Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Solution:
$S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$.
S is certainly nonempty.
Let $x, y \in S$. Then for some $r, s, u,v \in \R$,
$x = (r-2s,3r+s,s)$ and $y = (u-2v,3u+v,v)$.
Hence,
$x + y = (r-2s,3r+s,s) + (u-2v,3u+v,v)$
$= (r-2s+u-2v,3r+s+3u+v,s+v) = ((r+u)-2(s+v),3(r+u)+(s+v),(s+v))$
$= (k-2l,3k+l,l)$,
where $k = r + u, l=s+v$.
Consequently, $S$ is closed under addition.
Further, if $c \in \R$, then
$cx = c(r-2s,3r+s,s) = (c(r-2s),c(3r+s),cs) $
$= (cr-2cs,3cr+cs,cs) = (a-2b,3a+b,b)$,
where $a = cr,b=cs$.
Therefore $S$ is also closed under scalar multiplication.
It follows that $S$ is a subspace of $\R^3$.