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Unformatted text preview: Section 9.1 143 Chapter Nine ... Eigenvalues, Diagonalizatien and Special Matrices Section 9.1 Eigenvalues and Eigenvectors For Problems 1 through 16, we give the characteristic polyl'lomial of A as 33AM), the eigenvalues of A along with associated eigenvectors. Following Problem 20 are some sketches of Gerschgorin circles for a few selected (interesting) Problems. The circles are drawn solid and eigenvalues are plotted with an X mark. 1, 10):“pr m Al ...—_ A2 — 2A — 5 = 0 has roots A1=1+¢EA2 :1— \/G‘. Eigenvectors are V1 — ( ‘36 j , V2 3 ( “(2/6 )- 2 0 6 2. ﬁll/1005A —2A—8=0;)\1=4,V1= 4 l/\2:_21V2: _1 - r 2 7 {l d.pA(/\)=/\ +3Ar—10=U;z\}="5,vi= lA2321V2: - 4.3;A(A)=A2—10A+18=0;A1=5+x/'T’,V1=(1_Wﬁ);x\2=5 ----- W,V23(1+ﬁ) 5. mm = ){3—3,\+14 2 0m : 2 6. pA(/\) : A2 : 03A1I= A2 = 0. There is only one associated eigenvector, V1 2 ( 1 ). U 0 2 7.10.49): ,xa _5A2+e,\ : {”120}le 1 ;/\2 ___ 21V? 2 1 ”\3 2 3N3 _ 0 e U 2 3 0 2 2 - 1+ 29 1 A1211V1= 0 1)” Vl—IVE‘: 5+N/E ,/\3: “EA/3;- 5_\/§9 2 2 1 0 0 1 1 9. FA“) 3 ”(A + 3) Z 03)” : _3=V1 : 0 \$2 = A3 = 01V2 = {3 is only one eigenvector for A 2 O, 10» FAQ) '—"' A3 + 2A 2 U; 0 1 1 A1_ U’V] Z 1 9‘2 1 VELVQ m ""1 M3 3 — 2i,V3 :. ....1 U — 21' 21E 3+mi V1: ( _1+ We: ); 3‘— V’i’éa‘. (—1 ----- Mi) I44 ---~16 0 11. '1an -_». (A +14)(A — 2)2 = 0;)11 : —14,V, = 0 ;,\2 = A3 = 2N2 = a J 1 1 only 0110. cigenvector for A 2 2. 12- p11(A)w—-(A — 3m2 + A — 42) = a; o 30 0 A1=61V12 1 ;A2:3,V2: —2 ;A3=—7,V3= 8 _1 5 5 14 5 13- MA) = AW — 8A1?)=0,A1=0,V1— 7 mg -——— 1,V2 : 0 mg = avg 2 10 5 0 o 1 14. 33,401) = A2012 + 2A — 1) = [MA-1 2 A2 = U with two associated eigenvectors, V1 : 1 2 {I} ’ wl 0 1. _ 1 V2: i} ;/\3=“1+\/§,V3= 14—6/2 :A4:*1“\/§,Vq= 119/? U U 0 —2 O 2 _ —11 0 15.pA(A):(A~1)(A—2)(A +A—J.3):U;A1=1,V1: 0 ;A2-_—2,V2: 1 5 ' 1 U v’5—3—7 ' —\/5—3—7 “1111/5 0 —1~ 53 g :—~—-—Vr: ' :—v—-— 2 A3 2 3 J U 1A4 2 1V4 O 2 2 1 I 2 . —4 0 16.?)A()\):(/\—“1)(A—5)/\ :05A121,V1= 0 \$225,152 0 ; U 0 A3 = A4 I 0, V3 2 , only one eigenvector for A = 0. QI—‘D Section 9.1 145 Eiketches of Gersohgorin circles for some seiected Problems 1 through 20 Problem 1 Problem 4 Problem 5 P robiem 9 Problem 10 146 . Section 9.2 . 17. We have pA(A) 2 A2 — (a 1+ ’y)A + (an: — [12). The discriminant of this quadratic is B2 — 4A0 2 (e: + 7)? — 4(07 {32) 2 (or — 7):" + 462 2 0. Thus the quadratic characteristic equation has real roots. 18. To establish this result we define the inner product for complex entry 11 X 1 matrices :1: and y by( (:1), y) 2 219?. TWith this deﬁnition it is easy to establish the results (As), y) 2 A(3:I y) 71 and (:1: Ag): A(\$ y) for any complex scalai A, and the fact that( (33 2:)2 “93%|? 2 lezig. Let i=1 5' be the given 3 X 3 Sy!11111€3t1‘l( matrix having 1eal eritiies and note that 3123 and 5' = Finally let A be any eigenvalue 0115 with associated eigenveci or :1:. Then Ai|\$|12_ "2 A(:G :1?)': (As: :3) 2 (.951: 51:): (Sm) :c — :1: 5'51: — :1-‘f':S‘:'1', — :1: 3:122 (a: 311:): (:1: ALE)— — A(z1: (1:)— "2 Aua‘“2 From this it follows that A— ----- A: and hence A' is real. Note that this proof establishes that the eigenvalues of any n x a real symmetric matrix are real. 19. We are given that AB = AE. With E # 0. Now 151211") 2 A(AE) 2 A(AE) 2 MAE) = A213, so A2 is an eigenvalue of A2 with eigenvector E. T he general result follows by an induction proof to show AEE 2 AkE. 20. Since E and L are eigenvectors it foilows that neither is the zero vector. Assume now that E and L are linearly dependent. Then E — 19L for some nonzero scalar is. Now AE 2 A(kL) 2 kAL 2 11:11.1, 2 11E, and AB 2 AE, where we have used the fact that E and L are eigenvectors asscgiciated with A and 11 respectively. Subtract these two equations to get (11, — A)E 2 19. Since E 79 6‘ we must have 11 — A 2 {l or 11 2 A which is a contradiction. Thus E and L are linearly independent. 21. We have pA(A) 2 5A] -— Al. and the constant term of this poiynomial is found by putting A 2 U, which yields I — A| 2 (—1)”|/-l|. Recall that the constant. term of a. polynomial is zero if and only if A 2 U, is a root of the poiynoniiai. From this we have A 2 U is an eigenvalue of A if and only if IA] 2 l} and A is singular. Section 9.2 Diagonalization of Matrices 3 . _ _ 1. pA(A)— M A2 — iA + r1: 0 gives distinct roots A12 +2ﬁ3, with V1 2 ( B—Eﬁt ) ,_ . _(_ . —3+ﬁ1—3—\/11 and A2 2 '3 2X25?! with V2 2 ( '5 Sﬁ' ) Form P 2 8 8 , and 3+2 72‘ ' 0 13—11413 2 0 3—.2 71 2. MA 2 — 8A +12 2 0 gives distinct roots A; 2 2, with V] 2 ( :1 ) ,Ag — 6 V2 ----- /'_""\ x..___./ 2 U 1 diagonalizesK-l. P iAP— —— D 6 "lo Section 9.2 3. 10,400 2 A2 — 2)\ + 1 z 0 gives A} = A2 2 11 with only one eigenvector V; — ( 0 ), so A is not diagonaiizable. 4. pA(/\) '2 AZ-wii-A—rif) =0givesA1= —5,V1=(1);/\2 m9,V2 = ( 3 ),P = U 14 1 3 —5 0 O 14 diagonalizes A, P‘lAP : U 9 ' O 5 5. pA(/\)=A(/\—5)(A+2)=Ugives roots A1:O,V1 = 1 \$225}ng 1 ; U 0 0 {J 5 0 0 0 U _ m __ 1 1 —3 ‘ . _1 __ 0 5 0 A3 w —2, V3 — "23 . So P — U 0 2 dlagonahzes A, P AP — 0 0 _2 6. pA(/\) 2 My ~— 3A — 2) = U gives roots —2 0 - 0 3 17 3 — A1=0,V1= —3 #2: Jig/.572: 4 9‘3: 2mCV3: 4 ; 1 3 -+- m 3— x/ﬁ —2 U 0 0 0 D M —3 4 4 . . __1 _ {) 2111:2324? 0 P _ 1 3 + m 3 _ \/1—7 dwgonalues A, P AP _ 0 0 331,? U —3 '7. 10,100 = (A—1)()\+2)2 : 0 gives roots A} 2 1,V1 = 1 ;/\2 : A3 : ..... 2}V2 : 1 0 0 Since A has only two eigenvectors, A is not diagonalizable. 8. PAH) : (A — 2)()\2 m AA + 5) z 0 gives roots 1 0 U A132,V1= U ;)~222+i,V2z 1 ;/\3=2—i,V3= 1 U i —’i 1 [3 U 2 U 0 P z 3 i 3% diagonalizee A,P_1AP : 0 2 + 1 U 0 [3 ‘2—2‘ 9. mm x (A —1)(A — 4m? + 5A + 5) = 0 gives roots A1 :1,V1.-.—— OOCF—J 143 ' Section 9.2 U 0 0 1 —5+\/5 2“3 5 “EPA/5 2+3 5 A2W4’V2 0 M32 2 ’V3: #1’:11 5 ’A4# 2 1V4: _14_1 5 i 2 ? {J 1 1 1 U {l U 1 0 0 U 0 1 2,—fo M 0 4 0 0 P m 0 U ”12 5 3713—35 diagonalizes AID—IA? m 0 0 -5-23§ 0 0 0 1 l O O 0 —:3Ey:5 0 10- 10.400201 + 2)’1 z 0 gives roots )11 = A2 = )‘3 = A4 = ‘2. with V1 : 3) , O U . 0 V2 : g ’V3 m g ‘ Since A has Only three eigenvectors, A is not diagonalizable. 0 1 11' Since A2 is diagonalizable, A2 has n lineaﬂy independent eigenveotms X11X2, . . .‘X'n. with associated eigenvalues A1,)12,. . ”A” and (A2 — A3 In)X— — 0. Now (1423332,!) 2 (A "" V AjInM-A ‘5‘ V Ail) forj I 1, 2 n. We now have PM x/A jl—PA( \/—'l‘" ‘7pr? —01 so either ‘ M7- or #7017- is an eigenvalue of A and X7- will be an associated eigenveetor. )Thus A will have n linearly independent eigenveetors and be diagonalizable by Theorem 2. 12. SinceP diagonalizeSA, P 1241?: D or PDP 1“ A. Then Ak— (PEP-1k): (PDP )(PDP 1..) (PDP 1): PDkP-l, aswas to be shown. WWW k times 13. A = ( #11 B5 ) has 13,101): A2 + 6A+ 5, so eigenvalues are A1 = *1, )12 = — 18 P = ( i [1) ) diagoiializes A. Then A13 = P( ‘1 0 > P-1 = 0 —5 1/4 0 1 0 4 0 1 0 1 (11) (0 518) “1/41 3 55—3513 14‘A=(_3 :13)haSpA(A)=A2—A—18,SOA1=1+m7A221—m. — ' 2 2 Po —6 —6 73—1... 1 PM 6 "" Hm 7mm ’ ‘nm —7—m 7s - 1 7‘33” 16 15 15 16 17 ,. m: ( 2) 0 1 6(2 )e3 3(2 )—3 Ihen A P( O (7 _2 73)16 6—)}? ( _217 + 2(315) w216 _+_ 5(316) - 15. A z ( [1} 3 ) has eigenvalues A1 = 1/5,)12 : —\/§ With associated eigenvectors V1 — Section 9.3 149 elgenvectors V1 2 1 M m ) ,V2 2 ( 1 + m ). Thus 3 3 (10+x/T5)/60 ~x/ﬁ/20 Pz(1# 10),P_1: (lo—mm) «W20 Shem 31 m (—3 + NW 0 _1 _ km 5612 _ A — P ( 0 ("3 w VIEW P - 3421 1322 where F611“ — ”ﬂ [(1 ’1'" m)(— 3 + \/—0)31+(\/ﬁi "1)(W‘3 *— M)31]§ kn _ """L[(— 3 + W)31+3+ m )3113‘ 2m kglzrjmﬂ— 1+v’_0)( 3+¢15)31+ (M+1)(3+x/ﬁ)3ll; k22_23.1_0[(—~3+x/iﬁ)31+(3+\/1—0)].31 Section 9.3 Orthogonal and Symmetric Matrices 1.1014”): A2 ——5)\, 50 A1 301 A2 :5. Eigenvectors are V1 2 ( é ) 1V2 = ( “‘12 ). “1.. __ Tthkl thbl'VdZV Q €596 ,aecoumnso '0 e ,an ,30 z 7 . 0 0““ ‘ MI ‘ mu 2 w 0 0 Then (2‘1 : Q: and QtAQ 2 0 5 1+ V149 1~ M149 T’AQ 2 W' 10 10 W8C&ntdkev1m(7+\/ﬂ§)’vgm(7_V’@>‘ 2. pA(/\)EA2—A—37,SU A}: 150 Then (V1 -V2) 2 Vng 2 100 +49 w149 : B. 10 10 \/298 + 141/149 \/2gg ,. 141,421? Q = 7 +1 \/3_49 1’ — 1/149 11 \/298 +1.41/149 \/298 — 141/149 3. 1319):»? ----- 10Aw23,30/\1=5+\/§1,\2:5_\/§_ ’1‘akeV1: ( 1""'1V/§);v2= ( 1—ﬁ) 1 Then (V1 »V2) : V5162 2 1 — 2 +1: 0‘ 1+ 2* 1— 2 WHEN? 4—12ﬁ 4+2er 4—2v’i —9 + V293 /\ —9 ~ 1/293 —1 2 = —- 2 2 2 2 Take“ """ <17+m)‘vg_(17—¢2§§)‘ Then (V1»V2) = Vii/2 24+289 1. 293 = 0. 2 2 Q _ ( 13586—341/298 V586+34v298 ) 4- 11401) = A? + 9,1 53, so 11: 17 1/293 17+V293 \HSSG—de/QQS \/ 5864-3413293 5. 13110) 2 (A —3)()\2+2A— 1), so A123,)12 2 —1+\/§,)13 = —1 — VIE. 0 ‘ My”? 1—V/i Take v1 = 0 ,V2 : 1 ,V3 : 1 1 0 0 C1ear1y (V1 -V2) 2 (V1 -V3) : U,(V2-V3): 1— 2 +1 z 0. 0 1+3? 1—22 0 1 1 Q: 4+2fz 4—2\/§ 1 0 U 6. mm): (A—2)(A2—2,\—2), so A; =2,12=1+¢§,A3=1—¢§. a ~---1+ \/§ —1 — J3 Take V1 2 «1 ,V2 : 1 ,V3 2 1 . 1 1 1 Then(V1-V2)=(V1»V3): G, a11d(V2-V3):1—— 3 +1+1= 0. . 0 (—1.+¢E)/\/6 (—1—v’3)/¢6 Q2 —1/v’§ 1% 1N6 WE W6 1N5 Section 9.3 ' Section 9.3 r r 41 5— 41 7. p340}=A()13—5A—4),80/\1=0,)12:0+2\/—,,\3: f 0 5 + «E 5 m Take V1=1 V2— 0 ,Vg : O 0 4 ' 4 2:) Then (“w-v (“v1 v3)— 4 0 and (-V3 V3)— _ 25 — 41 + 16— — 0. 0 (5+Wﬁ) 1/1/2124r 10¢— (5— 3/41) N82 —10¢_ Q 3 WWWWW 0 4/\ 82+103/4_1 4/3/82— 10m 8. 312(2):)102 42-415) 5021:0233 T..1+v’1’*?,,\3=1—V/ﬁ 0 1+\/_ 1— W Take V1 2 0 ,V2 2 m4 ,V3 2 w.4 _ 1 0 0 Ther1(V1-V2)—---~(V1 -V3} 2 U, and (V2 AV3) 2 1 m 17 + 16 :2 O. D 42:0 1 9. 22(2) 2 2(22 4 2 4 4), 30 A1 x 0,22 = 1 0 0 Takevlw—(O),V2(1J1?),V31(1\/17) 0 4 4 Then (V1 -V2) = (V2 -V3) 2 0,311d(V2-v3): 1 — 17 +16 ..—_ 0. 1 0 Q: 0 (wlwx/ﬁv 34+2 (_1+‘/1—7./ {34:L'r2x/i7 U 4/ 34+2xfl—7 4/\/é4+2ﬂ 10- pA(r\)=(/\~1)(A2—A—10) 30212'1,,\2_1+21/-1 3_1—2\/éﬁ. 1 6 Take V3 = 0 ,V2 = —1 + bﬂ V,3— _. 41 -1 m . Then(V1-V2):(V13V3)=6—6:0,and(V2-V3):36+1—41+4:0. 1/3/16 5/ 82342342 5/[4 23/1“ Q: 0 (1+1/_)/ 82—23/_ (1*V—1//.§3.:.24 —3/\/1_0 2/ 82—2\/41 2/{82— 2m 151 152 11 m0) : A202 ~ 2A "— 3), so A] = A2 2 {M3 : —1,,\4_ = 3 1 U 0 0 ,1 0 0 1 _1 lake V1 3 0 ,V2 w 0 ,V3 2 1 ,V4 2 1 U 1 U U Weeasiiy see(Vg-Vj):0,zaé3 1 0 0 0 1 1 U “ 1Z5 "I" Q: 0- 0 E E 0 1 0 0 Section 9.4 Quadratic Forms Section 9.4 Section 9.4 153 1 a -«1/2 0 —1 2 6' A 2 —-1/2 2 0 —5 ‘2 "f. This form is X‘AX with A = 2 3 The characteristic polynomial of A is FAQ) 2 A2 + 2A — 19 with roots A1 = —1 + 2V5, A2 2 v1 — WE. By the Principal Axis Theorem, the standard form is (—1 + 2&3in + (—1 — 2'V’5Jyg- r _ 1 X1 _ r 8. 30,40) 2 A; — 5A m 32 with roots A1 = “2—53, A2 : 5—29—- Standard form is ' I Sid—vm 2 5—«153 2 T yi "i‘ f 292‘ 9.33/16): 1&2 — 4A — 25 with roots A1: 2 + m, A2 = 2 — \/2—9. Standard form is (2 + ﬁﬁhﬁ + (2 — ﬂag. I 1 ‘ — 1 10. 101400; A2 — 3A — 2 with roots A1: 5+2ﬂ, A2 2 d 3/7,. Standard form is 2 321 ‘1‘ 2 3’2- 11. pAO‘) : A2 # 4A — 9 with roots A; z 2 + m, A2 :2 2 m m. Standard form is (2 + mm + {2 — my; 12- PACK) = A2 — 7A ----- 6 with roots A1 = 1, A2 — 5_ Standard form is y? + (is? - U m1 0 13. This form is Xf'AX with A x —1 U U U U 2 The eigenvalues of A are A = —1,1,2. By the Principal Axis Theorem the standard form of this quadratic form is w? +113 + 2?}? i54 Section 9.4 14. This equation can be written XtAX 2 6 with A = ( 1 W1 ). —1 4 I’ 15 Eigenvalues of A are A = 3.3341 The standard form of the comic is 5 + m 2 . 5 — {1’3 2 ~— 2—m— yl + 2 3J2 = 6 Since both eigenvalues are positive, this comic is an ellipse. . . . f . 3 5/2 15. 1111s equation can be Wi‘ltten X 'AX : 5 With A = 5/? _3 . V61 Eigenvalues of A are A = :: ——2—-. The standard form of the sonic is «e 2 m 2 Tye???“ which is an hyperbole. 16. This equation can be written XtAX = 5 with A = < {/22 1é2 ). 2 j: m. rThe standard form of the conic is (243/279) y12+ (2 1/23) 3;; E 5’ Eigerwaiues of A are A : 17. This equation can be written XEAX : 8 with A : g 34 ). The eigenvalues of A are A = 5, —5. The standard form of the comic is 5y? m 5y3 = 8, which is an hyperbole. 18. This equation can be written X‘AX = 14 with A = ( 6 1 ). Eigelwaiues of A are A: inn/5 . The standard form of the comic is 11+\/5 . ﬂew/5 . which is an ellipse. 19. —2.’C% + 2:31:32 1‘" 612% 20. 14-56% — 6321332 “+- 2’53 ‘l" 2332163 + 711"; 21. 6.113 + 2mm; — 14\$1\$3 .+. 215% + 33% Section 9.5 ISS 22. 733% -l- 2:131:62 43513:;3 — 2332333 -l- 322% 2 O U 23. Take A x U D —-1 . Then PMA} : (A w 2)(A2 + 1) so A has eigenvalues 2,211. ------ 1i 0 1 U and is diagonalisable. But the standard form associated with X IAX would have complex coefﬁcients. Section. 9.5 Unitarm Hermitian and SkewuHermitian Matrices 1. The matrix is not liern'iitian, not skew~hermitian and not ui'lital‘y. pA(A) 2 A(A — 4) + 4 — (A — 2P, so A1 = A3 = 2. There is only one eigenvector, V1 2 < 1 ), so the matrix is not diagonelizable. 2. The matrix is not herniitian, not skew—hermitian, not unitary. pA(A) : A2 + 2A + 1 = [A + 1)2, so A; : A2 = —1. There is only one associated eigeiivector, V1 : ( j] ), so the matrix is not (liagonalizable. 3. The 1'11at1‘ix is skew—hermitian since 5‘ 2 _§‘ 333(A) : A(A2 + 3), 2 1 1 SO A]=O,V1: U ;A2£x/gi,V2= 3% ;}k3:~\/3t1V3= — 3?: 1 +i —1 —e' —1 ~—'i Take P = ('V1V2V3) and A1 0 0 P‘lSP 2- 0 A; e U 0 A3 4. Compute 30" = I, so U‘: 2 3—1 and U is unitary. pU(A) : (A —'- 1) (A2 —-— (12.1% +3), 0 1+v’3 .(I—Jé) 1+1 1— ‘ )VJ s A ILV = U ;)\ 2 4-? , 3 3'1— 3 ;,\‘ : m~m SO 1 1 1 2 Qﬁ Qﬁ ( Uvr) '3 2\/§ 1| o is ,9 3 T I V : ' _r‘ 2 . —1 I 2 z< NE ) 3 11(1l0x/3) IakeP (V1V2V3)andP UP 0 0 A3 5. The matrix is l'iel‘mitian since Ht : F. pH (A) = A3 — 3A"2 — 5A —[— 3, so eigenvalues are _ 1 1 (approximately) A1 : 4.0513?61,V1 = .525687 ;A2 : .482696,V2 : —1.258652 ; — .129755i 2.6075467: 156 Section 9.5 ' 1 A3 : —1.53407,V3 : —2.267035 . Take P : (V1V2V3) and ‘ w"1.47"?"r'91rg A1 0 U P‘1HP z 0 A2 0 {J 0 A3 6. The matrix is hermitian since H5 = H. pH(A) = (A - 1)(A2 + A m 10), so A; 2 1, U 6 — 2i — 6 — 22'; V1: 1 ;A2:W1:m,v2= 0 ;/\3= 1 Zﬂi,V3: 0 0 1 + E 1 —— WE A1 0 0 Take P = (V1V2V3) and P‘lHP : 0 A2 0 0 0 A3 7. The metrix is skew-hermitian since 8" 2 m3. p3(A) = A3 - iA2 + 5A — 4?}, so (approxi— —t 2 mateiy] A1 = ——2.164248i,V1 : —3.164248 ;A2 = 77286616, v2 = 227134 ;A3 = 2.924109 .587771 2' A1 0 U 2.3913822',V3 z w1.391382 . Take P : (V1V2V3) and P“'1SP : 0 A2 0 —1.163664 0 0 A3 8. The matrix is not hermitian, not skew—hermitian, not unitary. pA(A) = (A2 — 1)(A — 3i), 0 0 «105 50 A1 : 1,V1 2 i ;A2 : "‘1,V2 : ’6 ;A3 : 3i,V3 = 3 . Take P = 1 m1 —1 (V1V2V3) and A1 0 {1 P‘lAP = 0 A2 0 O 0 A3 9. The matrix is hermitian since Ht = F. pH(/\) = A(A2 -— 8A — 2), so A; 2 {no 2 0 4 + aﬁ 4 w 3x/5 1' ;A2=4+3\/§,V2= —1 ;A3:4——3\/§,V3=: ~1 .Tach: 1 ~—'£ . #i (V1V2V3) and A1 8 U P—1HP = 0 A2 L} U 0 A3 10. If A is hermitian, then A1 z X. Then A :_j = (2):, so (4(3)E : A2 = ("£th. If A is skewwhermitian, thee At = —Z. Then #A = “Z = (If, so {HZ}t = —A2 : —AA 2 (EVA. HA is unitary, then A‘1 2 (KY, so ACE): : ALA—1 : A“1A 3 (EPA. 11. If S is skew-hermitian then S" 2 .53: so 3”. : ~51"? for 1'" a 1,2,.. .,?1. But if 37‘1" 2 ow +11)” and s” : —§,.,. then arr z “or, for 7" 2 1, 2, . . . , n, hence a” 2 0. Hence 3” 2 ibw is either pure imaginary or zero. Section 9.5 157 12. If H is hermitian, then Ht 2 if so the diagonal elements satisfy h” : Em but then h“. is real for a" m 1,2, . . . ,n. 13. Let U and V be Elfin unitary matrices. Then U*1 = “th and 1V1 2 Vt. Compute (LN/)4 I V—lU—l = V U1: {T}— : V)‘ = (WV, so UV is unitary. ...
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