Tutorial MTH 3201Linear Algebras

Tutorial 1

1. Determine whether the given set V with the given operations is a vector space or not. For those that are NOT, list all axioms that fail to

hold.

Let

(a) V is the set of all 2 × 2 non-singular matrices. The operations of addition and scalar multiplication are the standard matrix operations.

V=2 × 2 non-singular matrices Operation=-standard matrix operations.

AXIOM 1:

Suppose 1 2 1 2

3 4 3 4

u u v vu and v

u u v v

1 2 1 2 1 1 2 2

3 4 3 4 3 3 4 4

u u v v u v u vu v

u u v v u v u v

V

Satisfied

AXIOM 2: u v v u

1 2 1 2 1 1 2 2

3 4 3 4 3 3 4 4

u u v v u v u vu v

u u v v u v u v

v u

1 2 1 2

3 4 3 4

u u v v

u u v v

1 1 2 2

3 3 4 4

v u v u

v u v u

VSatisfied

𝑉={[𝑎1 𝑎2

𝑎3 𝑎4]|𝑎1 ,𝑎2 ,𝑎3 ,𝑎4∈ℜ }

AXIOM 3:

1 2 1 2 1 2

3 4 3 4 3 4

( )u u v v w w

u v wu u v v w w

V

Satisfied

1 2 1 1 2 2

3 4 3 3 4 4

u u v w v w

u u v w v w

1 1 1 2 2 2

3 3 3 4 4 4

u v w u v w

u v w u v w

1 2 1 2 1 2

3 4 3 4 3 4

( )u u v v w w

u v wu u v v w w

1 1 2 2 1 2

3 3 4 4 3 4

u v u v w w

u v u v w w

1 1 1 2 2 2

3 3 3 4 4 4

u v w u v w

u v w u v w

V

( ) ( )u v w v u w

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )

(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

AXIOM 4:

1 2 1 2

3 4 3 4

0u u x x

uu u x x

Not

Satisfied

0 0 0Thereexist V such that u u u

1 1 2 2

3 3 4 4

u x u x

u x u x

1 1 1 1, 0u x u then x Then, 0 V

Let1 2

3 4

0 ;x x

x x

0u u

1 2

3 4

u u

u u

1 2

3 4

u u

u u

2 2 2 2, 0u x u then x

3 3 3 3, 0u x u then x

4 4 4 4, 0u x u then x

But 0 0 which is a singular matrix

AXIOM 5:

Not

Satisfied

s. t.u V u

Since, 0 V 0u u u u

AXIOM 7: k u v ku kv

Satisfied

AXIOM 6:

If is any scalar and then .k u V ku V

Satisfied 1 2

3 4

u uku k

u u

1 2

3 4

ku ku

ku ku

V

1 2 1 2

3 4 3 4

u u v vk u v k

u u v v

1 2 1 2

3 4 3 4

u u v vk ku u v v

ku kv V

AXIOM 8:

Satisfied

AXIOM 9:

Satisfied

k u ku u

k u k u

1 2

3 4

( )u u

ku u

k u

ku u

1 2 1 2

3 4 3 4

u u u uku u u u

1 1 2 2

3 3 4 4

ku u ku u

ku u ku u

1 2 1 2

3 4 3 4

ku ku u u

ku ku u u

1 2

3 4

u uk u k

u u

1 2

3 4

k u k u

k u k u

k u

CONCLUSION :

AXIOM 10:

Satisfied1 u

Hence, is is not a vector space

since Axiom 4 and Axiom 5 aren't satisfied.

V

1 2

3 4

1u u

u u

1 2

3 4

u u

u u

u

1 u u1

1

• i.e., one that has a matrix inverse (invertable)• A square matrix is nonsingular iff its

determinant is nonzero. Example:

V=2 × 2 non-singular matrices

0 0

0 0

0 V

31 2 3 1 2( , , )v v v v v v v (b)

AXIOM 1:

Suppose

1 2 3 1 2 3( , , ) and ( , , ) where ,u u u u v v v v u v V

1 2 3 1 2 3 1 1 2 2 3 3( , , ) ( , , ) ( , , )u v u u u v v v u v u v u v V Satisfied

AXIOM 2: u v v u

v u V

Satisfied1 2 3 1 2 3 1 1 2 2 3 3( , , ) ( , , ) ( , , )u v u u u v v v u v u v u v

1 1 2 2 3 3 1 2 3 1 2 3( , , ) ( , , ) ( , , )v u v u v u v v v u u u

ℜ 𝑣3

AXIOM 3:

1 2 3 1 2 3 1 2 3( ) ( , , ) ( , , ) ( , , )u v w u u u v v v w w w

V

Satisfied

( ) ( )u v w v u w

1 2 3 1 1 2 2 3 3( , , ) , ,u u u v w v w v w

1 1 1 2 2 2 3 3 3( , , )u v w u v w u v w

1 2 3 1 2 3 1 2 3( ) ( , , ) ( , , ) ( , , )u v w u u u v v v w w w

1 1 2 2 3 3 1 2 3, , ( , , )u v u v u v w w w

1 1 1 2 2 2 3 3 3( , , )u v w u v w u v w V

( ) ( )u v w v u w

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )

(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

AXIOM 4:

1 2 3 1 2 30 ( , , ) ( , , )u u u u x x x

Satisfied

0 0 0Thereexist V such that u u u

1 1 2 2 3 3( , , )u x u x u x

1 1 1 1, 0u x u then x

Then, 0 V

Let 1 2 30 ( , , );x x x 0u u

1 2 3( , , )u u u

1 2 3( , , )u u u

2 2 2 2, 0u x u then x 3 3 3 3, 0u x u then x

4 4 4 4, 0u x u then x

1 2 3( , , ) (0,0,0) 0x x x

AXIOM 5: s. t.u V u

1 2 3 1 2 3 1 2 3 1 2 3

1 1 2 2 3 3 1 1 2 2 3 3

1 2 3 1 2 3

0

( , , ) ( ( , , )) ( , , ) ( , , )

( , , ) ( , , )

( ( , , )) ( , , ) (0,0,0) 0

u u u u

u u u u u u u u u u u u

u u u u u u u u u u u u

u u u u u u

Since, u V Satisfied

AXIOM 7: k u v ku kv

Satisfied

AXIOM 6: If is any scalar and then .k u V ku V

Satisfied 1 2 3, ,ku k u u u 1 2 3( , , )ku ku ku V

1 2 3 1 2 3( , , ) ( , , )k u v k u u u v v v

ku kv V1 2 3 1 2 3( , , ) ( , , )k u u u k v v v

AXIOM 8:

Satisfied

AXIOM 9:

Satisfied

k u ku u

k u k u

1 2 3( ) , ,k u u u k u

ku u 1 2 3 1 2 3, , , ,k u u u u u u

1 1 2 2 3 3( , , )ku u ku u ku u

1 2 3 1 2 3( , , ) ( , , )ku ku ku u u u

1 2 3, ,k u k u u u

1 2 3( , , )k u k u k u

k u

CONCLUSION :

AXIOM 10:

Satisfied1 u

Hence, is a vector space.V

1 2 31 ( , , )u u u 1 2 3( , , )u u u u

1 u u 1 u u1

1~𝑢=1 (𝑢1,𝑢2 ,𝑢3 )

31 2 3 1 2( , , )v v v v v v v (c)

AXIOM 1:

Suppose

1 2 3 1 2 3( , , ) and ( , , ) where ,u u u u v v v v u v V

1 2 3 1 2 3 1 1 2 2 3 3( , , ) ( , , ) ( , , )u v u u u v v v u v u v u v V

Satisfied

AXIOM 2: u v v u

v u V Satisfied

1 2 3 1 2 3 1 1 2 2 3 3( , , ) ( , , ) ( , , )u v u u u v v v u v u v u v

1 1 2 2 3 3 1 2 3 1 2 3( , , ) ( , , ) ( , , )v u v u v u v v v u u u

ℜ

AXIOM 3:

1 2 3 1 2 3 1 2 3( ) ( , , ) ( , , ) ( , , )u v w u u u v v v w w w

V

Satisfied

( ) ( )u v w v u w

1 2 3 1 1 2 2 3 3( , , ) , ,u u u v w v w v w

1 1 1 2 2 2 3 3 3( , , )u v w u v w u v w

1 2 3 1 2 3 1 2 3( ) ( , , ) ( , , ) ( , , )u v w u u u v v v w w w

1 1 2 2 3 3 1 2 3, , ( , , )u v u v u v w w w

1 1 1 2 2 2 3 3 3( , , )u v w u v w u v w V

( ) ( )u v w v u w

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )

(~𝑢+~𝑣 )+~𝑤

~𝑢+ (~𝑣+~𝑤 )=(~𝑢+~𝑣 )+~𝑤

AXIOM 4:

1 2 3 1 2 30 ( , , ) ( , , )u u u u x x x

Not

Satisfied

0 0 0Thereexist V such that u u u

1 1 2 2 3 3( , , )u x u x u x

1 2Hence v .Then, 0 .v V

Let 1 2 30 ( , , );x x x 0u u

1 2 3( , , )u u u

1 2 3( , , )u u u

1 2 3( , , ) (0,0,0) 0x x x

AXIOM 5:

Since, 0 V Not

Satisfied

AXIOM 7: k u v ku kv

Satisfied

AXIOM 6: If is any scalar and then .k u V ku V

Satisfied 1 2 3, ,ku k u u u 1 2 3( , , )ku ku ku V

1 2 3 1 2 3( , , ) ( , , )k u v k u u u v v v

ku kv V1 2 3 1 2 3( , , ) ( , , )k u u u k v v v

AXIOM 8:

Satisfied

AXIOM 9:

Satisfied

k u ku u

k u k u

1 2 3( ) , ,k u u u k u

ku u 1 2 3 1 2 3, , , ,k u u u u u u

1 1 2 2 3 3( , , )ku u ku u ku u

1 2 3 1 2 3( , , ) ( , , )ku ku ku u u u

1 2 3, ,k u k u u u

1 2 3( , , )k u k u k u

k u

CONCLUSION :

AXIOM 10:

Satisfied1 u

Hence, is not a vector space because

Axiom 4 and Axiom 5 not satisfied.

V

1 2 31 ( , , )u u u 1 2 3( , , )u u u u

1 u u

2. 1

1 22

,x

v x x xx

1 3 2)

2 4 2i u v

) ( 1, 2) (3,4)a u v

1 1

2 2

1

1

u kuku k

u ku

1 1 1 1

2 2 2 2

u v u vu v

u v u v

1 2( 1) 1

11 1 3 3)2 1 13 3

(2) 13 3

ii u

1 3) 5 5 5 5

2 4

5( 1) 1 5(3) 1

5(2) 1 5( 4) 1

4 16 12

9 21 12

iii u v

ℜ

2 5(2) 1 11)5( ) 5

2 5( 2) 1 11iv u v

1 5( 1) 1 4) (2 5) 5 5

2 5(2) 1 9v u u

1 1) 2 3 2 3

2 2

2( 1) 1 5( 1) 1

2(2) 1 3(2) 1

3 2 1

5 5 0

vi u u

0 such that 0 for any .V u u u V (b) Find the object

1 1

2 2

Suppose 0 andx u

ux u

1 1 1

2 2 2

0x u u

ux u u

1 1 1

2 2 2

x u u

x u u

1 1 1 1

2 2 2 2

0

0.

00

0

Then x u u x

x u u x

V

0 in (b) exist,(c) If the object

1 1

2 2

0Suppose 0 , and

0

w uw u

w u

1 1

2 2

0

0

w uw u

w u

1 1

2 2

0

0

w u

w u

1 1 1 1

2 2 2 2

1

2

0

0 .

Then w u w u

w u w u

uw V

u

find the object such that 0 for any .w V u w u V

1 for each ?v v v V (d) Is

1

2

Supposev

vv

1

2

1 1v

vv

1 1

2 2

1( ) 1

1( ) 1

v v

v v

with the given two operation a vector space?v(e) Is

is not a vector space because

Axiom 8 is not satisfied.

V

11 2

2

,x

v x x xx

2 1 2(1) 2)

7 2 7 2 5i u v

) ( 2,7) (1, 2)a u v

1 1

2 2

u u kku k

u ku

1 1 1 1

2 2 2 2

u v u vu v

u v u v

1 3221 1 2 2)

7 1 72 2(7)

2 2

ii u

2 1) 3 3 3 3

7 2

3( 2)(1) 3 1

3(7) 3( 2)

1 4 4

21 6 15

iii u v

3. ℜ

2 3 2 2)3( ) 3( )

5 3(5) 15iv u v

2 2 7 5) (2 5) 7 7

7 7(7) 49v u u

2 2) 2 5 2 5

7 7

2 2 5 2

2(7) 5(7)

0 3 0

14 35 49

vi u u

1

0 such that 0 for any .V u u u V (b) Find the object

1 1

2 2

Suppose 0 andx u

ux u

1 1 1

2 2 2

0x u u

ux u u

1 1 1

2 2 2

x u u

x u u

1 1 1 1

2 2 2 2

1

0.

10

0

Then x u u x

x u u x

V

0 in (b) exist,(c) If the object

1 1

2 2

1Suppose 0 , and

0

w uw u

w u

1 1

2 2

1

0

w uw u

w u

1 1

2 2

1

0

wu

w u

1 1 11

2 2 2 2

1

2

11

0 .

1

Then wu wu

w u w u

uw V

u

find the object such that 0 for any .w V u w u V

( ) for each and , ?k v kv v v V k (d) Is

1

2

Supposev

vv

1 1

2 2

1 1 1 1 1 1

2 2 2 2 2 2

( )( ) ( )

( )

( )( )

v k vk v k

v k v

v v k v v k v vkv v k

v v kv v kv v

( )k v kv v

with the given two operation a vector space?v(e) Is

is not a vector space because

Axiom 8 is not satisfied.

V

1 2 1 2 1 2 1 2, ( ) , , ,v x u u v v u u v v

) ( 3, 2) ( 1,5)

( 3 ( 1),0) ( 4,0)

i u v

) ( 3,2) ( 1,5)a u v

1 2 1 2( , ) ( , )ku k u u ku ku 1 2 1 2 1 1( , ) ( , ) ( ,0)u v u u v v u v

1 1) ( 3,2)

2 21 1 3

( ( 3), (2)) ( ,1)2 2 2

ii u

) 2 2 2 3,2 2( 1,5)

( 6,4) ( 2,10)

( 8,0).

iii u v

4.ℜ

) 2( ) 2( 4,0) 8,0iv u v

) (2 5) 7 7( 3,2)

(7( 3),7(2)) ( 21,14)

v u u

) 2 5 2( 3,2) 5( 3,2)

( 6,4) ( 15,10)

( 21,0)

vi u u

0 such that 0 for any .V u u u V (b) Find the object

1 2 1 2Suppose 0 ( , ) and ( , )x x u u u

1 2 1 2 1 20 ( , ) ( , ) ( , )u x x u u u u

1 1 1 2( ,0) ( , )x u u u

1 1 1 1

2 2

0

0 0.

00

0

Then x u u x

u x

V

0 in (b) exist,(c) If the object

Since 0 not exist and then is not exist.w

find the object such that 0 for any .w V u w u V

( ) for each , and ?k v w kv kw v w V k (d) Is

1 2 1 2Suppose , ( , )v v v and w w w

1 2 1 2 1 2 1 2

1 1

1 2 1 2 1 2 1 2

1 1

( ) ( , ) ( , ) ( , ) ( , )

( ,0)

( , ) ( , ) ( , ) ( , )

( ,0)

k v w k v v k w w kv kv kw kw

kv kw

kv kw k v v k w w kv kv kw kw

kv kw

( )k v w kv kw

ℜ

with the given two operation a vector space?v(e) Is

is not a vector space since

0 is not exist.

V

From lecture note,5. Is the set W subspace of vector space V

5. Is the set W subspace of vector space V

2 3 2 3 2 3 2 3

2 3 2 3 2 3 2 3

2 2 3 3 2 2 3 3

2 3 2 3

( , , ) ( , , )

Axiom1

( ) ( , , ) ( , , )

(( ) ( ), , )

Axiom 6

k ( , , )

is subspace of

Let u u u u u v v v v v

u v u u u u v v v v

u v u v u v u v W

u k u u u u W

W V

31 2 3 1 2 3) , ,a W u u u u u u ℜℜ

5. Is the set W subspace of vector space V

1 2 1 2

1 2 1 2

1 1 2 2

1 2 1 2

1

( , ) ( , )

Axiom1

( ) ( , ) ( , )

( , )

Axiom 6

k ( , ) ( , )

If 0, then 0

is not subspace of

Let u u u v v v

u v u u v v

u v u v W

u k u u ku ku

k ku W

W V

2) , 0 ;b W x y x V ℜ ℜ

5. Is the set W subspace of vector space V

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 2

( , 2 ) ( , 2 )

Axiom1

( ) ( , 2 ) ( , 2 )

( , 2( ))

Axiom 6

k ( , 2 ) ( , 2 )

is subspace of

Let u u u v v v

u v u u v v

u v u v W

u k u u ku ku W

W V

2 2) , 2 ;c W x y y x V ℜ ℜ

5. Is the set W subspace of vector space V

11 1 1

12 2 2

1 11 2 1 1 2 2

11 2 1 2

1 11 1 1 1 1

( ) ... 0

( ) ... 0

Axiom1

( ) ( ) ... 0 ... 0

( ) ( ) ... 0

Axiom 6

k ( ) ( ... 0) ... 0

is subsp

n n

n n

n n n n

n n

n n n n

Let P x a x b x

P x a x b x

P x P x a x b x a x b x

a a x b b x W

P x k a x b x ka x kb x W

W

ace of V

2 2) (0) 0 ;d W p x P p V P

5. Is the set W subspace of vector space V

21 1

22 2

2 21 2 1 2

21 2

2 21 1 1

( ) 3 2

( ) 3 2

Axiom1

( ) ( ) 3 2 3 2

( ) 6 4

Axiom 6

k ( ) ( 3 2) 3 6

is not subspace of

Let P x a x x

P x a x x

P x P x a x x a x x

a a x x W

P x k a x x ka x kx W

W V

2) ( ) 3 2 ; ( )e W p x ax x a V P x ℜ

5. Is the set W subspace of vector space V

1 2 1 2

2 4 2 4

1 2 1 2 1 1 2 2

2 4 2 4 2 2 4 4

1 2 1 2

2 4 2 4

1

Axiom1

( )

Axiom 6

k

If 0, then 0

is not subspace o

u u v vLet u v

u u v v

u u v v u v u vu v W

u u v v u v u v

u u ku kuu k

u u ku ku

k ku W

W

f V

22) , ;a c

f W a c V Mc a

ℜ

5. Is the set W subspace of vector space V

1 2 1 2

2 2

1 2 1 2 1 1 2 2

2 2 2 2

1 2 1 2

2 2

0 0

Axiom1

0 0 (0 0)

Axiom 6

k0 0

is subspace of

u u v vLet u v

u v

u u v v u v u vu v W

u v u v

u u ku kuu k W

u ku

W V

22) , , ;0

a bg W a b c V M

c

ℜ

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