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?
Convergence of a pair linearly independent elements of a vector space
3 Answers
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.
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0It could've been stated clearer, but Henning is pretty straight on. Basically, if they were linearly dependent, and you still "tried" to make them into a basis, you would have that for instance $a_n = n, b_n = k\cdot n$ satisfies the rest of the hypothesis for some $k$, but clearly not the conclusion. – 2011-10-19
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 $v_2 = \sqrt{2}$.
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0Well, if that is a counter example or not depends on what norm you use. Of course you could give any 2-dim vector space over $\mathbb{R}$ or $\mathbb{Q}$ a ridiculously "skewed" norm, and that would make the argument difficult. But if you just use the standard Pythagoras norm, it works out fine. Of course, the Pythagoras norm doesn't comply with the multiplicative structure of the vector space, but that is not an issue in this specific setting. – 2011-10-17
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.
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1@ARA: I'm assuming $\mathbb{R}$ as the base field. The statement that "a linear operator on a finite-dimensional vector space is continuous" breaks down when the base field is not locally compact. – 2011-10-17