Respuesta :
Answer:
(a) Let [tex]U\subset V[/tex] be a subset of a vector space [tex]V[/tex]. [tex]U[/tex] is a subspace of [tex]V[/tex] if and only if the following two conditions hold:
i) [tex]U[/tex] is closed under the sum operation. That is to say, [tex]u_{1}+u_{2}\in U[/tex] whenever [tex]u_{1},u_{2}[/tex] are elements of [tex]U[/tex].
ii) [tex]U[/tex] is closed under the scalar multiplication. That is to say, [tex]\lambda u\in U[/tex] whenever [tex]u\in U[/tex] and [tex]\lambda \in \mathbb{R}[/tex]
Step-by-step explanation:
For the part (b) we have the set of all sequences of the form[tex]\bar{u}=(v,1,0,1,0,1,...)[/tex], where [tex]v\in \mathbb{R}[/tex]. Observe the if you multiply any sequence of this form by and scalar [tex]\lambda\neq 1[/tex] then the sequence stops being like the given form. For example, let [tex]\lambda=5[/tex]. Then:
[tex]5 \bar{u}=(5v,5,0,5,0,5,...)[/tex]
This implies that the set under consideration is not closet under scalar multiplication, which implies that the set is not a subspace of the vector space of all sequences.
