Is $W$ a subspace of the vector space? Question: $W$ is the set of all vectors in $\mathbb{R}^3$ whose first component is $-3$. Is $W$ a subspace of the vector space? If not, state why.
My Answer: $W$ is not a subspace of the vector space because it is not closed under addition and neither is it closed under scalar multiplication.
But how do I provide a counterexample to this question? I’m unsure of how to do it.
 A: As you note, given vector space $V$ over ground field $F$, the requirements for $W \subset V$ to be a subspace are as follows:

*

*$\vec{0} \in W$

*$w_1+w_2 \in W$ for all $w_1, w_2 \in W$

*$\lambda w \in W$ for all $w \in W$ and $\lambda \in F$

With $V = \mathbb{R}^3$ and $W \subset \mathbb{R}^3$ such that the first component of $w \in W$ is $-3$, it will suffice to show that any one of these three subspace axioms do not hold.
For example, it is easy to see that $\vec{0} \in \mathbb{R}^3 \notin W$ since the first component is not -3.
Alternatively, one could argue that with $w_1 = w_2 = \begin{bmatrix} -3 \\ 0 \\ 0 \end{bmatrix}$, we have $w_1 + w_2 = \begin{bmatrix} -6 \\ 0 \\ 0 \end{bmatrix} \notin W$.
As a final alternative, one could argue that with $w$ as above and $\lambda = 2$, we have $\lambda w = \begin{bmatrix} -6 \\ 0 \\ 0 \end{bmatrix} \notin W$.
Of course, any one of these examples (and there are many more) is sufficient to show that this $W$ is not a vector space, but perhaps they provide a useful illustration of how each of the axioms can fail.
