I am reading a book on Operator Algebras and at a point the authors seem to make the following implication:
Let $A$ be a $C^*$-algebra and $\{a_i\}_{i=1}^n$ a linearly independent subset. If $(u_\lambda)_{\lambda\in\Lambda}$ is an approximate unit for $A$, there exists a $\lambda_0\in\Lambda$ such that $\{a_iu_{\lambda_0}\}_{i=1}^n$ is also linearly independent.
I have doubt about the validity of this, or at least that it is immediate. My work is the following:
There is an elementary lemma in functional analysis (mostly used to show that all norms over finite dimensions are equivalent) that says that "whenever $\{a_i\}_{i=1}^n\subset X$ are linearly independent in a normed space, then there is a constant $C>0$ such that for all $(x_i)_{i=1}^n\subset\mathbb{C}$ it is $\|\sum_{i=1}^nx_ia_i\|\geq C\cdot\sum_{i=1}^n|x_i|$.
Let $C>0$ be the constant for $\{a_i\}_{i=1}^n$. Suppose that for each $\lambda\in\Lambda$ we have complex numbers $\{x_i^\lambda\}_{i=1}^n$ such that $\sum_{i=1}^nx_i^\lambda a_iu_\lambda=0$. Suppose furthermore that $\{x_i^\lambda\}_{i,\lambda}$ is bounded, say by a constant $M>0$.
Let $\varepsilon>0$ and pick $\lambda_0\in\Lambda$ such that for all $\lambda\geq\lambda_0$ it is $\|\sum_{i=1}^na_iu_\lambda-\sum_{i=1}^na_i\|<\varepsilon C/M$. Then for such $\lambda$ it is $$\sum_{i=1}^n|x_i^\lambda|\leq \frac{1}{C}\|\sum_{i=1}^nx_i^\lambda a_i\|=\frac{1}{C}\|\sum_{i=1}^nx_i^\lambda a_i-\sum_{i=1}^nx_i^\lambda a_iu_\lambda\|\leq \frac{M}{C}\|\sum_{i=1}^na_i-\sum_{i=1}^na_iu_\lambda\|<\varepsilon.$$ So we obtain that for all $i$ it is $x_i^\lambda\xrightarrow[\lambda\in\Lambda]{}0$. But that's the best I can get, and, don't forget I have added a hypothesis (the boundedness of the coefficients).
Am I missing out the obvious? This question refers to the same thing (the same book too). OP answers to himself, but It doesn't make any sense to me, they pick an index large enough such that an approximation holds for all scalars? This seems wrong to me.
I think the idea of the answer to the other question is fine, though potentially a little unclear so let me try to tidy it up here.
First, fix an $\varepsilon$ such that $0 < \varepsilon < \operatorname{dist}(a_n, \operatorname{span}\{a_1, \dots, a_{n-1}\})$. This is possible by linear independence of the set $\{a_1, \dots, a_n\}$.
Now note that for any $\lambda$ and for any scalars $\mu_i$, $$\|a_n u_\lambda - a_n - \sum_{i=1}^{n-1} (\mu_i a_i u_\lambda - \mu_i a_i) \| \leq (1 + \sum_{i=1}^{n-1} |\mu_i|) \max_{i=1, \dots, n} \|a_i u_\lambda - a_i\|$$
Fix a $\lambda_0$ such that $\max_i\|a_i u_{\lambda_0} - a_i\| < \frac{\varepsilon}{n}$. Now suppose that $\{a_i u_{\lambda_0}\}$ is not linearly independent. Then there exist scalars $\nu_i$ which are not all $0$ such that $\sum_{i=1}^n \nu_i a_i u_{\lambda_0} = 0$. Without loss of generality, $\nu_n$ can be assumed to satisfy $|\nu_n| \geq |\nu_i|$ for all $i$. We then have $$a_n u_{\lambda_0} = \sum_{i=1}^{n-1} \frac{\nu_i}{\nu_n} a_i u_{\lambda_0}$$ By our earlier bound with $\mu_i = \frac{\nu_i}{\nu_n}$, we also get that $$\|a_n u_{\lambda_0} - a_n - \sum_{i=1}^{n-1} \frac{\nu_i}{\nu_n} a_i u_{\lambda_0} - \frac{\nu_i}{\nu_n} a_i \| \leq (1 + \sum_{i=1}^{n-1} 1) \max_i \|a_i u_{\lambda_0} - a_i \| < \varepsilon.$$
This implies that $$ 0 = \|a_n u_{\lambda_0} - \sum_{i=1}^n \frac{\nu_i}{\nu_n} a_i u_{\lambda_0}\| > \|a_i - \sum_{i=1}^{n-1} \frac{\nu_i}{\nu_n} a_i \| - \varepsilon > 0$$ where the last inequality follows by the choice of $\varepsilon$. This is a contradiction and so the desired result follows.