let u and v be distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a and b are nonzero scalrs, then both {u+v,au} and {au,bv} are also bases for V.
Please help, i dont even know where to start!!
well, i'll do the first one for you and you do the second one..
i) suppose $\displaystyle u+v$ and $\displaystyle au$ are linearly dependent. then there exists a nonzero scalar $\displaystyle k$ such that $\displaystyle u+v = kau$
then, $\displaystyle u+v - kau = (1-ka)u + v = 0$
this means that $\displaystyle v = (ka-1)u$ which contradicts the fact that {$\displaystyle u,v$} is a basis for $\displaystyle V$.
hence, $\displaystyle u+v$ and $\displaystyle au$ must be linearly independent.
ii) Let $\displaystyle w\in V$. since {$\displaystyle u,v$} is a basis for $\displaystyle V$, we have $\displaystyle w = mu + nv$ for some scalars $\displaystyle m$ and $\displaystyle n$ in $\displaystyle V$. In particular, let $\displaystyle m=M+Na$ and $\displaystyle n=M$ where $\displaystyle M$ and $\displaystyle N$ are scalars in $\displaystyle V$. Hence, $\displaystyle w = mu + nv = (M+Na)u + Mv = Mu + Nau + Mv = M(u+v) + Nau$. Thus, $\displaystyle w\in \mbox{span}(\{u+v, au\})$ implies $\displaystyle V\subseteq \mbox{span}(\{u+v, au\})$ since we have written $\displaystyle w$ as a linear combination of $\displaystyle \{u+v, au\}$. Clearly, $\displaystyle \mbox{span}(\{u+v, au\}) \subseteq V$. Therefore, $\displaystyle \mbox{span}(\{u+v, au\}) = V$.