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Let $(X,\|\cdot\|)$ be a normed linear space and $x_{1}, x_{2}, \ldots, x_{n}$ be $n$ linearly independent vectors in $X$. Show that there exists $\epsilon > 0$ such that if $y_{1}, y_{2}, \cdots, y_{n} \in X$ with $\|y_{i}\| < \epsilon$, $i = 1,2,\ldots,n$, then $x_{1} + y_{1}, x_{2} + y_{2},\ldots,x_{n} + y_{n}$ are also linearly independent vectors in $X$.

I've been thinking over this for a few days and I'm not really getting anywhere.

I've tried to start with the definition of linear independence for the $x$-vectors and then working towards the linear independence of $x+y$ given the condition on $y$, but I'm not getting anywhere.

May someone offer a hint or a solution?

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    $\begingroup$ Consider the map $L \colon \mathbb{K}^n \to X$ given by $c \mapsto \sum c_k\cdot x_k$. The unit sphere in $\mathbb{K}^n$ is compact. $\endgroup$ – Daniel Fischer Mar 6 '16 at 16:32
  • $\begingroup$ Assume the notion of compactness is not available. What could you say? $\endgroup$ – clocktower Mar 6 '16 at 19:28
  • $\begingroup$ What's the objection to compactness? You only need it insofar as "the image of a compact set under a continuous function is compact", which is standard fare for a basic real analysis or topology course, which is probably assumed for functional analysis. $\endgroup$ – Jon Warneke Aug 11 '16 at 17:52
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EDIT:

What I've written below the line is not correct. One cannot conclude that $\|T\|_\text{op} \leq n \varepsilon$, because the preceding only holds for one particular nonzero $a \in \mathbb F^n$ provided by dependence, not all nonzero $a \in \mathbb F^n$.

It's bothering me that there are no correct proofs now, so here I'll give the proof that Daniel Fischer described in the hopes that, after seeing it, you don't think it's so bad. Feel free to comment otherwise, though!

Let $\mathbb F$ be the field over which the vector space $X$ is defined. To each $n$-tuple $\mathbf v = (v_1, \dots, v_n)$ of vectors $v_i \in X$, define the map $T_{\mathbf v} : \mathbb F^n \longrightarrow X$ given by $$ \mathbb F^n \ni a = (a_1, \dots, a_n) \longmapsto \sum_{i=1}^n a_i v_i \in X. $$ $T_{\mathbf v}$ is a bounded linear map (this is easy to check), and hence it's continuous as a map between metric spaces. If we let $S = \left\{ a \in \mathbb F^n : \|a\| = 1 \right\}$ be the unit sphere in $\mathbb F^n$, observe that \begin{align*} \mathbf v \text{ is linearly independent } &\iff T_{\mathbf v}(S) \text{ is disjoint from } 0 \in X. \end{align*} (This follows directly from the definition of linear independence.) Since $S$ is compact and $T_{\mathbf v}$ is continuous, the set $T_{\mathbf v}(S)$ is compact, and therefore closed, in $X$. Since $X$ is a metric space, it's regular as a topological space. (A quick way to see this is metric $\implies$ normal $\implies$ regular.) By linear independence, the closed set $T_{\mathbf x}(S)$ is disjoint from $0$, and by regularity it has a neighborhood $U$ disjoint from $0$. We'll show that $T_{\mathbf x + \mathbf y}(S)$ is contained in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$; from compactness of $T_{\mathbf x}(S)$, it follows that taking $\varepsilon$ sufficiently small puts $T_{\mathbf x + \mathbf y}(S) \subseteq U$. In particular, $T_{\mathbf x + \mathbf y}(S)$ is disjoint from $0$, and hence $\mathbf x + \mathbf y$ is a linearly independent list.

To see that $T_{\mathbf x + \mathbf y}(S)$ is contained in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$, consider a general point $p = \sum_{i=1}^n a_i (x_i + y_i)$ of $T_{\mathbf x + \mathbf y}(S)$ (so $\|a\| = 1$). Then $q = \sum_{i=1}^n a_i x_i$ is in $T_{\mathbf x}(S)$, and $$ d(p, q) = \left\| \sum_{i=1}^n a_i (x_i + y_i) - \sum_{i=1}^n a_i x_i \right\| \leq \sum_{i=1}^n \left| a_i \right| \left\| y_i \right\| < \varepsilon \|a\| = \varepsilon, $$ where we've arbitrarily assigned $\mathbb F^n$ the $1$-norm $\|a\| = \sum_{i=1}^n |a_i|$, which is fine since all norms on a finite-dimensional vector space are topologically equivalent. Hence every point $p \in T_{\mathbf x + \mathbf y}(S)$ is within distance $\varepsilon$ from some point $q$ of $T_{\mathbf x}(S)$, establishing that $T_{\mathbf x + \mathbf y}(S)$ lies in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$.


At the risk of stealing Sheldon Axler's thunder, here's a solution without determinants. It also avoids the problem that @Svetoslav points out: we allow the $y_1, \dots, y_n$ to be any vectors in $X$, rather than restrcting them to be in $L = \text{span}(x_1, \dots, x_n)$.

Let $\mathbb{F}$ be the field of our vector space. We're given that $x_1, \dots, x_n \in X$ is independent, which is to say that the linear map $T :\mathbb F^n \to X$ given by $$ T(a_1, \dots, a_n) = \sum a_i x_i $$ is injective. In particular, $T$ is not the zero-map, so it has operator norm $$ \|T\|_{\text{op}} = \sup_{0 \neq a \in \mathbb{F}^n} \frac{\|T a \|}{\| a \|_\infty} $$ which is positive (see this by taking any $a$ with $Ta \neq 0$). (Here, we've given $\mathbb{F}^n$ the sup norm $\| a \|_\infty = \sup\{|a_1|, \dots, |a_n|\}$.)

We prove your claim by contradiction: suppose that for all $\varepsilon > 0$, there exist $y_1, \dots, y_n \in X$ with $\|y_i\| < \varepsilon$ such that $x_1 + y_1, \dots, x_n + y_n$ is dependent. Then there exists $a = (a_1, \dots, a_n) \in \mathbb{F}^n \backslash \{0\}$ such that \begin{align*} 0 = \sum a_i (x_i + y_i) &\implies \sum a_i x_i = - \sum a_i y_i \\ &\implies \|Ta\| = \left\| \sum a_i y_i \right\| && \text{take norms} \\ &\implies \|Ta\| \leq \| a \|_\infty n \varepsilon && \text{triangle inequality} \\ &\implies \| T \|_{\text{op}} \leq n \varepsilon. && a \neq 0 \end{align*} Since $\varepsilon > 0$ was arbitrary, this implies that $\|T\|_\text{op} = 0$, i.e., $T$ is the zero-map, giving us our contradiction.

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  • $\begingroup$ Yes, I think your solution does not suffer from l.o.g and also feels a bit more straightforward. Btw, I upvoted it together with Norbert's answer :) $\endgroup$ – Svetoslav Mar 15 '16 at 16:00
  • $\begingroup$ Yes, you are right it doesn't follow that $\|T\|_{op}\leq n\epsilon$. I missed this in my previous read. $\endgroup$ – Svetoslav Oct 5 '16 at 13:32
  • $\begingroup$ @Svetoslav where is the loss of generality in my proof? $\endgroup$ – Norbert Oct 5 '16 at 18:12
  • $\begingroup$ @Norbert what happened to your answer? I had a notification of a reply to my comment, but when I came it was not here... $\endgroup$ – ziggurism Oct 6 '16 at 12:08
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Linear independence is an open condition, which is most easily seen by observing that $x_1, \dotsc, x_n$ are independent iff $x_1\wedge \dotsb\wedge x_n \neq 0$. So for any neighborhood $U$ of $x_1\wedge \dotsb\wedge x_n$ in $\Lambda^n X$, there is a neighborhood $V$ in $X\times\dotsb\times X$ mapping into $U$ under the wedge map. If we choose $U$ away from the origin, then any $\epsilon$ keeping in the neighborhood $V$ will do the job.

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  • $\begingroup$ I like the idea, but i'm not sure i understand it completely. "So for any neighborhood $U$ of $x_1 \wedge \cdots \wedge x_n$ in $\Lambda^n X$"... how do we topologize $\Lambda^n X$? Looking at en.wikipedia.org/wiki/Topological_tensor_product there seem to be many norms on the tensor product. Then it seems that, if $X$ were finite dimensional, then they'd all be equivalent since $\dim \Lambda^n X = \binom{\dim X}{n}$. Is this sound? And, in our case, $X$ may be infinite dimensional, so what to do? $\endgroup$ – Jon Warneke Oct 6 '16 at 15:34
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    $\begingroup$ @JonWarneke it's a valid concern. The exterior product is constructed as a quotient of the tensor product. For a general normed space, there are many different choices of norm to put on the tensor product. But no matter what choice you make, it must be the case that the map $(x_1,x_2)\mapsto x_1\otimes x_2$ is continuous, as this is part of the definition of tensor product. The wedge map is the composition with another canonical quotient map, so it is continuous. $\endgroup$ – ziggurism Oct 6 '16 at 15:53

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