How do I show that the following set is a vector space? Question
Determine if the following set is a vector space:
$$ \left\{ \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} : \begin{matrix} a - 2b & = & 4c \\ 2a & = & c + 3d\end{matrix}\right\}$$
I have a clue that I'm suppose to use the axioms to prove that it is a vector space, like that there exists the zero vector, its closed under addition and multiplication but I have no clue how to go about it
 A: I am going to show a part of it just to give you the idea. You can do the rest. If $V_{1}$ and $V_{2}$ belongs to the vector space (S), we want to show that $V = V_{1}+V_{2}$ also belongs to that vector space. In other words, we want to show $V$ has the properties defined in the set.
$$ V_{1} = \begin{bmatrix} a_{1} \\ b_{1} \\ c_{1} \\ d_{1}\end{bmatrix}, \space  V_{2} = \begin{bmatrix} a_{2} \\ b_{2} \\ c_{2} \\ d_{2}\end{bmatrix} \longrightarrow V = \begin{bmatrix} a=a_{1}+a_{2} \\ b=b_{1}+b_{2} \\ c=c_{1}+c_{2} \\ d=d_{1}+d_{2}\end{bmatrix} $$
Since the vectors $V_{1}$ and $V_{2}$ belong to S, we can say:
$$ \text{Because $V_{1} \in S$}: \space a_{1} - 2b_{1} = 4c_{1} $$
$$ \text{Because $V_{2} \in S$}: \space a_{2} - 2b_{2} = 4c_{2} $$
Now, let's add Left-hand side together, and Right-hand side together:
$$ (a_{1}+a_{2}) - 2(b_{1}+b_{2}) = 4(c_{1}+c_{2}) $$
$$ a - 2b = 4c $$
Similarly, you can show $2a=c+3d$ can be satisfied as well.
Good Luck.
A: Consider the function $f(a,b,c,d)=(a-2b-4c,\,2a-c-3d)$ from $\Bbb R^4$ to $\Bbb R^2.$
Confirm that if $v_1,v_2\in \Bbb R^4$ and $r\in \Bbb R$
then $(**)\;f(v_1+rv_2)=f(v_1)+rf(v_2).$
Let $V=\{v\in \Bbb R^4: f(v)=(0,0)\}.$ Then
(i). $V\ne \emptyset$ (e.g.$(0,0,0,0)\in V$).
(ii). If  $v_1,v_2\in V$ and $r\in \Bbb R$ then $f(v_1+rv_2)=f(v_1)+rf(v_2)=(0,0)$ by $(** )$ so $v_1+rv_2\in V.$
And you're done.
A: Presumably you are using the usual operations of addition in the vector space and multiplication by real scalars.  You now need to verify the axioms of a vector space with this subset of $\Bbb R^4$ and these operations.  The same $0$ vector works.  You need to show that it is in your subset.  Once you do that, the fact that it is the additive identity is inherited from $\Bbb R^4$.  Now you need to show that it is closed under multiplication by scalars, so show that any vector in your set can be multiplied by a real number and stay in the set.  Addition works the same way:  you are to take two vectors in your set and show their sum is still in the set.  Just write it out and see where it takes you.
