If $U$ is a subspace of a vector space $V$, then it is also a vector space? If we define $U$ as a subspace of a vector space $V$ then does that mean that $U$ is also a vector space?
 A: Let $V$ be a vector space and let $W$ be a subset of $V$.  We have two equivalent definitions of a "subspace":
Definition 1:  If $W$ is a vector space, then $W$ is a subspace of $V$.
Definition 2:  If $W$ satisfies the following


*

*$W$ is non-empty;

*$W$ is closed under vector addition;

*$W$ is closed under scalar multiplication,


then $W$ is a subspace of $V$.
[A third definition replaces "$W$ is non-empty" with "$W$ contains the zero vector" in Definition 2.]
We can check these are equivalent:  If $W$ is a vector space, then it satisfies the conditions of Definition 2, by definition of a vector space.  Conversely, if $W$ satisfies the conditions of Definition 2, we can check that all of the vector space axioms are hold.
A: Yes, this is trivially true if you use the usual definition of a subspace of a vector space $V$, which is a subset $U$ which is also a vector space under the vector addition and scalar multiplication of $V$.
If you are using some other definition (for instance 'a subspace $U$ of a vector space $V$ is a span of some subset of $V$'), then you may have some work to do to show that such a thing is a vector space. For instance in the above definition in terms of spans, one would need to show that a span of elements in a vector space is a vector space.
Suppose our definition of a subspace is the following:

Let $V$ be a vector space. $U$ is a subspace of $V$ if there exists a linear transformation $T\colon V\to V$ such that $U=\ker T=\{u\in V\mid T(u)=0\}$.

This is one possible definition of a subspace (for finite dimensional vector spaces anyway) which you might not have seen, but it can be shown to be equivalent to the usual definition. From this definition we can show that a subspace is a vector space with the inherited vector addition and scalar multiplication.
Indeed, $T(0)=0$ because $T$ is linear and so $0\in U$. Also, for all $u,v\in U$ we have $T(u)=T(v)=0$ so and so $T(u+v)=0$ because $T$ is linear. It follows that $u+v\in\ker T$ and so $u+v\in U$.
Similarly, for any $\alpha\in K$ the base field, we have $T(\alpha u)=\alpha T(u)$ because $T$ is linear and so $T(\alpha u)=\alpha.0=0$ so $\alpha u\in\ker T\implies \alpha u\in U$.
Hence, $U$ satisfies all the properties of being a subspace of $V$ under the usual definition.
Now, to further show that $U$ is a vector space, we simply mention that $+|_U$ is associative, commutative and distributive over scalar multiplication because $+$ is associative, commutative and distributive over scalar multiplication in $V$. We also have that $1\in F$ is an identity of scalar multiplication restricted to $U$ because it is an identity of scalar multiplication on $V$, and scalar multiplication is compatible with field multiplication because it is in $V$. Similarly, because $0$ is in $U$, $U$ has an identity under addition given by the same zero. This proves that $U$ is a vector space itself.
