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).
Your answer contains quite a few leaps of logic I'm afraid. Let's go over your arguments in detail.
$\rightarrow$ If $b=0$ then the system has at least one solution
which is a trivial one
This is correct.
or, moreover, infinite number of solutions (including the trivial one).
While this is true for infinite fields, it is not true for a finite field.
And what's worse is that you are just asserting this without any proof. You'll have to give an argument for why this is the case.
Clearly, in both cases the solutions set is a linear subspace of $\mathbb R^n$
True (and obvious) if $0$ is the only solution. But there are plenty of infinite subsets of $\mathbb R^n$ that are not subspaces.
(it includes the null space and closed under additivity and scalar multiplication).
This is the part you actually need to show.
Alternatively,
If $b\neq 0$ then has no solution or has a unique solution, both way is a vector of constant terms which don't satisfies the condition of a linear subspace.
In general there might be a lot of solutions of $Ax=b$ even if $b\neq 0$. For example, if $A=\left(\begin{matrix} 1 & 0\end{matrix}\right)$ and $b=1$ then all vectors whose first component is $1$ are solutions. In general the solutions of such an equation form a so called affine subspace.
So how do we prove it? Let us call the set of solutions $S$. We want to show that $S$ is a linear subspace if and only if $b=0$.
$\Rightarrow$: Let us assume that $S$ is a linear subspace. Then $0\in S$ which means that $A0=b$. Therefore $0=A0=b$.
$\Leftarrow$: Suppose $b=0$ we want to show that $S$ is a linear subspace. So we need to show that $S$ is closed under addition and scalar multiplication. For addition assume that $x,y\in S$. This means that $Ax=0$ and $Ay=0$. Therefore $A(x+y) = Ax +Ay=0+0=0$, which means that $x+y\in S$.
I'll leave it up to you to show that $S$ is also closed under scalar multiplication.
Best Answer
From the question, we know that $W$ is closed under addition and scalar multiplication. So all we have to do to prove that $W$ is a subspace is to show that $W$ contains the zero vector. So, if $x$ is in $W$, and $ax$ is in $W$, then we can take $a=-1$. By this, we have -$x$ is in $W$ (Closure under scalar multiplication). Now, because we know that $W$ is closed under addition, we have $x+(-x)=x-x=0$, which is in $W$, thus showing that $W$ must contain the zero vector.