# 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?

• Yes. Do you know why? (Can you show linear independence?) – Cameron Williams May 27 '15 at 19:18
• As the question was posed, the answer is no. You have to add the hypothesis that $k$ and $m$ are both nonzero. – Jack Lee May 27 '15 at 19:19
• I think so. Is it because the system $|{kv_1 v_2}|$ $|{x}|=0$ will always have a unique solution i.e 0? (There two are matrices. I don't know how to write matrices in latex so sorry about writing them like this) – user140161 May 27 '15 at 19:24
• Also, suppose the subspace is question is the column space of a matrix. Then does it make any difference if I write the basis from the reduced row echelon form or from the original matrix? Or is it correct to take them from either one? – user140161 May 27 '15 at 19:32

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.

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 $V$.

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.