Tutorial MTH 3201 Linear Algebras

Tutorial 5

𝐴𝑥 = 0

3 4−1 2

𝑥1

𝑥2=

00

3 4−1 2

00

1 00 1

00

𝐸𝑅𝑂

∴ 𝑥1 = 𝑥2 = 0, 𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒, 𝑥 =00

𝐴𝑥 = 0

1 0 3−1 1 −12 4 0

𝑥1

𝑥2

𝑥3

=000

1 0 3−1 1 −12 4 0

000

𝐸𝑅𝑂 1 0 30 1 20 0 1

000

∴ 𝑥1 = 𝑥3 = 𝑥2 = 0, 𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒, 𝑥 =000

𝐴𝑥 = 0

−2 61 −3

𝑥1

𝑥2=

00

−2 61 −3

00

1 −30 0

00

𝐸𝑅𝑂

𝑥1 − 3𝑥2 = 0, 𝑥1 = 3𝑥2

Let 𝑥2 = 𝑡, 𝑥1 = 3𝑡

𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒, 𝑥 =3𝑡𝑡

= 𝑡31

𝐴𝑥 = 0

1 5 10 2 −36 −2 4

𝑥1

𝑥2

𝑥3

=000

1 5 10 2 −36 −2 4

000

𝐸𝑅𝑂 1 5 1

0 1 −3

20 0 1

000

∴ 𝑥1 = 𝑥3 = 𝑥2 = 0, 𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒, 𝑥 =000

𝐴𝑥 = 0

1 −2 32 1 −11 −3 4

𝑥1

𝑥2

𝑥3

=000

1 −2 32 1 −11 −3 4

000

𝐸𝑅𝑂 1 −2 3

0 1 −7

50 0 1

000

𝑥3 = 0,

𝑥2 −7𝑥3

5= 0 → 𝑥2 = 0,

𝑥1 − 2𝑥2 + 3𝑥3 → 𝑥1 = 0

𝑥 =000

, So there are no basis.

Dimension=0

1 −3 1−1 −5 13 1 −2

1

−1−2

000

𝐸𝑅𝑂

1 −3 10 1 −1/40 0 1

102

000

𝑥3 + 2𝑥4 = 0 → 𝑥3 = −2𝑥4 𝑥2 − 𝑥3/4 = 0 → 𝑥3 = 4𝑥2

𝑥1 − 3𝑥2 + 𝑥3 + 𝑥4 = 0 → 𝑥1 = 𝑥3/4

Let 𝑥3 = 𝑡, 𝑥1 =𝑡

4, 𝑥2 =

𝑡

4, 𝑥4 = −𝑡/2

𝑥 = 𝑡

1/41/41

−1/2

= 𝑡

−1/2−1/2−21

∴ 𝑣 =

−1/2−1/2−21

span solution space, dimension = 1

𝐴𝑥 = 𝑏

−5 23 7

𝑥1

𝑥2=

8−1

−5 23 7

8

−1

1 −2/50 1

−8/519/41

𝐸𝑅𝑂

𝑥2 =19

41, 𝑥1 = −

58

41

𝑏 is in column space of A.

𝑏 =81

=−58

41−53

+19

4127

linear combination of column space of A.

𝐴𝑥 = 𝑏 7 2 12 3 41 1 0

𝑥1

𝑥2

𝑥3

=28

−1

7 2 12 3 41 1 0

28

−1 𝐸𝑅𝑂 1 1 0

0 1 40 0 1

−110

59/21

𝑥3 =59

21, 𝑥2 =

−26

21, 𝑥1 =

5

21

𝑏 =28

−1=

5

21

721

−26

21

231

+59

21

140

𝐴𝑥 = 0

3 −1 01 7 02 4 0

000

𝐸𝑅𝑂 1 7 0

0 1 00 0 0

000

𝑥3 = 𝑡, 𝑥2 = 0, 𝑥1 = 0

𝑥 = 𝑡001

001

is a basis for nullspace

−1 3 5−6 4 1−3 −2 1

000

𝐸𝑅𝑂 1 −3 −5

0 1 29/140 0 1

000

𝑥3 = 0, 𝑥2 = 0, 𝑥1 = 0

𝑥 =000

There is no basis

1 2 −1−3 3 45 1 1

−212

000

𝐸𝑅𝑂

1 2 −10 1 1/90 0 1

−2

−5/91

000

𝑥4 = 𝑡, 𝑥3 = −𝑡, 𝑥2 =2𝑡

3, 𝑥1 = −𝑡/3

𝑥 =1

3𝑡

−12

−33

𝑣1 =

−12

−33

is the basis for the nullspace

7 1 61 4 52 2 −1

000

𝐸𝑅𝑂 1 0 0

0 1 00 0 1

000

𝑥 =000

There is no basis

𝐴𝑥 = 0

3 −1 01 7 02 4 0

𝐸𝑅𝑂 1 0 0

0 1 00 0 0

∴ r 1 and r 2 are bases for row space of A

−1 3 5−6 4 1−3 −2 1

𝐸𝑅𝑂 1 0 0

0 1 00 0 1

𝑟 1 = 1 0 0 , 𝑟 2 = 0 1 0

𝑐 ′1 =100

𝑎𝑛𝑑 𝑐 ′2 =010

𝑐 1 =312

𝑎𝑛𝑑 𝑐 2 =−174

∴ c 1 and c 2 are bases for column space of A

∴ r 1, r 2and r 3 are bases for row space of A

𝑟 1 = 1 0 0 , 𝑟 2 = 0 1 0 , 𝑟 3 = 0 0 1

𝑐 1 =−1−6−3

, 𝑐 2 =34

−2𝑎𝑛𝑑 𝑐 3 =

511

∴ c 1 , c 2 and c 3 are bases for column space of A

1 2 −1−3 3 45 1 1

−212

𝐸𝑅𝑂

1 0 00 1 00 0 1

1/3

−2/31

7 1 61 4 52 2 −1

𝐸𝑅𝑂 1 0 0

0 1 00 0 1

∴ r 1, r 2and r 3 are bases for row space of A

𝑟 1 = 1 0 0 1/3 , 𝑟 2 = 0 1 0 − 2/3 , 𝑟 3 = 0 0 1 1

𝑐 1 =1

−35

, 𝑐 2 =231

𝑎𝑛𝑑 𝑐 3 =−141

∴ c 1 , c 2 and c 3 are bases for column space of A

∴ r 1, r 2and r 3 are bases for row space of A

𝑟 1 = 1 0 0 , 𝑟 2 = 0 1 0 , 𝑟 3 = 0 0 1

𝑐 1 =712

, 𝑐 2 =142

𝑎𝑛𝑑 𝑐 3 =65

−1

∴ c 1 , c 2 and c 3 are bases for column space of A

𝐴𝑇 =3 1 2

−1 7 40 0 0

𝐸𝑅𝑂 1 15 10

0 1 −14/220 0 0

𝑐1 ′ =

100

, 𝑐2 ′ =

1510

𝑐1 =3

−10

, 𝑐2 =170

c1 and c2 form bases for column space ofAT

basis of C. S of AT= basis of R. S of AT

𝑟1 = 3 −1 0 , 𝑟2 = 1 7 0

r1 and r2 form bases of row space consisting entirely of row vector 𝐴

𝐴𝑇 =−1 −6 −33 4 −25 1 1

𝐸𝑅𝑂

1 6 30 1 11/40 0 1

𝑐1 ′ =

100

, 𝑐2 ′ =

610

, 𝑐3 ′ =

311/4

1

𝑐1 =−135

, 𝑐2 =−641

, 𝑐2 =−3−21

c1 , c2 and c3 form bases for column space of AT

basis of C. S of AT= basis of R. S of AT

𝑟1 = −1 3 5 , 𝑟2 = −6 4 1

r1 , r2 and r3 form bases of row space consisting entirely of row vector 𝐴

𝑟3 = −3 −2 1

𝐴𝑇 =

1 −3 52 3 1

−1−2

41

12

𝐸𝑅𝑂 1 −3 50 1 −100

00

10

𝑐1 =

12

−1−2

, 𝑐2 =

−3341

, 𝑐2 =

5112

c1 , c2 and c3 form bases for column space of AT

basis of C. S of AT= basis of R. S of AT

𝑟1 = 1 2 −1 −2 , 𝑟2 = −3 3 4 1

r1 , r2 and r3 form bases of row space consisting entirely of row vector 𝐴

𝑟3 = 5 1 1 2

𝐴𝑇 =7 1 21 4 26 5 −1

𝐸𝑅𝑂

1 4 20 1 4/90 0 1

𝑐1 =716

, 𝑐2 =145

, 𝑐2 =22

−1

c1 , c2 and c3 form bases for column space of AT

basis of C. S of AT= basis of R. S of AT

𝑟1 = 7 1 6 , 𝑟2 = 1 4 5

r1 , r2 and r3 form bases of row space consisting entirely of row vector 𝐴

𝑟3 = 2 2 −1

Row space based on reduced matrix form

Column space based on original matrix

3 0 33 −4 −15 −4 1

−2−11

𝐸𝑅𝑂 1 0 1

0 1 10 0 0

−2/3−1/4

1

Bases for row space A

𝑤1 = 1 0 1 −2/3 𝑤2 = 0 1 1 −1/4

𝑤3 = 0 0 0 1

−2 1 33 −1 3

−1 −5 2

42

−3 𝐸𝑅𝑂 1 5 −2

0 1 9/160 0 1

3

7/161

Bases for row space A

𝑤1 = 1 5 −2 3 𝑤2 = 0 1 9/16 7/16

𝑤3 = 0 0 1 1

0 1 01 1 0

−1 1 0

101

𝐸𝑅𝑂

1 1 00 1 00 0 0

011

Bases for row space A

𝑤1 = 1 1 0 0 𝑤2 = 0 1 0 1 𝑤3 = 0 0 0 1

Let 𝐴 =

−3 5 1−3 3 −130

−11

31

−5−791

, 𝐴𝑇 =

−3 −3 35 3 −11

−5−1−7

39

0111

𝐸𝑅𝑂

1 0 10 1 −200

00

00

1/2−1/2

00

𝑤1 =

1000

, 𝑤2 =

0100

w1, and w2 form bases for column space of R

𝑣1 =

−351

−5

, 𝑣2 =

−33

−1−7

v1, and v2 form bases for column space of𝐴𝑇

Vectors in the basis

For vectors not in the basis:

𝑤3 =

1−200

=

1000

+ (−2)

1100

𝑤3 = 𝑤1 − 2𝑤2

𝑤4 =

0−1/2

00

= 1/2

1000

+ (−1/2)

1100

𝑤4 = (1/2)𝑤1 − (1

2)𝑤2

∴ 𝑣3 = 𝑣1 − 2𝑣2 , 𝑣4 = (1/2)𝑣1 − (1/2)𝑣2

Let 𝐴 =

1 1 0−1 3 2−13

50

35

16

−2−1

, 𝐴𝑇 =

1 −1 −11 3 501

26

3−2

305

−1

𝐸𝑅𝑂

1 0 00 1 000

00

10

0001

∴ v1, v2, v3, v4 is a subset of S that forms a basis for the space spanned by the vector in S

= 𝑟𝑎𝑛𝑘 𝐴 = 2 = 𝑟𝑎𝑛𝑘 𝐴 = 2

= 𝑛 − 𝑟𝑎𝑛𝑘 𝐴 = 4 − 2 = 2

= 𝑚 − 𝑟𝑎𝑛𝑘 𝐴 = 4 − 2 = 2

= 𝑟𝑎𝑛𝑘 𝐴 = 3 = 𝑟𝑎𝑛𝑘 𝐴 = 3

= 𝑛 − 𝑟𝑎𝑛𝑘 𝐴 = 5 − 3 = 2

= 𝑚 − 𝑟𝑎𝑛𝑘 𝐴 = 7 − 3 = 4

= 𝑟𝑎𝑛𝑘 𝐴 = 0 = 𝑟𝑎𝑛𝑘 𝐴 = 0

= 𝑛 − 𝑟𝑎𝑛𝑘 𝐴 = 3 − 0 = 2

= 𝑚 − 𝑟𝑎𝑛𝑘 𝐴 = 3 − 0 = 4

𝑚 × 𝑛

𝑛𝑢𝑙𝑙𝑖𝑡𝑦 + 𝑟𝑎𝑛𝑘 = 𝑛

𝑛𝑢𝑙𝑙𝑖𝑡𝑦 + 𝑟𝑎𝑛𝑘 = 𝑚

𝐴 =

𝑎11

𝑎21

𝑎31

𝑎12

𝑎22

𝑎32

… … . . 𝑎16

… … . . 𝑎26

… … . . 𝑎36𝑎41 𝑎42 … … . . 𝑎46

𝑎51 𝑎52 … … . . 𝑎56

𝑐𝑜𝑙𝑢𝑚𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝐴:

𝑐1 =

𝑎11

𝑎21𝑎31

𝑎41

𝑎51

, 𝑐2 =

𝑎12

𝑎22𝑎32

𝑎42

𝑎52

,……., 𝑐6 =

𝑎16

𝑎26𝑎36

𝑎46

𝑎56

Consider the equation 𝐾1𝑐 1 + 𝐾2𝑐 2+ 𝐾3𝑐 3+….+ 𝐾6𝑐 6 = 0

Then, express both sides of these equation in term of components:

𝐾1𝑎11 + 𝐾2𝑎12 + 𝐾3𝑎13 + ⋯ + 𝐾6𝑎16 = 0

𝐾1𝑎21 + 𝐾2𝑎22 + 𝐾3𝑎23 + ⋯ + 𝐾6𝑎26 = 0

𝐾1𝑎31 + 𝐾2𝑎32 + 𝐾3𝑎33 + ⋯ + 𝐾6𝑎36 = 0

𝐾1𝑎41 + 𝐾2𝑎42 + 𝐾3𝑎43 + ⋯ + 𝐾6𝑎46 = 0

𝐾1𝑎51 + 𝐾2𝑎52 + 𝐾3𝑎53 + ⋯ + 𝐾6𝑎56 = 0

This is a homogenous system with 5 equations and 6 unknowns.

Then, there exists nontrivial solution. Therefore c1 , c2 , … , c6 are linearly dependent.

𝑟𝑎𝑛𝑘 7𝐴 = 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑜𝑤 𝑠𝑝𝑎𝑐𝑒 7𝐴 = 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑜𝑤 𝑠𝑝𝑎𝑐𝑒 𝐴 = 𝑟𝑎𝑛𝑘 𝐴

𝐿𝑒𝑡 𝐴 = 𝑛 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 𝐴 + 𝑟𝑎𝑛𝑘 𝐴 = 𝑛𝑜. 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 = 𝑛

𝑛𝑢𝑙𝑙𝑖𝑡𝑦 𝐴 = 𝑛 − 𝑟𝑎𝑛𝑘 𝐴 = 𝑛 − 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑜𝑤 𝑠𝑝𝑎𝑐𝑒 𝐴

= 𝑛 − 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑜𝑤 𝑠𝑝𝑎𝑐𝑒 𝐴𝑇

= 𝑛 − 𝑟𝑎𝑛𝑘 𝐴𝑇

= 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 𝐴𝑇

-dr Radz