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 $\R^3$.
Comments
Post a Comment