Linear independence under weird condition This is a problem from a linear algebra textbook. Given a finite dimensional inner product space $V$ with orthonormal basis $e_1, \ldots, e_n$, show that if a list of vectors $v_1, \ldots, v_n$ satisfies $\|e_j - v_j\| < \frac{1}{\sqrt{n}}$ for all $j$ in $\{1, \ldots, n\}$, then $v_j$'s form a basis of $V$.
I have no idea. It's intuitive when I think about $\mathbb{R}^2$, looking at little spheres at the tips of the $e_j$'s. 
I thought about looking at prefixes, like it should be true that
$\|e_j - v_j\| < \frac{1}{\sqrt{i}}$ for all $j$ in $\{1, \ldots, n\}$.
So now if $v_i \in \operatorname{span}(v_1, \ldots, v_i)$ then it should violate the inequality. It just looked ugly. Is this even a good direction?
Edit: I see it's ridiculous now...
Is this something hard (it's Axler's book, 3rd edition, and the problems aren't marked by difficulty, so I don't want to waste time), or am I just being silly?
 A: Define $u_i := v_i - e_i$ so $v_i = e_i + u_i$ and $||u_i||^2 < \frac{1}{n}$. It is enough to show that $(e_1 + u_1, \ldots, e_n + u_n)$ are linearly independent. Let $a_i \in \mathbb{C}$ be scalars such that $\sum_{i=1}^n a_i (e_i + u_i) = 0$. Then
$$ \sum_{i=1}^n a_i e_i = -\sum_{i=1}^n a_i u_i. $$
Taking norms and using the fact that $(e_i)$ is an orthonormal basis, we have
$$ \sum_{i=1}^n |a_i|^2 = \left| \left| \sum_{i=1}^n a_i e_i \right| \right|^2 = \left| \left| \sum_{i=1}^n a_i u_i \right| \right|^2. $$
Using the triangle inequality and Cauchy-Schwartz inequality (in $\mathbb{R}^n$), we have
$$ \sum_{i=1}^n |a_i|^2  = \left| \left| \sum_{i=1}^n a_i u_i \right| \right|^2 \leq \left( \sum_{i=1}^n |a_i| ||u_i|| \right)^2 \leq \left( \sum_{i=1}^n |a_i|^2 \right) \left( \sum_{i=1}^n ||u_i||^2 \right). $$
Thus, we obtain
$$ \left(\sum_{i=1}^n |a_i|^2 \right) \left( 1 - \sum_{i=1}^n ||u_i||^2 \right) \leq 0.$$
Since $\sum_{i=1}^n ||u_i||^2 < \sum_{i=1}^n \frac{1}{n} = 1$, we must have $\sum_{i=1}^n |a_i|^2 = 0$ showing that $a_i = 0$ for all $1 \leq i \leq n$.
A: May  be just use the definition of linear dependence of vectors. Let $\alpha=(\alpha_1,\cdots,\alpha_m)$ such that $\sum_{i=1}^n\alpha_iv_i=0$
then
$\sum_{i=1}^n|\alpha_i|^2=\|\sum_{i=1}^n\alpha_ie_i-\sum_{i=1}^n\alpha_iv_i\|^2=\|\sum_{i=1}^n\alpha_i(e_i-v_i)\|^2$
$\leq (\sum_{i=1}^n|\alpha_i|\|(e_i-v_i)\|)^2<(\sum_{i=1}^n|\alpha_i|\frac1{\sqrt{n}})^2<\sum_{i=1}^n|\alpha_i|^2$ 
then $\alpha=0$
A: This scribble is split in two: The first half answers the OP,
while $(2)$ offers a general symmetrical example to demonstrate that the assumptions cannot be relaxed from $\,$"$<$"$\,$ to $\,$"$\leq\,$".
$\mathbf{(1)}\:\;$ It is shown that $L\in\mathscr L(V)$, defined by $Le_j =v_j\,$ for $1\leqslant j\leqslant n$, is injective. Then $\{v_1,\ldots,v_n\}\subset V$ is linear independent, hence a basis.
Pick any $\,x=\sum_{j=1}^nx_je_j\in V$. Now a trio of inequalities, namely  triangle, Cauchy-Bunyakovsky-Schwarz, and the assumed ones, implies
$$\begin{align}
\|x\|_2 & \:=\: \left\|\,Lx + \sum\nolimits_{j=1}^nx_j(\mathbb 1 -L)e_j\right\|_2\\[2ex]
 &\:\leqslant\:\|Lx\|_2\: +\: \sum\nolimits_{j=1}^n|x_j|\cdot\|e_j -Le_j\|_2\\[2ex]
&\:\leqslant\:\|Lx\|_2\: +\: \left(\sum\nolimits_{j=1}^n|x_j|^2\right)^{1/2}
\underbrace{\left(\sum\nolimits_{j=1}^n\|e_j -v_j\|^2\right)^{1/2}}_
{\qquad\;\displaystyle =c<1}
\end{align}$$
Thus $\,\|Lx\|_2\geqslant(1-c)\,\|x\|_2\;\forall x\in V$, proving injectivity of $L\,$.
$\mathbf{(2)}\:\;$ There is an orthogonal projector $P\in\mathscr L(V)$ of rank $\,n-1\,$ such that the $v_j:=Pe_j$ fulfill
$\,\|e_j-v_j\|=\tfrac{1}{\sqrt n}$ for all $j\,$. But
$\{v_1,v_2,\ldots,v_n\}\subset \operatorname{Im}P\,$ is certainly linear dependent.
Let $\,e=\sum_{j=1}^ne_j\,$ and $\,E=\operatorname{span}\{e\}\,$. The announced $P=P_{E^\perp}$ is the orthogonal projector onto $E^\perp\subset V\,$, and $\,\dim E^\perp=n-1$. We have
$$P_{E^\perp}\:=\;\mathbb 1-P_E\;=\;\mathbb 1-\frac1n(\,\cdot\mid e)\,e\,,$$
and setting $\,v_j=P_{E^\perp}(e_j)\,$ gives
$$\,\|e_j-v_j\|_2\;=\;\big\|(\mathbb 1 -P_{E^\perp})\,e_j\big\|_2\;=\;\frac1n \|e\|_2
 \;=\;\frac{1}{\sqrt n}\quad\forall\:j=1,\ldots,n\,.$$
