tutorial 4 mth 3201
TRANSCRIPT
Tutorial MTH 3201 Linear Algebras
Tutorial 4
Let π’ = π’1, π’2, π’3 and π£ = π£1, π£2, π£3 are vectors in π π
IMPORTANT!
π’ , π£ = π’ β π£ = π’1π£1 + π’2π£2 + π’3π£3
Euclidean inner product on π π
*Please develop your writing skills in mathematics
Symmetry
Additivity
Homogeneity
Positivity
Axiom
π’ , π€ = π’ β π€ = π’2π€2 + π’3π€3
= π€2π’2 + π€3π’3
= π€ β π’
= π€ , π’
π’ , π€ = π€ , π’
π’ + π€ , π£ = π’ , π£ + π€ , π£
Since π’ , π€ = π’2π€2 + π’3π€3
π’ + π€ , π£ = π’2 + π€2 π£2 + π’3 + π€3 π£3
= π’2π£2 + π€2 π£2 + π’3π£3 + π€3π£3
= π’2π£2 + π’3π£3 + π€2 π£2 + π€3π£3
= π’ , π£ + π€ , π£
= π’ β π£ + π€ β π£
Since π’ , π€ = π’2π€2 + π’3π€3
= π π€2π’2 + π€3π’3
ππ’ , π€ = ππ’2π€2 + ππ’3π€3
= π π€ , π’
= ππ€2 π’2 + ππ€3π’3
ππ’ , π€ = π π€ , π’
Let π’ = π’1, π’2, π’3 and π€ = π€1, π€2, π€3
π€ ,π€ β₯ 0 πππ π€ ,π€ = 0 πππ π€ = 0
π€ ,π€ = π€ β π€ = π€2π€2 + π€3π€3
π€ ,π€ = 0 πππ π€2 = π€3= 0
π΅π’π‘, π€1 πππ ππ πππ¦ π£πππ’π, β 0
β΄ π’ , π€ = π’2π€2 + π’3π€3 are not inner product on π 3
Symmetry
Additivity
Homogeneity
Positivity
Axiom
π’ , π€ = π’ β π€ = π’1π€1 + 2π’2π€2 + 3π’3π€3
= π€1π’1 + 2π€2π’2 + 3π€3π’3
= π€ β π’
= π€ , π’
π’ , π€ = π€ , π’
π’ + π€ , π£ = π’ , π£ + π€ , π£
Since π’ , π€ = π’1π€1 + 2π’2π€2 + 3π’3π€3 π’ + π€ , π£ = π’1 + π€1 π£1
+2 π’2 + π€2 π£2 + 3 π’3 + π€3 π£3
= π’ , π£ + π€ , π£ = π’ β π£ + π€ β π£
Since π’ , π€ = π’1π€1 + 2π’2π€2 + 3π’3π€3
= π π€1π’1 + 2π€2π’2 + 3π€3π’3
ππ’ , π€ = ππ’1π€1 + 2ππ’2π€2 + 3ππ’3π€3
= π π€ , π’
ππ’ , π€ = π π€ , π’
Let π’ = π’1, π’2, π’3 and π€ = π€1, π€2, π€3
π€ ,π€ β₯ 0 πππ π€ ,π€ = 0 πππ π€ = 0
π€ ,π€ = π€ β π€ = π€1π€1 + 2π€2π€2 + 3π€3π€3
π€ ,π€ = 0 ππππ€1 = π€2 = π€3= 0
β΄ π’ , π€ = π’1π€1 + 2π’2π€2 + 3π’3π€3 are inner product on π 3
= π’1π£1 + π€1 π£1 + 2π’2π£2 + 2π€2 π£2 +3π’3π£3 + 3π€3π£3
= π’1π£1 + 2π’2π£2 + 3π’3π£3 + π€1π£1 + 2π€2 π£2 + 3π€3π£3
= ππ€1π’1 + 2ππ€2π’2 + 3ππ€3π’3
Symmetry
Additivity
Homogeneity
Positivity
Axiom
π’ , π€ = π’ β π€ = π’1π€1 + π’2π€2 β π’3π€3
= π€1π’1 + π€2π’2 β π€3π’3
= π€ β π’
= π€ , π’
π’ , π€ = π€ , π’
π’ + π€ , π£ = π’ , π£ + π€ , π£
Since π’ , π€ = π’1π€1 + π’2π€2 β π’3π€3
π’ + π€ , π£ = π’1 + π€1 π£1 + π’2 + π€2 π£2 β π’3 + π€3 π£3
= π’1π£1 + π€1π£1 + π’2π£2 + π€2π£2 β π’3 π£3 β π€3π£3
= π’ , π£ + π€ , π£
= π’ β π£ + π€ β π£
Since π’ , π€ = π’1π€1 + π’2π€2 β π’3π€3
= π π€1π’1 + π€2π’2 β π€3π’3
ππ’ , π€ = ππ’1π€1 + ππ’2π€2 β ππ’3π€3
= π π€ , π’
ππ’ , π€ = π π€ , π’
Let π’ = π’1, π’2, π’3 and π€ = π€1, π€2, π€3
π€ ,π€ β₯ 0 πππ π€ ,π€ = 0 πππ π€ = 0
π€ ,π€ = π€ β π€ = π€1π€1 + π€2π€2 β π€3π€3
π€ ,π€ β₯ 0,
π€12 + π€2
2 π βππ’ππ ππ πππππ‘ππ π‘βππ π€32
β΄ π’ , π€ = π€1π€1 + π€2π€2 β π€3π€3 are not inner product on π 3
= π’1π£1 + π’2π£2 β π’3 π£3 + π€1π£1 + π€2π£2 β π€3π£3
= ππ€1 π’1 + ππ€2π’2 β ππ€3π’3
π€12 + π€2
2 β π€32 β₯ 0
π€12 + π€2
2 β₯ π€32
π€12 + π€2
2 < π€32
π€ = π€ ,π€ 1/2 = π€12 + π€2
2 = 22 + (β5)2 = 29
π€ = 5(2) 2 + 2(β5)2 = 70
π΄π€ =β2 32 7
2β5
=β19β31
π΄π€ = (β19) 2 + (β31)2 = 1322
π π’ , π€ = π’ β π€ = β1 β 3 2 + 3 β 5 2 = 20
π π’ , π€ = π’ β π€ = 5 β1 β 3 2 + 2 3 β 5 2
= π’ β π€ , π’ β π€ 1/2
= π’ β π€ , π’ β π€ 1/2 = 88
π’ β π€ = β1,3 β 3,5 = β4, β2
π΄ β π’ β π€ =β2 32 7
β4β2
=2
β22
π΄ β π’ β π€ = (2) 2 + (β22)2 = 488
π΄ = π΄, π΄ 1/2 =2 4β3 1
,2 4β3 1
1/2
= 22 + 42 + (β3)2+1 = 30
π π΄, π΅ = π΄ β π΅ = π΄ β π΅, π΄ β π΅ 1/2 = (2 + 4)2+(4 β 2)2+(β3 β 5)2+ 1 β 1 = 104
π΄ = π΄, π΄ 1/2 =6 β17 4
,6 β17 4
1/2
= 62 + (β1)2+72 + 42 = 102
π π΄, π΅ = π΄ β π΅ = π΄ β π΅, π΄ β π΅ 1/2 = (6 + 1)2+(β1 β 8)2+72 + (4 β 2)2
= 183
π = π , π 1/2 = π π₯ π π₯ ππ₯1
0
1/2
= 3π₯2 β 2 3π₯2 β 2 ππ₯1
0
= 9π₯4 β 12π₯2 + 4 ππ₯1
0
=9π₯5
5β
12π₯3
3+ 4π₯
0
1
=9
5
π β π = 2π₯2 β x β 2
π π , π = π β π
= π β π , π β π 1/2 = (2π₯2 β x β 2) 2π₯2 β x β 2 ππ₯1
0
1/2
= 4π₯4 β 4π₯3 β 7π₯2 + 4π₯ + 4 ππ₯1
0
=52
15
π = π , π 1/2 = π π₯ π π₯ ππ₯1
0
1/2
= π₯2 + π₯ + 1 π₯2 + π₯ + 1 ππ₯1
0
= π₯4 + 2π₯3 + 3π₯2 + 2π₯ + 1 ππ₯1
0
=π₯5
5+
2π₯4
4+
3π₯3
3+
2π₯2
2+ π₯
0
1
=37
10
π β π = β4π₯2 + 2x β 2
π π , π = π β π
= π β π , π β π 1/2 = (β4π₯2 + 2x β 2 ) β4π₯2 + 2x β 2 ππ₯1
0
1/2
=16π₯5
5β
16π₯4
4+
20π₯3
3β
8π₯2
2+ 4π₯
0
1
=88
15
2π’ , π£ β π€ + π£ , π£ β π€ = 2π’ , π£ β 2π’ , π€ + π£ , π£ β π£ β π€
= 2 π’ , π£ β 2 π’ , π€ + π£ 2 β π£ β π€ = 2 3 β 2 7 + 25 + 2 = 19
π’ , π£ + 3π€ β π£ , π£ + 3π€ + 2π€ , π£ + 3π€ = π’ , π£ + π’ , 3π€ β π£ , π£ β π£ , 3π€ + 2π€ , π£ + 2π€ , 3π€
= 3 + 3 7 β 52 β 3(β2) + 2(β2) + 6(8)2= 385
2π’ + π€ , 2π’ + π€ 1/2 = 2π’ , 2π’ + π€ + π€ , 2π’ + π€ 1/2 = 4 π’ , π’ + 2 π’ , π€ + 2 π€ , π’ + π€ ,π€ 1/2
= 4(2)2 + 2(7) + 2(7) + 82 1/2 = 108
π’ β 3π£ + π€ , π’ β 3π£ + π€ 1/2
= π’ , π’ β 3π£ + π€ β 3π£ , π’ β 3π£ + π€ + π€ , π’ β 3π£ + π€ 1/2 = ( π’ , π’ β 3 π’ , π£ + π’ , π€ β 3 π£ , π’ + 9 π£ , π£ β 3 π£ , π€ + π€ , π’ β 3 π€ , π£
+ π€ ,π€ )1/2
= (2)2β3 3 + 7 β 3 3 + 9 5 2 β 3 β2 + 7 β 3(β2) + 82 1/2 = 301
π’ , ππ£ = ππ£ , π’ = π π£ , π’ = π π£ , π’
symmetry homogeneity symmetry
π’ β π£ , π€ = π€ , π’ β π£ = π€ , π’ + π€ ,βπ£
= π€ , π’ + (β1) π€ , π£ = π’ , π€ β π£ , π€
symmetry additivity homogeneity symmetry
π’ , π£ = 5π’1π£1 + 2π’2π£2 = 5 2 0 + 2 3 β1 = β6 = 6 = 36
π’ = π’ , π’ 1/2 = 5(2)2 + 2(3)2 = 38
π£ = π£ , π£ 1/2 = 5(0)2 + 2(β1)2 = 2
π’ β π£ = 38 2 = 76 compare
36 < 76
β΄ π’ , π£ β€ π’ β π£
Cauchy-Schwarz inequality
π’ , π£ =2 61 β3
,β3 14 2
= β6 + 6 + 4 β 6 = β2 = 2 = 4
π’ = π’ , π’ 1/2 =2 61 β3
,2 61 β3
= 4 + 36 + 1 + 9 = 50
π£ = π£ , π£ 1/2 =β3 14 2
,β3 14 2
= 9 + 1 + 16 + 4 = 30
π’ β π£ = 50 30 = 1500
compare
4 < 1500
β΄ π’ , π£ β€ π’ β π£
Cauchy-Schwarz inequality
β’ Do as your exercise
πβπππππ 1.5.3(π)
π£ =0 iff π£ = 0
π£ = π£ , π£ 1/2 = π£12 + π£2
2 + β―+ π£π2 β₯ 0
π£ = π£ , π£ 1/2 = π£12 + π£2
2 + β―+ π£π2 = 0 iff π£1 = π£2 = β― = π£π = 0 β π£ = 0
β΄ π£ =0 iff π£ = 0
πβπππππ 1.5.3(π)
ππ£ = π π£
ππ£ = ππ£ , ππ£ 1/2 = π π£ , ππ£ 1/2 = π2 π£ , π£ 1/2 = π π£ , π£ 1/2 = π π£
πβπππππ 1.5.3(π)
π π’ , π£ =0 iff π’ = π£
= π’1 β π£12 + π’2 β π£2
2 + β―+ π’π β π£π2 β₯ 0
π’1 = π£1, β¦ π’π = π£π β π’ = π£
πβπππππ 1.5.3(β)
π π’ , π£ β€ π π’ , π€ + π π€ , π£
π π’ , π£ = π’ β π£ = π’ β π£ , π’ β π£ 1/2
π π’ , π£ = π’ β π£ = π’1 β π£12 + π’2 β π£2
2 + β―+ π’π β π£π2 = 0 iff
β΄ π π’ , π£ =0 iff π’ = π£
π π’ , π£ = π’ β π£ = π’ + π€ β π€ β π£ = π’ β π€ + π€ β π£ β€ π’ β π€ + π€ β π£
= π π’ , π€ + π π€ , π£
β΄ π π’ , π£ β€ π π’ , π€ + π π€ , π£
-dr Radz