Respuesta :
Answer:
The additive identity of [tex]V[/tex], denoted here by [tex]0_{V}[/tex], must be an element of [tex]U[/tex]. With this in mind and the provided properties you can prove it as follows.
Step-by-step explanation:
In order to a set be a vector space it is required that the set has two operations, the sum and scalar multiplication, and the following properties are also required:
- Conmutativity.
- Associativity
- Additive Identity
- Inverse additive
- Multiplicative identity
- Distributive properties.
Now, if you have that [tex]V[/tex] is a vector space over a field [tex]\mathbb{K}[/tex] and [tex]U\subset V[/tex] is a subset that contains the additive identity [tex]e=0_{V}[/tex] then [tex]U[/tex] and [tex]cv+w \in U[/tex] provided that [tex]u,v\in U, c\in \mathbb{K}[/tex], then [tex]U[/tex] is a closed set under the operations of sum and scalar multiplicattion, then it is a vector space since the properties listed above are inherited from V since the elements of [tex]U[/tex] are elements of V. Then [tex]U[/tex] is a subspace of [tex]V[/tex].
Now if we know that [tex]U[/tex] is a subspace of [tex]V[/tex] then [tex]U[/tex] is a vector space, and clearly it satisfies the properties [tex]cv+w\in U[/tex] whenever [tex]v,w\in U, c\in \mathbb{K}[/tex] and [tex]0_{V}\in U[/tex].
This is an useful criteria to determine whether a given set is subspace of a vector space.
