Determine the set is a subspace of vector space Determine whether the following sets $U$ is a subspace of the indicated vector space $V$. Justify your answer.
$$V =\mathbb{R}^3 , U =\{(a,b,c)|a,c \in\mathbb{R}\}.$$
How do we determine and justify? Is about finding the zero vector? 
 A: You want to find for which values of $b, U$ is a subspace of $V$. So what are the conditions to be a subspace? $U$ must verify:
$0 \in U \\
U \subset V\\
\forall x,y \in U, x+y \in U \\
\forall \lambda \in R, \forall x \in U, \lambda.x \in U$
Now what are the possible values of $b$ that makes U verify all those properties?
A: If $U = \{ (a,b,c) \mid a,b,c \in \mathbb{R} \}$ then $U = V$, trivial.
So I suppose what you are considering is the set $U := \{ (a,0,c) \mid a,c \in \mathbb{R} \}$.
Take a linear algebra book, take a field, and take a vector space over the field. Then in the book you will find that it can be shown that a subset of the vector space is a subspace if and only if (i) the zero of the vector space lies in the subset; (ii) the sum of any two points of the subset still lies in the subset; and (iii) the product of any point of the field and any point of the subset again lies in the subset.
Equipped with the above result, we can now manage to prove that $U$ is a subspace of $V$. Here we have $0_{V} \in U$. Moreover, if $(a,0,c), (b,0,d) \in U$ then $(a+b, 0, c+d) \in U$; and if $t \in \mathbb{R}$ and $(a,0,c) \in U$ then $t(a,0,c) \in U$. Thus we are done.
