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2 answers:
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Answer:
a) No
b) No
c) No
d) No
Step-by-step explanation:
Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:
1. u+v∈V
2. u+v=v+u
3. (u+v)+w=u+(v+w).
4. Exist 0∈V such that u+0=u
5. For each u∈V exist −u∈V such that u+(−u)=0.
6. if c is an escalar and u∈V, then cu∈V
7. c(u+v)=cu+cv
8. (c+d)u=cu+du
9. c(du)=(cd)u
10. 1u=u
let's check each of the properties for the respective operations:
Let
Observe that
1. u+v∈V
2. u+v=v+u, because the adittion of reals is conmutative
3. (u+v)+w=u+(v+w). because the adittion of reals is associative
4.
5.
then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.
a)
6.
7.
8.
Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product
b) 6.
7.
8.
9.
10
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product
c) Observe that
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .
d) Observe that
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .