How to find a subspace in $\mathbb{R}^3$ I am having trouble finding if there is a subspace in $\mathbb{R}^3$ for the following question:
$$\{(x,1,0) \, | \, x \, \text{arbitrary numbers}\}$$
What are the steps to solve such a question?
Thanks!
EDIT: 
I am having trouble understanding why this question is being closed. Can someone please explain the reason?
 A: The way to test if a subset of a vector space is a subspace is to test:


*

*Is it closed under vector addition?

*Is it closed under scalar multiplication?

*Is it nonempty?


To be a subspace it must pass both tests, for all vectors and scalars.  This particular subset  fails both tests catastrophically, i.e. finding an example is very easy because anything you try will be an example.  Here are two, one for each (although only one is sufficient).
$(1,1,0)$ and $(2,1,0)$ are both in your set, but their sum $(3,2,0)$ is not.
$(2,1,0)$ is in your set, and $3$ is a real number, but the product $3(2,1,0)=(6,3,0)$ is not in your set.
A: Hint: Subspaces must contain the zero vector. Does this set do that?
To see why it contains the zero vector, notice that if $\vec a$ lies in the subspace, so does $-\vec{a}$. Hence we get $\vec a - \vec a = 0$ as an element.
A: A subspace of a vector space is a subset that is itself a vector space (using the addition and scalar multiplication operations inherited from larger vector space).  One of the parts of the definition of a vector space is that it must contain a zero vector, which is an additive identity.  It can be proven from the properties of "zero vector" that a vector space has exactly one zero vector.  So it is easy to show that a subspace must contain the zero vector from the larger vector space. In your example, your subset of $\mathbb{R}^3$ does not contain the zero vector, so it's not a subspace.
There is a standard test to determine whether a subset of a vector space is a subspace.  It contains 3 conditions.  Sometimes people say "the set is nonempty" is one of the conditions, and sometimes people say "the set contains the zero vector" is one of the conditions instead.  You can use either one because they work out to be equivalent if the subset must satisfy all three conditions.  It is probably easier in general to check whether the zero vector belongs to the set.
The point of my first paragraph is to emphasize that a subspace cannot be the empty set, and that to justify why a subspace must contain the zero vector, you do not need to use the fact that a subspace must be closed under addition and scalar multiplication.
A: Say $X = \{(x,1,0)|x \in \mathbb{R}\}$ and $Y = \{(y,1,0)|y \in \mathbb{R}\}$.
If $X,Y \in \mathbb{V}$, then we must test if the subset is closed under addition which means that $X+Y \in \mathbb{V}$.
In this case, $X + Y = \{(x+y,2,0):x,y \in \mathbb{R}\}$, which demonstrates that the subset is not closed under addition. If you look at the second entry of the sum of $X$ and $Y$, you can see that it is is 2 which clearly isn't 1. So, $X + Y \notin \mathbb{V}$. This condition alone means that the subset in not subspace of $\mathbb{V}$, but let's test scalar multiplication as well.
Let $a \in \mathbb{R}$ and $X \in \mathbb{V}$. In order for the set to be closed under scalar multiplication, there must be an $aX \in \mathbb{V}$.
$aX= :\{a(x,1,0):x \in \mathbb{R}\} = (a\cdot x, a\cdot1,a\cdot0)$.
It is clear by the above that if $a \neq 1$, then scalar multiplication will fail to remain in the subset, because $a\cdot1 = 1$, if and only if $a = 1$. Therefore, $aX \notin \mathbb{V}$.
The subset is not closed under addition, or scalar multiplication, therefore it is not a subspace.
