How do I show that the following set is a vector space? Question
Determine if the following set is a vector space:
$$ \left\{ \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} : \begin{matrix} a - 2b &amp; = &amp; 4c \\ 2a &amp; = &amp; c + 3d\end{matrix}\right\}$$
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
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.
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.