Uniqueness of basis vectors Say I have 2 vectors $v_1$ and $v_2$ as basis of a subspace. Then is it true that $kv_1$ and $mv_2$ where $k$ and $m$ are real numbers, are also basis for that subspace? 
 A: Hint: Considering $k,m \neq 0$ then yes. Just consider this $$a (k v_1) + b (m v_2) = 0$$
and use the fact that $v_1$ and $v_2$ are L.I.
A: Start with the definition of a basis of a vector space. A basis is a set of linearly independent vectors that spans the vector space. 
Now  $\{v_1,v_2\}$ is a spanning set of vectors for a vector space $V$ over a field $F$ if and only if every vector $v\in V$ can be written as a linear combination of the basis vectors. That is, for all $v\in V$ there exist some scalars $a_v,b_v \in F$ such that
$$
v=a_vv_1+b_vv_2.
$$
Then, if $k$ and $m$ in $F$ are non zero,
$$
v=(a_v(1/k))(kv_1)+(b_v(1/m)(mv_2)
$$
so that the set $\{kv_1, mv_2\}$ spans $V$.
On the other hand, $\{v_1,v_2\}$ is a linearly independent set of vectors if and only if the equation
$$
av_1+bv_2=0
$$
implies that $a=b=0$.
Then
$$
a(kv_1)+b(mv_2)=(ak)v_1+(bm)v_2=0
$$
implies that $a$ and $b$ are zero (since $k$ and $m$) are assumed to be nonzero.
It follows that $\{kv_1, mv_2\}$ is  basis for $V$.
A: If $k,m \neq 0$, you can write: $$v = av_1 + bv_2 = \frac{a}{k}(kv_1)+\frac{b}{m}(mv_2),$$ for $v \in \Bbb Rv_1 + \Bbb Rv_2$. Since $a$ and $b$ are unique for $v$, then you got a unique linear combination of $kv_1$ and $mv_2$, so we're done.
