Prove subset is not a subspace of $ \mathbb{R}^4$

Solution 1:

Without dividing, I think it is pretty clear that $(0,0,0,0)$ is in $S$. So that's not the problem. You can also check directly from the definition that if $(a, b, c, d)$ is in $S$, then the same is $t(a, b, c, d)$ for any real number $t$.

So the only problem is the additivity. You will need to find two elements $(a_1, b_1, c_1, d_1)$, $(a_2, b_2, c_2, d_2)$ in $S$, so that the addition

$$(a_1 +a_2, b_1+b_2, c_1+c_2, d_1+d_2)$$ is NOT in $S$: that is, you do not have

$$ -(d_1+d_2)(a_1+a_2) + (b_1+b_2)(c_1+c_2) = 0.$$

Solution 2:

A counterexample would be sufficient proof to show that this is not a subspace.

Take the case of: $$\begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix} \& \begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}$$ Both of these vectors would be in $S$ but their sum will not be since $-(1)(1)+(0)(0)\neq 0$. Since the addition property is violated, $S$ is not a subspace.