To show a subset is a subspace, you need to show three things:
- Show it is closed under addition.
- Show it is closed under scalar multiplication.
- Show that the vector $0$ is in the subset.
To show 1, as you said, let $w_{1} = (a_{1}, b_{1}, c_{1})$ and $w_{2} = (a_{2}, b_{2}, c_{2})$. Suppose $w_{1}$ and $w_{2}$ are in our subset. Then they must satisfy $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$. We need to check that $w_{1} + w_{2}$ is in our subset, that is, we need to check that if
$$w_{1} + w_{2} = (a_{1} + a_{2}, b_{1} + b_{2},c_{1} + c_{2})$$
then it satisfies that the first coordinate is greater than or equal to the second coordinate. Is $a_{1} + a_{2} \geq b_{1} + b_{2}$? Yes, by adding the two inequalities $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$ together. So $w_{1} + w_{2}$ is in our subset.
For 2, we need to show that if $\alpha$ is any scalar, then if $w_{1} = (a_{1}, b_{1}, c_{1})$ is in our subset, so is $\alpha w_{1}$. But $\alpha w_{1} = (\alpha a_{1}, \alpha b_{1}, \alpha c_{1})$. We know since $w_{1}$ is in our subset that $a_{1} \geq b_{1}$. We need to check that for any $\alpha$, $\alpha a_{1} \geq \alpha b_{1}$. This inequality is definitely true if $\alpha \geq 0$, but we need it to be true for all $\alpha$, and it's not, because if $\alpha$ is negative, then multiplying the inequality $a_{1} \geq b_{1}$ on both sides by a negative number makes the inequality flip. So we would get $\alpha a_{1} \leq \alpha b_{1}$ if $\alpha < 0$, which means the inequality doesn't hold. Thus, if $\alpha < 0$, then $\alpha w_{1}$ is not in our subset because it doesn't satisfy $\alpha a_{1} \geq \alpha b_{1}$.
So our subset is not a subspace because it doesn't satisfy 2 (it is not closed under scalar multiplication, because any negative scalar would cause this problem).
Let $W = \{ (x,y,z) \in \mathbb{R}^3 \;:\; x+2y-z=0\}$.
So geometrically your $W$ is the set of all point in $\mathbb{R}^3$ which lie on the plane whose equation is $x+2y-z=0$. This plane passes through the origin: $0+2(0)-0=0$ (i.e., $(0,0,0) \in W$) and is perpendicular to $[1\;2\;-1]$.
To show $W$ is a subspace of $\mathbb{R}^3$ you simply need to verify that $W$ is a subset of $\mathbb{R}^3$ (this is obvious). Then you need to make sure it is non-empty (this is accomplished by noting that $(0,0,0) \in W$). Then the real task is to show it is closed under vector addition and scalar multiplication.
Take $(x,y,z), (a,b,c) \in W$ and $s \in \mathbb{R}^3$. Since $(x,y,z) \in W$, you have $x+2y-z=0$ similarly for $(a,b,c)$. Notice that adding the equation involving $x,y,z$ and the equation involving $a,b,c$ yields an equation for the vector $(x+a,y+b,z+c)$. Thus $(x,y,z)+(a,b,c)=(x+a,y+b,z+c) \in W$. Similarly, scaling the equation involving $x,y,z$ by $s$ yields an equation for $(sx,sy,sz)$. Thus $s(x,y,z)=(sx,sy,sz) \in W$.
This will show that $W$ is a subspace of $\mathbb{R}^3$.
An easier way to see that $W$ is a subspace is to note that it is the nullspace (or kernel) of a matrix/transformation whose matrix is $[1\;2\;-1]$. Notice that
$$\begin{bmatrix} 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} =0$$
is exactly the criterion for belonging to $W$.
In the end, every subspace can be recognized to be a nullspace of something (or the column space/span of something).
Geometrically, subspaces of $\mathbb{R}^3$ can be organized by dimension:
- Dimension 0: The only 0-dimensional subspace is $\{(0,0,0)\}$
- Dimension 1: The 1-dimensional subspaces are lines through the origin. They are the span of a single non-zero vector. Or you can view these as the set of solutions of a pair of homogeneous equations (equations set equal to $0$).
- Dimension 2: The 2-dimensional subspaces are planes through the origin. They are the span of a pair of (linearly independent) vectors. Or you can view these are the set of solutions of a single homogeneous equation (your $W$ is exactly this).
- Dimension 3: The only 3-dimensional subspace of $\mathbb{R}^3$ is $\mathbb{R}^3$ itself.
Basically subspaces of $\mathbb{R}^n$ are "flat" things through the origin. Since your $W$ is one of those kinds of things, it's a subspace. :)
Best Answer
The number of axioms is subject to taste and debate (for me there is just one: A vector space is an abelian group on which a field acts). You should not want to distinguish by noting that there are different criteria. Actually, there is a reason why a subspace is called a subspace: It is also a vector space and it happens to be (as a set) a subset of a given space and the addition of vectors and multiplicataion by scalars are "the same", or "inherited" from that other space. So this way there is no real difference, and one should better introduce and define the notion of subspace per "vectorspace that is contained (the way I describe above) in a vector space" instead of "subset with operations that have some magical other properties". Rather the fact that "nonempty and closed under multiplication and addition" are (necessary and) sufficient conditions for a subset to be a subspace should be seen as a simple theorem, or a criterion to see when a subset of a vector space is in fact a subspace. It gives you a simple recipe to check whether a subset of a vector space is a supspace.