Problem 55: Let $S = \{x \in \R^2 : x = (2k,−3k), k \in \R \}$. Establish that $S$ is a subspace of $\R^2$.

Problem 55: Let $S = \{x \in \R^2 : x = (2k,−3k), k \in \R \}$. Establish that $S$ is a subspace of $\R^2$.

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