Question

Suppose U is a nonempty subset of the vector space V over eld F. Prove that U is a

subspace if and only if cv + w 2 U for any c 2 F and any v;w 2 U

Answer 1

The additive identity of
`V`, denoted here by
`0_(V)`, must be an element of
`U`. With this in mind and the provided properties you can prove it as follows.

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:

1. Conmutativity.
2. Associativity
3. Additive Identity
4. Inverse additive
5. Multiplicative identity
6. Distributive properties.

Now, if you have that
`V` is a vector space over a field
`\mathbb{K}` and
`U\subset V` is a subset that contains the additive identity
`e=0_(V)` then
`U` and
`cv+w \in U` provided that
`u,v\in U, c\in \mathbb{K}`, then
`U` 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
`U` are elements of V. Then
`U` is a subspace of
`V`.

Now if we know that
`U` is a subspace of
`V` then
`U` is a vector space, and clearly it satisfies the properties
`cv+w\in U` whenever
`v,w\in U, c\in \mathbb{K}` and
`0_(V)\in U`.

This is an useful criteria to determine whether a given set is subspace of a vector space.