Answer:
Proof
Step-by-step explanation:
(a)
- We first have to show that [tex]0 \in W[/tex]. Since [tex]W_1, W_2[/tex] are subspaces, it holds that [tex]0 \in W_1, W_2[/tex], hence [tex]0 \in W[/tex].
- Second, we have to show that [tex]W[/tex] is closed under sum and scalar multiplcation. First, let us take [tex]a, b \in W[/tex].
Since a, b belong to [tex]W=W_1+W_2[/tex], we can find [tex]a_1,b_1 \in W_1[/tex] and [tex]a_2, b_2 \in W_2[/tex] such that
[tex]a=a_1+a_2\\\\b=b_1+b_2[/tex]
Therefore,
[tex]a+b=a_1+a_2+b_1+b_2=(a_1+b_1)+(a_2+b_2)[/tex]
since [tex]W_1, W_2[/tex] are vector subspaces of [tex]V[/tex], it holds that
[tex]a_1+b_1 \in W_1 \quad\text{and}\quad a_2 + b_2 \in W_2[/tex], which shows us that W is closed under sum.
On the other hand, let us take [tex]a\in M[/tex] and [tex]\lambda \in \mathbb{R}[/tex]. We already know that
[tex]a=a_1+a_2[/tex]
where [tex]a_1 \in W_1[/tex] and [tex]a_2 \in W_2[/tex]. Moreover,
[tex]\lambda a_1 \in W_1 \quad \text{and} \quad \lambda a_2 \in W_2[/tex],
hence
[tex]\lambda a =\lambda(a_1 + a_2) =\lambda a_1 + \lambda a_2[/tex]
belongs to [tex]W[/tex], which shows that [tex]W[/tex] is closed under scalar multiplication.
(b) It is clear that [tex]W_1, W_2 \subseteq W[/tex]. Moreover, since [tex]W_1, W_2[/tex] are subespaces of [tex]V[/tex], they are closed under sum and scalar multplication, this propertie is remains true for [tex]W_1,W_2[/tex] as subsets of [tex]W[/tex], which tells us that [tex]W_1,W_2[/tex] are subspaces of [tex]W[/tex].
