TITLE: Show linear independence of a set of vectors close to an orthonormal basis
QUESTION [11 upvotes]: This is a question from Linear Algebra Done Right.
Suppose $e_1, \dots, e_n$ is an orthonormal basis of $V$ and $v_1, \dots, v_n$ are vectors in $V$ such that
$$ \| e_j - v_j \| < \frac{1} {\sqrt{n}} $$
for each $j$. Prove that $v_1, \dots, v_n$ is a basis of $V$.
I tried to show this if $V$ is a two dimensional space by contradiction. However, it seems too messy to use the same idea for a general case.
Does anyone have hints for this problem?

REPLY [3 votes]: For each $i,$ let $v_i=e_i-u_i.$ Suppose there are scalars $s_1,.., s_n,$ not all $0,$ such that $\sum_{i=1}^ns_iv_i=0.$ $$\text {Let }\quad e= \sum_{i=1}^ns_ie_i,\quad u=\sum_{i=1}^ns_iu_i.$$ Then $e=u.$ And $\|e\|^2=\sum_{i=1}^n|s_i|^2.$ We have $$\|u\|^2\leq \left(\sum_{j=1}^n\|s_iu_i\|\right)^2<\left(\sum_{i=1}^n|s_i|/\sqrt n\right)^2.$$ The second inequality is strict because not all $s_i$ are $0$, and because $\|u_i\|<1\sqrt n$ for every $i$. 
Consider the vectors $s=(|s_1|,..,|s_n|)$ and $t=(t_1,..,t_n)$ where $t_i=1/\sqrt n$ for every $i.$ We have  $$\|u\|^2<\left(\sum_{i=1}^n|s_i|/\sqrt n\right)^2=(s\cdot t)^2\leq \|s\|^2\; \|t\|^2=\|s\|^2=\sum_{i=1}^n|s_i|^2=\|e\|^2=\|u\|^2,$$ a contradiction.
Remarks: I thought of this before reading the other answers. It seems to be sufficiently different, at least in its style. I think the one from jing007 is the best (including mine)... At one point I wrote $|s_i|^2$ rather than $s_i^2$, so that it is valid for $\mathbb C^n$ as well as for $\mathbb R^n.$