Question

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

Answer 1

Answer:

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.

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:

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.