(b). Prove that axiom 2 is fails and justify your claims. eVector Space: A nonem

Question

(b). Prove that axiom 2 is fails and justify your claims.

eVector Space: A nonempty set V with operations of vector addition and scalar multiplication such that all vectors u, v, w in V and all scalars c, d satisfy 1. u+v is in V; 2. u+v=v+u; 3. uv+w) (u+v) w; 4. V has a zero vector such that u 0 u for all u in V; 5. Every u in V has an additive inverse-u in V such that u(-u 6. cu is in V 7, c(u + v) = cu + cv; 8, (c + d)u-: cu + du ; 9. c(du) (cd)u; 10, lu= u. 0; 4. Consider M22 with standard scalar multiplication, but addition redefined as

Explanation / Answer

a). Let A =

a

b

c

d

be an arbitrary vector/matrix in M2,2. Also, let M =

0

-3

5

0

Then A+M = M+A = A so that M is the zero vector. Thus, axiom 4 is satisfied.

Let B =

-a

-b-6

-c+5

-d

Then A+B = B+A =

0

-3

5

0

so that B is the additive inverse of A. Thus, axiom 5 is satisfied.

b). Let A1 =

a1

b1

c1

d1

and A2 =

a2

b2

c2

d2

Since addition is commutative, hence a1+a2= a2+a1, b1+b2+3 = b2+b1+3, c1+c2-5= c2+c1-5 and d1+d2 = d2+d1. Hence, A1+A2 = A2+A1 so that addition is commutative in M2,2. Thus, axiom 2 is also satisfied.

However, axiom 7 fails, as k(A1+A2) is not equal to kA1+kA2 since kb1+kb2+3 is not equal to k(b1+b2+3). Also, kc1+kc2-5 is not equal to k(c1+c2-5).

a

b

c

d

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