# Thread: Plz.. help with linear algebra

1. ## Plz.. help with linear algebra

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.

2. Originally Posted by mathlovet
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.

i) suppose $u+v$ and $au$ are linearly dependent. then there exists a nonzero scalar $k$ such that $u+v = kau$
then, $u+v - kau = (1-ka)u + v = 0$
this means that $v = (ka-1)u$ which contradicts the fact that { $u,v$} is a basis for $V$.
hence, $u+v$ and $au$ must be linearly independent.
ii) Let $w\in V$. since { $u,v$} is a basis for $V$, we have $w = mu + nv$ for some scalars $m$ and $n$ in $V$. In particular, let $m=M+Na$ and $n=M$ where $M$ and $N$ are scalars in $V$. Hence, $w = mu + nv = (M+Na)u + Mv = Mu + Nau + Mv = M(u+v) + Nau$. Thus, $w\in \mbox{span}(\{u+v, au\})$ implies $V\subseteq \mbox{span}(\{u+v, au\})$ since we have written $w$ as a linear combination of $\{u+v, au\}$. Clearly, $\mbox{span}(\{u+v, au\}) \subseteq V$. Therefore, $\mbox{span}(\{u+v, au\}) = V$.