Pinter's Abstract Algebra, Chapter 28, Exercise B2.

Question

Prove that the set of all $(x,y,z)\in\Bbb{R}^3$ which satisfy the pair of equations $ax+by+c=0,dx+ey+f= 0$ is a subspace of $\Bbb{R}^3$.

Is there a typo, in that the pair of equations should be $ax+by+cz=0,dx+ey+fz= 0$ instead? Otherwise, the subset is not closed with respect to scalar multiplication and therefore not a subspace.

Answer 1

Yes, it's likely there's a typo and Pinter in fact meant to include a $z$ in the last term. Notice however, that even if they really meant the original equations (without the $z$) these could still be solved simultaneously in $\mathbb{R}^3$. We would simply need to acknowledge the lack of dependence on $z$. In other words, we could set this up as a linear equation:

$$ \left [ \begin{array}{ccc} a & b & 0 \\ d & e & 0 \\ \end{array} \right ] \left [ \begin{array}{c} x\\y\\z\\ \end{array} \right ] \;\; = \;\; \left [ \begin{array}{c} -c \\ -f \\ \end{array} \right ]. $$

This would impose constraints on $x$ and $y$, but would leave $z$ unconstrained.

Answer 2

Another way of putting it is that unless $c=f=0$, in which case we have the intersection of two planes through the origin, we don't have a subspace at all.

That is, the planes must go through the origin, since every vector space contains zero.