Let V= the set of all ordered triples of real numbers with the operations define
ID: 1948776 • Letter: L
Question
Let V= the set of all ordered triples of real numbers with the operations defined below.(i) For u =(u1,u2,u3), and v =(v1,v2,v3) in V, define u + v =(u1,u2,u3) + (v1,v2,v3) = (u1+v1,u2+v2,u2+v3)
(ii) For u =(u1,u2,u3) in v and real number k, define k*u = (k^2*u1, k^2*u2, k^2*u3)
Determine if V is a vector space under the given operations. If you determine that V is not a vector space, then list all the axioms that fail to hold. YOU MUST GIVE AN EXAMPLE TO SHOW THAT THE PROPERTY FAILS! SHOW REASONING FOR ALL FAILS!
Explanation / Answer
According to the definition of vector spacea,the set V is a vector space over the field F of real numbers if the following axioms are satisfied.They are as follows:
1.u + (v + w) = (u + v) + w
2.u + v = v + u
3.There exists an element 0 V, called the zero vector, such that v + 0 = v for all v V.
4.For every v V, there exists an element v V, called the additive inverse of v, such that v + (v) = 0
5.a(u + v) = au + av
6.(a + b)v = av + bv
7.a(bv) = (ab)v
8.1v = v, where 1 denotes the multiplicative identity in F.
All the axioms are satisfied except 6 since (a+b)^2=a^2+b^2+2ab which when multiplied to v according to the definition of multiplication in V is not equal to av+bv.
Also if we change the field F V might not be a vector space over F because in that case kv may not belong to V for some real number k.
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