I was asked the following question:
Determine if the following set is a vector space:
$$W=\left\{\left[\begin{matrix}p\\q\\r\\s\\\end{matrix}\right]:\begin{matrix}-3p+2q=-s\\p=-s+3r\\\end{matrix}\right\}$$
I know the answer is yes and you can show it by showing that W is a subspace of $\mathbb{R}_4$. But, I have no idea how to show that, or in general how to determine if a set is a vector space. I am interested in understanding so that I can apply it to future questions, not just so that I can answer this question.
Best Answer
A general approach to show that a set $M$ is a vector space is to find an already known vector space $(X,K,+,\cdot)$, where $X$ is the set of vectors, $K$ a field and $M\subset X$.
Then $M$ is a vector subspace (and thus a vector space), under the operations of $X$, if for every $x,y\in M$ and any $\alpha , \beta \in K$ we have $\alpha \cdot x + \beta \cdot y\in M$. You can check that if this is true then $M$ is vector space.
So to your specific example, and I will refer to $W$ as the set you have given:
We have $W\subset \mathbb{R}^{4}$, so let us endow $W$ with the operations of the vector space $\mathbb{R}^{4}$. Pick any $\alpha,\beta \in \mathbb{R}$ and let $$x = \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix} \quad y = \begin{bmatrix}y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\end{bmatrix}$$ be two arbitrary members of $W$, then: $$\alpha x + \beta y = \begin{bmatrix}\alpha x_{1} + \beta y_{1}\\ \alpha x_{2} + \beta y_{2}\\ \alpha x_{3} + \beta y_{3}\\ \alpha x_{4}+\beta y_{4}\end{bmatrix}.$$
We need to check that this new vector is indeed in $W$. From the description of $W$ this would mean that $$-3(\alpha x_{1} + \beta y_{1})+2(\alpha x_{2} + \beta y_{2})=-(\alpha x_{4} + \beta y_{4}) \tag{1}$$ and also $$(\alpha x_{1} + \beta y_{1})=-(\alpha x_{4} + \beta y_{4})+3(\alpha x_{3} + \beta y_{3})\tag{2}.$$ Starting with $(1)$ we know that since $x,y\in W\subset \mathbb{R}^{4}$ that \begin{align*}-3(\alpha x_{1} + \beta y_{1})+2(\alpha x_{2} + \beta y_{2})&=\alpha(-3x_{1}+2x_{2})+\beta(-3y_{1}+2y_{2})\\ &= \alpha(-x_{4})+\beta (-y_{4})\\ &= -(\alpha x_{4}+ \beta y_{4}). \end{align*}
We now go on to $(2)$: \begin{align*}(\alpha x_{1} + \beta y_{1}) &= \alpha(x_{1})+ \beta(y_{2})\\ &=\alpha(-x_{4}+3x_{3})+\beta(-y_{4}+3y_{3})\\ &=-(\alpha x_{4} + \beta y_{4})+3(\alpha x_{3}+\beta y_{3}).\end{align*}
We therefore conclude, since $(1)$ and $(2)$ are satisfied, that $\alpha x + \beta y \in W$, $W$ is a vector subspace of $\mathbb{R}^{4}$ and therefore $W$ is a vector space.
I will end on that you can check that $\alpha \cdot x\in W$ and then $x+y\in W$ respectively as long as you let the vectors and scalars be arbitrary members of their respective sets. For more examples I recommend the wikipedia page on Linear subspace.