Convergence of a pair linearly independent elements of a vector space Let $v_1$ and $v_2$ be linearly independent elements of some normed vector space $V$. We have a sequence $(a_n,b_n)$ and are told that $a_nv_1+b_nv_2$ converges to the zero element. Is it the case that $(a_n,b_n)$ converges to $(0,0)$ (I strongly believe so), and how can this be shown?
 A: Let $U$ be the subspace of $V$ spanned by $v_1, v_2$, and let those two vectors be the basis for $U$. Then, since in $V$, we have that $a_nv_1 + b_nv_2$ converges towards the zero vector, it will converge towards the zero vector in $U$ too. But in $U$, the vectors are of the form $(a_n, b_n)$, so we have that $(a_n, b_n)$ converges towards $(0, 0)$

In this argument there are assumptions about the vector space $V$, most notably that it accepts any and all of the elements of $a_n$ and $b_n$ as scalars. $U$ is spanned by all linear combinations of $v_1$ and $v_2$ with coefficients in whatever space the sequences $a_n$ and $b_n$ lie. Adding these details should make the proof strict enough, although I doubt it is necessary on this level.
A: Here's an argument via compactness. This part of the answer assumes that the underlying field is $\mathbb R$.
Let $K \subseteq \mathbb R^2$ be the compact set $\{ (x,y) \,:\, |x| + |y| = 1 \}$. Then since the function $(x,y) \mapsto \| x v_1 + yv_2 \|$ is continuous everywhere, it attains its maximum and minimum values in $K$ (by the extreme value theorem). Let $m$ denote the minimum value of $\| x v_1 + yv_2 \|$ for all $(x,y) \in K$; note that $m > 0$. 
Now, 
$$
\| a_nv_1 + b_n v_2  \| = (|a_n|+|b_n|) \left\| \frac{a_n}{|a_n|+|b_n|} v_1 + \frac{b_n}{|a_n|+|b_n|} v_2  \right\| \geqslant (|a_n|+|b_n|) \cdot m,
$$
since $\left(\frac{a_n}{|a_n|+|b_n|}, \frac{b_n}{|a_n|+|b_n|} \right) \in K$. Therefore, as $n \to \infty$, we have $$|a_n| + |b_n| \leqslant \frac{\| a_n v_1 + b_n v_2 \|}{m} \to 0 ,$$ and hence $(a_n, b_n) \to (0,0)$. $\quad \diamond$

If the underlying field is different from $\mathbb R$, the conclusion might be false. For example, $$\mathbb Q[\sqrt{2}] := \{ m + n \sqrt{2} \,:\, m,n \in \mathbb  Q \} $$ is a normed linear space over $\mathbb Q$, with the usual absolute value function $x \mapsto | x |$ as a norm. It is an easy exercise that for any $\varepsilon> 0$, there exist $a_n, b_n \in \mathbb Z$ such that $|a_n|, |b_n| \geq 1$ and $|a_n + b_n \sqrt{2}| < \varepsilon$, which provides a counter-example with $v_1 = 1$ and $v_2 = \sqrt{2}$.
A: Note there is no loss of generality in assuming $V = \mathrm{span}\{v_1, v_2\}$ since no other vectors enter into the problem.  Define $T : V \to \mathbb{R}^2$ by $T (a v_1 + b v_2) = (a,b)$.  Since $v_1, v_2$ are linearly independent, $T$ is well defined and linear.  Also, since $V$ is finite dimensional, $T$ is automatically continuous (with respect to the usual topology on $\mathbb{R}^2$).  Thus if $a_n v_1 + b_n v_2 \to 0$ in the norm of $V$, by applying $T$ we have $(a_n, b_n) \to 0$ in $\mathbb{R}^2$; in particular $a_n \to 0$ and $b_n \to 0$.
Edit: This assumes $V$ is a vector space over $\mathbb{R}$.  Over $\mathbb{C}$ the proof is similar.  Over a general field it is false.  The statement "$T$ is automatically continuous" is implicitly using the local compactness of the base field.
