Tutorial MTH 3201 Linear Algebras

Tutorial 4

Let 𝑢 = 𝑢1, 𝑢2, 𝑢3 and 𝑣 = 𝑣1, 𝑣2, 𝑣3 are vectors in 𝑅𝑛

IMPORTANT!

𝑢 , 𝑣 = 𝑢 ∙ 𝑣 = 𝑢1𝑣1 + 𝑢2𝑣2 + 𝑢3𝑣3

Euclidean inner product on 𝑅𝑛

*Please develop your writing skills in mathematics

Symmetry

Additivity

Homogeneity

Positivity

Axiom

𝑢 , 𝑤 = 𝑢 ∙ 𝑤 = 𝑢2𝑤2 + 𝑢3𝑤3

= 𝑤2𝑢2 + 𝑤3𝑢3

= 𝑤 ∙ 𝑢

= 𝑤 , 𝑢

𝑢 , 𝑤 = 𝑤 , 𝑢

𝑢 + 𝑤 , 𝑣 = 𝑢 , 𝑣 + 𝑤 , 𝑣

Since 𝑢 , 𝑤 = 𝑢2𝑤2 + 𝑢3𝑤3

𝑢 + 𝑤 , 𝑣 = 𝑢2 + 𝑤2 𝑣2 + 𝑢3 + 𝑤3 𝑣3

= 𝑢2𝑣2 + 𝑤2 𝑣2 + 𝑢3𝑣3 + 𝑤3𝑣3

= 𝑢2𝑣2 + 𝑢3𝑣3 + 𝑤2 𝑣2 + 𝑤3𝑣3

= 𝑢 , 𝑣 + 𝑤 , 𝑣

= 𝑢 ∙ 𝑣 + 𝑤 ∙ 𝑣

Since 𝑢 , 𝑤 = 𝑢2𝑤2 + 𝑢3𝑤3

= 𝑘 𝑤2𝑢2 + 𝑤3𝑢3

𝑘𝑢 , 𝑤 = 𝑘𝑢2𝑤2 + 𝑘𝑢3𝑤3

= 𝑘 𝑤 , 𝑢

= 𝑘𝑤2 𝑢2 + 𝑘𝑤3𝑢3

𝑘𝑢 , 𝑤 = 𝑘 𝑤 , 𝑢

Let 𝑢 = 𝑢1, 𝑢2, 𝑢3 and 𝑤 = 𝑤1, 𝑤2, 𝑤3

𝑤 ,𝑤 ≥ 0 𝑎𝑛𝑑 𝑤 ,𝑤 = 0 𝑖𝑓𝑓 𝑤 = 0

𝑤 ,𝑤 = 𝑤 ∙ 𝑤 = 𝑤2𝑤2 + 𝑤3𝑤3

𝑤 ,𝑤 = 0 𝑖𝑓𝑓 𝑤2 = 𝑤3= 0

𝐵𝑢𝑡, 𝑤1 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑣𝑎𝑙𝑢𝑒, ≠ 0

∴ 𝑢 , 𝑤 = 𝑢2𝑤2 + 𝑢3𝑤3 are not inner product on 𝑅3

Symmetry

Additivity

Homogeneity

Positivity

Axiom

𝑢 , 𝑤 = 𝑢 ∙ 𝑤 = 𝑢1𝑤1 + 2𝑢2𝑤2 + 3𝑢3𝑤3

= 𝑤1𝑢1 + 2𝑤2𝑢2 + 3𝑤3𝑢3

= 𝑤 ∙ 𝑢

= 𝑤 , 𝑢

𝑢 , 𝑤 = 𝑤 , 𝑢

𝑢 + 𝑤 , 𝑣 = 𝑢 , 𝑣 + 𝑤 , 𝑣

Since 𝑢 , 𝑤 = 𝑢1𝑤1 + 2𝑢2𝑤2 + 3𝑢3𝑤3 𝑢 + 𝑤 , 𝑣 = 𝑢1 + 𝑤1 𝑣1

+2 𝑢2 + 𝑤2 𝑣2 + 3 𝑢3 + 𝑤3 𝑣3

= 𝑢 , 𝑣 + 𝑤 , 𝑣 = 𝑢 ∙ 𝑣 + 𝑤 ∙ 𝑣

Since 𝑢 , 𝑤 = 𝑢1𝑤1 + 2𝑢2𝑤2 + 3𝑢3𝑤3

= 𝑘 𝑤1𝑢1 + 2𝑤2𝑢2 + 3𝑤3𝑢3

𝑘𝑢 , 𝑤 = 𝑘𝑢1𝑤1 + 2𝑘𝑢2𝑤2 + 3𝑘𝑢3𝑤3

= 𝑘 𝑤 , 𝑢

𝑘𝑢 , 𝑤 = 𝑘 𝑤 , 𝑢

Let 𝑢 = 𝑢1, 𝑢2, 𝑢3 and 𝑤 = 𝑤1, 𝑤2, 𝑤3

𝑤 ,𝑤 ≥ 0 𝑎𝑛𝑑 𝑤 ,𝑤 = 0 𝑖𝑓𝑓 𝑤 = 0

𝑤 ,𝑤 = 𝑤 ∙ 𝑤 = 𝑤1𝑤1 + 2𝑤2𝑤2 + 3𝑤3𝑤3

𝑤 ,𝑤 = 0 𝑖𝑓𝑓𝑤1 = 𝑤2 = 𝑤3= 0

∴ 𝑢 , 𝑤 = 𝑢1𝑤1 + 2𝑢2𝑤2 + 3𝑢3𝑤3 are inner product on 𝑅3

= 𝑢1𝑣1 + 𝑤1 𝑣1 + 2𝑢2𝑣2 + 2𝑤2 𝑣2 +3𝑢3𝑣3 + 3𝑤3𝑣3

= 𝑢1𝑣1 + 2𝑢2𝑣2 + 3𝑢3𝑣3 + 𝑤1𝑣1 + 2𝑤2 𝑣2 + 3𝑤3𝑣3

= 𝑘𝑤1𝑢1 + 2𝑘𝑤2𝑢2 + 3𝑘𝑤3𝑢3

Symmetry

Additivity

Homogeneity

Positivity

Axiom

𝑢 , 𝑤 = 𝑢 ∙ 𝑤 = 𝑢1𝑤1 + 𝑢2𝑤2 − 𝑢3𝑤3

= 𝑤1𝑢1 + 𝑤2𝑢2 − 𝑤3𝑢3

= 𝑤 ∙ 𝑢

= 𝑤 , 𝑢

𝑢 , 𝑤 = 𝑤 , 𝑢

𝑢 + 𝑤 , 𝑣 = 𝑢 , 𝑣 + 𝑤 , 𝑣

Since 𝑢 , 𝑤 = 𝑢1𝑤1 + 𝑢2𝑤2 − 𝑢3𝑤3

𝑢 + 𝑤 , 𝑣 = 𝑢1 + 𝑤1 𝑣1 + 𝑢2 + 𝑤2 𝑣2 − 𝑢3 + 𝑤3 𝑣3

= 𝑢1𝑣1 + 𝑤1𝑣1 + 𝑢2𝑣2 + 𝑤2𝑣2 − 𝑢3 𝑣3 − 𝑤3𝑣3

= 𝑢 , 𝑣 + 𝑤 , 𝑣

= 𝑢 ∙ 𝑣 + 𝑤 ∙ 𝑣

Since 𝑢 , 𝑤 = 𝑢1𝑤1 + 𝑢2𝑤2 − 𝑢3𝑤3

= 𝑘 𝑤1𝑢1 + 𝑤2𝑢2 − 𝑤3𝑢3

𝑘𝑢 , 𝑤 = 𝑘𝑢1𝑤1 + 𝑘𝑢2𝑤2 − 𝑘𝑢3𝑤3

= 𝑘 𝑤 , 𝑢

𝑘𝑢 , 𝑤 = 𝑘 𝑤 , 𝑢

Let 𝑢 = 𝑢1, 𝑢2, 𝑢3 and 𝑤 = 𝑤1, 𝑤2, 𝑤3

𝑤 ,𝑤 ≥ 0 𝑎𝑛𝑑 𝑤 ,𝑤 = 0 𝑖𝑓𝑓 𝑤 = 0

𝑤 ,𝑤 = 𝑤 ∙ 𝑤 = 𝑤1𝑤1 + 𝑤2𝑤2 − 𝑤3𝑤3

𝑤 ,𝑤 ≥ 0,

𝑤12 + 𝑤2

2 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑤32

∴ 𝑢 , 𝑤 = 𝑤1𝑤1 + 𝑤2𝑤2 − 𝑤3𝑤3 are not inner product on 𝑅3

= 𝑢1𝑣1 + 𝑢2𝑣2 − 𝑢3 𝑣3 + 𝑤1𝑣1 + 𝑤2𝑣2 − 𝑤3𝑣3

= 𝑘𝑤1 𝑢1 + 𝑘𝑤2𝑢2 − 𝑘𝑤3𝑢3

𝑤12 + 𝑤2

2 − 𝑤32 ≥ 0

𝑤12 + 𝑤2

2 ≥ 𝑤32

𝑤12 + 𝑤2

2 < 𝑤32

𝑤 = 𝑤 ,𝑤 1/2 = 𝑤12 + 𝑤2

2 = 22 + (−5)2 = 29

𝑤 = 5(2) 2 + 2(−5)2 = 70

𝐴𝑤 =−2 32 7

2−5

=−19−31

𝐴𝑤 = (−19) 2 + (−31)2 = 1322

𝑑 𝑢 , 𝑤 = 𝑢 − 𝑤 = −1 − 3 2 + 3 − 5 2 = 20

𝑑 𝑢 , 𝑤 = 𝑢 − 𝑤 = 5 −1 − 3 2 + 2 3 − 5 2

= 𝑢 − 𝑤 , 𝑢 − 𝑤 1/2

= 𝑢 − 𝑤 , 𝑢 − 𝑤 1/2 = 88

𝑢 − 𝑤 = −1,3 − 3,5 = −4, −2

𝐴 ∙ 𝑢 − 𝑤 =−2 32 7

−4−2

=2

−22

𝐴 ∙ 𝑢 − 𝑤 = (2) 2 + (−22)2 = 488

𝐴 = 𝐴, 𝐴 1/2 =2 4−3 1

,2 4−3 1

1/2

= 22 + 42 + (−3)2+1 = 30

𝑑 𝐴, 𝐵 = 𝐴 − 𝐵 = 𝐴 − 𝐵, 𝐴 − 𝐵 1/2 = (2 + 4)2+(4 − 2)2+(−3 − 5)2+ 1 − 1 = 104

𝐴 = 𝐴, 𝐴 1/2 =6 −17 4

,6 −17 4

1/2

= 62 + (−1)2+72 + 42 = 102

𝑑 𝐴, 𝐵 = 𝐴 − 𝐵 = 𝐴 − 𝐵, 𝐴 − 𝐵 1/2 = (6 + 1)2+(−1 − 8)2+72 + (4 − 2)2

= 183

𝑝 = 𝑝 , 𝑝 1/2 = 𝑝 𝑥 𝑝 𝑥 𝑑𝑥1

0

1/2

= 3𝑥2 − 2 3𝑥2 − 2 𝑑𝑥1

0

= 9𝑥4 − 12𝑥2 + 4 𝑑𝑥1

0

=9𝑥5

5−

12𝑥3

3+ 4𝑥

0

1

=9

5

𝑝 − 𝑞 = 2𝑥2 − x − 2

𝑑 𝑝 , 𝑞 = 𝑝 − 𝑞

= 𝑝 − 𝑞 , 𝑝 − 𝑞 1/2 = (2𝑥2 − x − 2) 2𝑥2 − x − 2 𝑑𝑥1

0

1/2

= 4𝑥4 − 4𝑥3 − 7𝑥2 + 4𝑥 + 4 𝑑𝑥1

0

=52

15

𝑝 = 𝑝 , 𝑝 1/2 = 𝑝 𝑥 𝑝 𝑥 𝑑𝑥1

0

1/2

= 𝑥2 + 𝑥 + 1 𝑥2 + 𝑥 + 1 𝑑𝑥1

0

= 𝑥4 + 2𝑥3 + 3𝑥2 + 2𝑥 + 1 𝑑𝑥1

0

=𝑥5

5+

2𝑥4

4+

3𝑥3

3+

2𝑥2

2+ 𝑥

0

1

=37

10

𝑝 − 𝑞 = −4𝑥2 + 2x − 2

𝑑 𝑝 , 𝑞 = 𝑝 − 𝑞

= 𝑝 − 𝑞 , 𝑝 − 𝑞 1/2 = (−4𝑥2 + 2x − 2 ) −4𝑥2 + 2x − 2 𝑑𝑥1

0

1/2

=16𝑥5

5−

16𝑥4

4+

20𝑥3

3−

8𝑥2

2+ 4𝑥

0

1

=88

15

2𝑢 , 𝑣 − 𝑤 + 𝑣 , 𝑣 − 𝑤 = 2𝑢 , 𝑣 − 2𝑢 , 𝑤 + 𝑣 , 𝑣 − 𝑣 − 𝑤

= 2 𝑢 , 𝑣 − 2 𝑢 , 𝑤 + 𝑣 2 − 𝑣 − 𝑤 = 2 3 − 2 7 + 25 + 2 = 19

𝑢 , 𝑣 + 3𝑤 − 𝑣 , 𝑣 + 3𝑤 + 2𝑤 , 𝑣 + 3𝑤 = 𝑢 , 𝑣 + 𝑢 , 3𝑤 − 𝑣 , 𝑣 − 𝑣 , 3𝑤 + 2𝑤 , 𝑣 + 2𝑤 , 3𝑤

= 3 + 3 7 − 52 − 3(−2) + 2(−2) + 6(8)2= 385

2𝑢 + 𝑤 , 2𝑢 + 𝑤 1/2 = 2𝑢 , 2𝑢 + 𝑤 + 𝑤 , 2𝑢 + 𝑤 1/2 = 4 𝑢 , 𝑢 + 2 𝑢 , 𝑤 + 2 𝑤 , 𝑢 + 𝑤 ,𝑤 1/2

= 4(2)2 + 2(7) + 2(7) + 82 1/2 = 108

𝑢 − 3𝑣 + 𝑤 , 𝑢 − 3𝑣 + 𝑤 1/2

= 𝑢 , 𝑢 − 3𝑣 + 𝑤 − 3𝑣 , 𝑢 − 3𝑣 + 𝑤 + 𝑤 , 𝑢 − 3𝑣 + 𝑤 1/2 = ( 𝑢 , 𝑢 − 3 𝑢 , 𝑣 + 𝑢 , 𝑤 − 3 𝑣 , 𝑢 + 9 𝑣 , 𝑣 − 3 𝑣 , 𝑤 + 𝑤 , 𝑢 − 3 𝑤 , 𝑣

+ 𝑤 ,𝑤 )1/2

= (2)2−3 3 + 7 − 3 3 + 9 5 2 − 3 −2 + 7 − 3(−2) + 82 1/2 = 301

𝑢 , 𝑘𝑣 = 𝑘𝑣 , 𝑢 = 𝑘 𝑣 , 𝑢 = 𝑘 𝑣 , 𝑢

symmetry homogeneity symmetry

𝑢 − 𝑣 , 𝑤 = 𝑤 , 𝑢 − 𝑣 = 𝑤 , 𝑢 + 𝑤 ,−𝑣

= 𝑤 , 𝑢 + (−1) 𝑤 , 𝑣 = 𝑢 , 𝑤 − 𝑣 , 𝑤

symmetry additivity homogeneity symmetry

𝑢 , 𝑣 = 5𝑢1𝑣1 + 2𝑢2𝑣2 = 5 2 0 + 2 3 −1 = −6 = 6 = 36

𝑢 = 𝑢 , 𝑢 1/2 = 5(2)2 + 2(3)2 = 38

𝑣 = 𝑣 , 𝑣 1/2 = 5(0)2 + 2(−1)2 = 2

𝑢 ∙ 𝑣 = 38 2 = 76 compare

36 < 76

∴ 𝑢 , 𝑣 ≤ 𝑢 ∙ 𝑣

Cauchy-Schwarz inequality

𝑢 , 𝑣 =2 61 −3

,−3 14 2

= −6 + 6 + 4 − 6 = −2 = 2 = 4

𝑢 = 𝑢 , 𝑢 1/2 =2 61 −3

,2 61 −3

= 4 + 36 + 1 + 9 = 50

𝑣 = 𝑣 , 𝑣 1/2 =−3 14 2

,−3 14 2

= 9 + 1 + 16 + 4 = 30

𝑢 ∙ 𝑣 = 50 30 = 1500

compare

4 < 1500

∴ 𝑢 , 𝑣 ≤ 𝑢 ∙ 𝑣

Cauchy-Schwarz inequality

• Do as your exercise

𝑇ℎ𝑒𝑜𝑟𝑒𝑚 1.5.3(𝑏)

𝑣 =0 iff 𝑣 = 0

𝑣 = 𝑣 , 𝑣 1/2 = 𝑣12 + 𝑣2

2 + ⋯+ 𝑣𝑛2 ≥ 0

𝑣 = 𝑣 , 𝑣 1/2 = 𝑣12 + 𝑣2

2 + ⋯+ 𝑣𝑛2 = 0 iff 𝑣1 = 𝑣2 = ⋯ = 𝑣𝑛 = 0 → 𝑣 = 0

∴ 𝑣 =0 iff 𝑣 = 0

𝑇ℎ𝑒𝑜𝑟𝑒𝑚 1.5.3(𝑐)

𝑘𝑣 = 𝑘 𝑣

𝑘𝑣 = 𝑘𝑣 , 𝑘𝑣 1/2 = 𝑘 𝑣 , 𝑘𝑣 1/2 = 𝑘2 𝑣 , 𝑣 1/2 = 𝑘 𝑣 , 𝑣 1/2 = 𝑘 𝑣

𝑇ℎ𝑒𝑜𝑟𝑒𝑚 1.5.3(𝑓)

𝑑 𝑢 , 𝑣 =0 iff 𝑢 = 𝑣

= 𝑢1 − 𝑣12 + 𝑢2 − 𝑣2

2 + ⋯+ 𝑢𝑛 − 𝑣𝑛2 ≥ 0

𝑢1 = 𝑣1, … 𝑢𝑛 = 𝑣𝑛 → 𝑢 = 𝑣

𝑇ℎ𝑒𝑜𝑟𝑒𝑚 1.5.3(ℎ)

𝑑 𝑢 , 𝑣 ≤ 𝑑 𝑢 , 𝑤 + 𝑑 𝑤 , 𝑣

𝑑 𝑢 , 𝑣 = 𝑢 − 𝑣 = 𝑢 − 𝑣 , 𝑢 − 𝑣 1/2

𝑑 𝑢 , 𝑣 = 𝑢 − 𝑣 = 𝑢1 − 𝑣12 + 𝑢2 − 𝑣2

2 + ⋯+ 𝑢𝑛 − 𝑣𝑛2 = 0 iff

∴ 𝑑 𝑢 , 𝑣 =0 iff 𝑢 = 𝑣

𝑑 𝑢 , 𝑣 = 𝑢 − 𝑣 = 𝑢 + 𝑤 − 𝑤 − 𝑣 = 𝑢 − 𝑤 + 𝑤 − 𝑣 ≤ 𝑢 − 𝑤 + 𝑤 − 𝑣

= 𝑑 𝑢 , 𝑤 + 𝑑 𝑤 , 𝑣

∴ 𝑑 𝑢 , 𝑣 ≤ 𝑑 𝑢 , 𝑤 + 𝑑 𝑤 , 𝑣

-dr Radz