Tutorial MTH 3201 Linear Algebras

Tutorial 6

𝐴, 𝐵 =1 00 1

,3 65 −3

= 1 3 + 0 6 + 0 5 + 1(−3) = 0

1.

∴ A and B are orthogonal

𝐴, 𝐵 =1 23 4

,4 32 1

= 1 4 + 2 3 + 3 2 + 4(1) = 20 ≠ 0

∴ A and B are not orthogonal

𝐴, 𝐵 =

= 0

≠ 0

orthogonal

not orthogonal

𝑢 , 𝑤 1 = 1 0 − 1 0 + 2 0 + 0 3 = 0

𝑢 , 𝑤 2 = 1 −3 − 1 3 + 2 3 + 0 −8 = 0

𝑢 , 𝑤 3 = 1 2 − 1 −2 + 2 3 + 0 1 = 10

∴ 𝑢 is not orthogonal to set vectors W

≠ 0

𝑢 , 𝑤 1 = 1 3 − 1 −3 + 2 −3 + 0 7 = 0

𝑢 , 𝑤 2 = 1 0 − 1 2 + 2 1 + 0 5 = 0

𝑢 , 𝑤 1 = 1 6 − 1 0 + 2 −3 + 0 6 = 0

∴ 𝑢 is orthogonal to set vectors W

The space W spanned by v 1, v 2 and v 3 is the same as the row space of the matrix

A =3 1 1

−4 4 52 6 7

and Ax = 0

3 1 1−4 4 52 6 7

000

𝐸𝑅𝑂 1 0 −1/16

0 1 19/160 0 0

𝑥2 +

19𝑥3

16= 0

𝑥1 −𝑥3

16= 0

→ 𝑥2 = −19𝑥3/16

→ 𝑥1= 𝑥3/16

Let 𝑥3 = 𝑡, 𝑥2 = −19𝑡/16, 𝑥1 = 𝑡/16, x = 𝑡

1/16−19/16

1

∴ Basic orthogonal component =1/16

−19/161

The space W spanned by v 1 and v 2 is the same as the row space of the matrix

A =1 2 − 33 6 − 9

and Ax = 0

1 2 − 33 6 − 9

00

𝐸𝑅𝑂

𝑥1 + 2𝑥2 − 3𝑥3 = 0 → 𝑥1 = −2𝑥2 + 3𝑥3

Let 𝑥3 = 𝑡, 𝑥2 = 𝑠, 𝑥1 = 3𝑡 − 2𝑠

x =3𝑡 − 2𝑠

𝑠𝑡

= 𝑡301

+ 𝑠−210

∴ Basic orthogonal components =301

and −210

1 2 − 30 0 0

00

𝑢 , 𝑣 = −1/ 2 1/ 6 + 0 −2/ 6 + 1/ 2 1/ 6 = 0

∴ 𝑢 𝑎𝑛𝑑 𝑣 are orthogonal on 𝑅3

𝑢 , 𝑣 = 0 0 + 1 0 + 0 1 = 0

∴ 𝑢 , 𝑣 𝑎𝑛𝑑 𝑤 are not orthogonal on 𝑅3

𝑢 , 𝑤 = 0 1/ 2 + 1 1/ 2 + 0 0 = 1/ 2

𝑣 , 𝑤 = 0 1/ 2 + 0 1/ 2 + 1 0 = 0

𝐴, 𝐵 =

= 0

≠ 0

Orthogonal and if 𝐴 = 𝐵 = 1

not orthogonal

orthonormal

not orthonormal

𝑢 , 𝑣 = 0 , 𝑢 𝑎𝑛𝑑 𝑣 are orthogonal on 𝑅3

𝑢 = 𝑢 , 𝑢 1/2 = −1/ 22

+ 0 + 1/ 22

= 1

𝑣 = 𝑣 , 𝑣 1/2 = 1/ 62

+ −2/ 62

+ 1/ 62

= 1

𝑢 𝑎𝑛𝑑 𝑣 are orthonormal on 𝑅3

∴ 𝑢 , 𝑣 𝑎𝑛𝑑 𝑤 are not orthogonal on 𝑅3, so u , v and w are not orthonormal on R3

𝑢 1, 𝑢 2 = 2 0 + 0 6 + 3 0 = 0

∴ u 1, u 2 and u 2 are orthogonal on R3

𝑢 1, 𝑢 3 = 2 −3 + 0 0 + 3 2 = 0

𝑢 2, 𝑢 3 = 0 −3 + 6 0 + 0 2 = 0

𝑣 1 =𝑢 1

𝑢 1=

2,0,3

22 + 02 + 32=

2

13 , 0,

3

13

𝑣 2 =𝑢 2

𝑢 2=

0,6,0

02 + 62 + 02= 0 , 1,0

𝑣 3 =𝑢 3

𝑢 3=

−3,0,2

32 + 02 + 22=

−3

13 , 0,

2

13

∴ The sets of v 1, v 2, v 3 are orthonormal sets

𝑢 1, 𝑢 2 = −1/5 −1/3 + −1/5 −1/3 + −1/5 2/3 = 0

∴ u 1, u 2 and u 2 are orthogonal on R3

𝑢 1, 𝑢 3 = −1/5 1/2 + −1/5 −1/2 + −1/5 0 = 0

𝑢 2, 𝑢 3 = −1/3 1/2 + −1/3 −1/2 + 2/3 0 = 0

𝑣 1 =𝑢 1

𝑢 1=

−1

3 ,

−1

3,−1

3

𝑣 2 =𝑢 2

𝑢 2=

−1

6 ,

−1

6,

2

3

𝑣 3 =𝑢 3

𝑢 3=

1

2 ,

−1

2, 0

∴ The sets of v 1, v 2, v 3 are orthonormal sets

𝑢 1, 𝑢 2 = 0 −3/5 + 0 4/5 + 1 0 = 0

𝑢 1, 𝑢 3 = 0 4/5 + 0 3/5 + 1 0 = 0

𝑢 2, 𝑢 3 = −3/5 4/5 + 4/5 3/5 + 0 0 = 0

∴ u 1, u 2 and u 2 are orthogonal on R3

𝑢1 = 𝑢1, 𝑢11/2 = 0 2 + 0 + 1 2 = 1

𝑢 2 = 𝑢 2, 𝑢 21/2 = −3/5 2 + 4/5 2 + 0 2 = 1

∴ S = u 1, u 2, u 3 is an orthonormal basis for the Euclidean space R3

𝑢 3 = 𝑢 3, 𝑢 31/2 = 4/5 2 + 3/5 2 + 0 2 = 1

𝑢 1 = 𝑢 2 = 𝑢 3 = 1

Then, express w = 2,1, −3 as a linear combination of vector S

w , 𝑢 1 = 2 0 + 1 0 + −3 1 = −3

w , 𝑢 2 = 2 −3/5 + 1 4/5 + −3 0 = −2/5

w , 𝑢 3 = 2 4/5 + 1 3/5 + 0 0 = 11/5

∴ w = −3u 1 −2

5u 2

+(11

5)u 3

∴ 𝐶𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑤 𝑠 = (−3, −2

5,11/5)

𝑣 1, 𝑣 2 = 1 2 + (−2)(1) + 3 −4 + −4 3 = 0

𝑣 1, 𝑣 3 = 1 −3 + (−2)(4) + 3 1 + −4 −2 = 0

𝑣 1, 𝑣 4 = 1 4 + (−2)(3) + 3 2 + −4 1 = 0

𝑣 2, 𝑣 3 = 2 −3 + (1)(4) + −4 1 + −3 −2 = 0

𝑣 2, 𝑣 4 = 2 4 + (1)(3) + −4 2 + −3 1 = 0

𝑣 3, 𝑣 4 = −3 4 + (4)(3) + 1 2 + −2 1 = 0

∴ S = v 1, v 2, v 3, v 4 is an orthogonal basis for the Euclidean space R4

w , 𝑣 1 = −5

w , 𝑣 3 = −5 w , 𝑣 4 = 15

w , 𝑣 2 = −5

𝑣 12 = 𝑣1, 𝑣1

1/2 2= 30

𝑣 22 = 𝑣2, 𝑣2

1/2 2= 30

𝑣 32 = 𝑣3, 𝑣3

1/2 2= 30

𝑣 42 = 𝑣4, 𝑣4

1/2 2= 30

𝑤 =𝑤 , 𝑣 1𝑣 1

2 𝑣 1 +𝑤 , 𝑣 2𝑣 2

2 𝑣 2

+𝑤 , 𝑣 3𝑣 3

2 𝑣 3 +𝑤 , 𝑣 4𝑣 4

2 𝑣 4

𝑤 =−5

30𝑣 1 −

5

30𝑣 2 −

5

30𝑣 3 +

15

30𝑣 4

∴ 𝑤 =−1

6𝑣 1 −

1

6𝑣 2 −

1

6𝑣 3 +

1

2𝑣 4

w , 𝑣 1 = −3

w , 𝑣 3 = 7 w , 𝑣 4 = 29

w , 𝑣 2 = 29

𝑣 12 = 𝑣1, 𝑣1

1/2 2= 30

𝑣 22 = 𝑣2, 𝑣2

1/2 2= 30

𝑣 32 = 𝑣3, 𝑣3

1/2 2= 30

𝑣 42 = 𝑣4, 𝑣4

1/2 2= 30

𝑤 =−3

30𝑣 1 +

29

30𝑣 2 +

7

30𝑣 3 +

29

30𝑣 4

𝑤 =𝑤 , 𝑣 1𝑣 1

2 𝑣 1 +𝑤 , 𝑣 2𝑣 2

2 𝑣 2

+𝑤 , 𝑣 3𝑣 3

2 𝑣 3 +𝑤 , 𝑣 4𝑣 4

2 𝑣 4

𝑏𝑎𝑠𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 𝑜𝑟𝑡ℎ𝑜𝑛𝑜𝑟𝑚𝑎𝑙 𝑏𝑎𝑠𝑖𝑠

𝑞 𝑛 =𝑣 𝑛𝑣 𝑛

Gram-Schmidt process

𝑆𝑡𝑒𝑝 1 ∶ 𝐿𝑒𝑡 𝑣 1 = 𝑢 1 , 𝑣 1 = 1,1, −1

𝑆𝑡𝑒𝑝 2 ∶ 𝐿𝑒𝑡 𝑣 2 = 𝑢 2 − 𝑝𝑟𝑜𝑗𝑤1𝑢 2 = 𝑢 2 −

𝑢 2, 𝑣 1𝑣 1

2 𝑣 1 = −1,1,0 − 0

𝑣 2 = −1,1,0

𝑆𝑡𝑒𝑝 3 ∶ 𝐿𝑒𝑡 𝑣 3 = 𝑢 3 − 𝑝𝑟𝑜𝑗𝑤2𝑢 3 = 𝑢 3 −

𝑢 3, 𝑣 1𝑣 1

2𝑣 1 −

𝑢 3, 𝑣 2𝑣 2

2𝑣 2

= 1, −2,1 − −2

3, −

2

3,2

3−

3

2, −

3

2, 0 =

1

6,1

6,1

3

v 1, v 2, v 3 is an orthogonal basis

Gram-Schmidt process:

𝑞 1 =𝑣 1𝑣 1

=1,1, −1

3=

1

3 ,

1

3,−1

3

𝑞 2 =𝑣 2𝑣 2

=−1,1,0

2=

−1

2,

1

2, 0

𝑞 3 =𝑣 3𝑣 3

=

16 ,

16 ,

13

1/6=

3

6 ,

1

6,

6

3

∴ q 1, q 2, q 3 is an orthonormal basis

𝑞 𝑛 =𝑣 𝑛𝑣 𝑛

Do as your exercise

𝑣 1 = 𝑢 1 = (1,1,1)

𝑢 , 𝑣 = 𝑢1𝑣1 + 2𝑢2𝑣2 + 3𝑢3𝑣3 , 𝑢 2, 𝑣 1 = 1 1 + 2 −1 1 + 3 0 1 = −1

𝑣 12 = 1 + 2 + 3 = 6

𝑆𝑡𝑒𝑝 1 ∶ 𝐿𝑒𝑡 𝑣 1 = 𝑢 1 ,

𝑆𝑡𝑒𝑝 2 ∶ 𝐿𝑒𝑡 𝑣 2 = 𝑢 2 − 𝑝𝑟𝑜𝑗𝑤1𝑢 2 = 𝑢 2 −

𝑢 2, 𝑣 1𝑣 1

2 𝑣 1 = 1, −1,0 +1

61,1,1

𝑣 2 =7

6, −

5

6,1

6

Gram-Schmidt process:

𝑢 3, 𝑣 2 = 17

6+ 2 0 −

5

6+ 3 0

1

6=

7

6

𝑢 3, 𝑣 1 = 1(1) + 2 0 1 + 3 0 1 = 1

𝑣 22 = 17/6

𝑆𝑡𝑒𝑝 3 ∶ 𝐿𝑒𝑡 𝑣 3 = 𝑢 3 − 𝑝𝑟𝑜𝑗𝑤2𝑢 3 = 𝑢 3 −

𝑢 3, 𝑣 1𝑣 1

2 𝑣 1 −𝑢 3, 𝑣 2𝑣 2

2 𝑣 2

= 1,0,0 −7/6

17/6

7

6, −

5

6,1

6−

1

61,1,1 =

6

17,

3

17,−4

17

v 1, v 2, v 3 is an orthogonal basis

𝑞 𝑛 =𝑣 𝑛𝑣 𝑛

𝑞 1 =𝑣 1𝑣 1

=1,1,1

6=

1

6 ,

1

6,

1

6

𝑞 2 =𝑣 2𝑣 2

=7

102,

−5

102,

1

102

𝑞 3 =𝑣 3𝑣 3

=6

102,

3

102,

−4

102

∴ q 1, q 2, q 3 is an orthonormal basis

-dr Radz