .1. win-F15 ﬂirt—15 n {h} We have s: = ﬁHIl—§?F+{Ij—i]2+---+{£n-§f}2] ED. and sizﬂifandenlyifrr] = i. it: = 5,. ..,.'r.1 =.t, i.e.1ifand only ifml :3; =--- = 1.3,. Thus a: is positive semideﬁnite. D1. {a} False. For example the matrix A = [1: has eigem'alues —1 and 3; thus A is indeﬁnite. (h) Fla-lee. The term 4.113233 is not quadratic in the variables x11x2,z3. {a} True. When expanded, each ef the terms of the rmulting expression is quadratic {uf degree 2] in the variables. {:1} True. The eigenvalues of a positive deﬁnite matrix A are strictl}r positive; in particular, ﬂ is not an eigenvalue of A and an A is invertible. {e} False. For example the matrix A a [a 3] is positive semi—deﬁnite. {f} True. If the eigenvalues of A. are pctsjtiee+ then the eigenvalues of —A are negative. D2. [a] True. When written in matrix form, we have x - x = TAX where A = I. [h] True. If A has pasiti‘lm eigenvalues, then an does A“. {c} True. See Theorem 8.4.3ia}. (:1) True. Both of the principal submatﬁnen of A will have a positive determinant. l l. [e] False. II' A = [_1 I]. then xTAx = I! + y”. 011 the other hand, the statement is true if A in 353th to be symmetric. P1. Rats-ting the coordinate axes through an angle 9 corresponds to the change nf variable 3! = Px’ what-e P = 1:9]. i.e.I a: = renal? — y'ainii and y = :r'ainﬂ + y'eneﬂ. Substituting these ELIE. expressions into the quadratic form at:2 + ﬂirty + cg” leada to Ax” + Br'gf + Cy”. where the ceeﬂ'icdent ef the cross product term in a = -2amasme +2lil[unnad' - an2a1+ Ecumdaim? = {-u + c}ainﬂﬂ + Ebeneﬂﬂ Thus the resulting quadratic form in the variables a“ and if has no creaa product term if and only if (—a + clainEE + 2hooa29 = (I. or [equivalently] eat 23 = 95—59. ...
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## This note was uploaded on 05/19/2008 for the course MS 4032 taught by Professor Anony during the Spring '08 term at A.T. Still University.

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13th - EXERCISE SET 3.! 12. The characteristic polynomial...

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