Why do these vectors not belong to the same vector space? I'm trying to verify that $W$ (being the set of all vectors in $\mathbb R^3$ whose third component is $-1$) is not a subspace of the vector space.
You can have a vector $(0,0,-1)$ and through a scalar of $4$, you can obtain vector $(0,0,-4)$.
I understand to prove a subspace one only needs to prove that set is closed under both addition and multiplication.
However, I don't understand, why $(0,0,-1)$ and $(0,0,-4)$ are not in the same vector space.
 A: A subspace has to contain the $0$-vector. All vectors of $W$ have $-1$ as their third component.
A: I think that by your definition of $W$, $W$ is not a vector space, although it is s subset of a vector space. It is not a vector space, since the set must be closed under addition, and this is clearly not closed, since two vectors $(a_1, a_2, -1)$ and $(b_1, b_2, -1)$ add to $(a_1 + b_1, a_2 + b_2, -2)$ thus the sum of any two elements of $W$ is not in $W$ since $W$ contains only elements of the form $(x_1, x_2, -1)$. Thus it is not closed under addition, nor as you point out, under scalar multiplication.
A: The differences of any two elements of $W$ is a sub-vector space $W_0$. $W$ itself is but a translated of this vector space with equation $z=0$ by the vector $(0,0,-1)$, and is called  an affine subspace with direction $W_0$.
A: Recall the axioms of a vector space. In particular, for any vector space $V$, there is an element $0 \in v$, such that $v+0=0+v=v$ for any $v \in V$. Obviously, your set $W = \{ (x_1,x_2,x_3) \in \Bbb R^3 \mid x_3 = -1\}$, doesn't contain such an element, thus can't be a vector space.
