Proving a basis for inner product space V when $||e_j-v_j||< \frac{1}{\sqrt{n}}$. Suppose $(e_1,e_2,...,e_n)$ is an orthonormal basis of the inner product space $V$ and $v_1,v_2,...,v_n$ are vectors of $V$ such that $$||e_j-v_j||< \frac{1}{\sqrt{n}}$$ for each $j \in \left\{1,2,...,n \right \}$. Prove that $(v_1,v_2,...,v_n)$ is a basis of $V$.
I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?
 A: Note that it's enough to show that the $v_i$ are linearly independent. Suppose that there are scalars $c_1,\dots,c_n$ not all zero such that $\sum_ic_iv_i=0$. Then
$$ 0=\sum_{i}c_iv_i=\sum_{i}c_i(v_i-e_i)+\sum_ic_ie_i$$
hence
$$ \Big|\Big|\sum_{i}c_i(v_i-e_i)\Big|\Big|=\Big|\Big|\sum_{i}c_ie_i\Big|\Big|$$
However,
$$\Big|\Big|\sum_{i}c_i(v_i-e_i)\Big|\Big|\leq \sum_i|c_i|\cdot ||v_i-e_i||<\frac{1}{\sqrt{n}}\sum_{i=1}^n|c_i|\leq \Big[\sum_{i=1}^n|c_i|^2\Big]^{\frac{1}{2}} $$
(with the last step using Cauchy-Schwarz), while since the $e_i$ are orthonormal we have
$$ \Big|\Big|\sum_{i}c_ie_i\Big|\Big|^2=\sum_{i=1}^n|c_i|^2 $$
so that
$$ \Big|\Big|\sum_{i}c_ie_i\Big|\Big|=\Big[\sum_{i=1}^n|c_i|^2\Big]^{\frac{1}{2}} $$
This contradiction shows that the $v_i$ are linearly independent.
A: Hint (not sure if this will work or not):
Since there are $n$ vectors in the set $v_1, \ldots, v_n$, you only need to show that they are linearly independent. So, suppose $\lambda_1 v_1 + \cdots + \lambda_n v_n = 0$. We need to show that this implies $\lambda_1 = \cdots = 0$. Write $v_i = e_i + d_i$, where the difference vectors $d_i$ are small. Try to deduce that $\lambda_1 e_1 + \cdots + \lambda_n e_n = 0$.
A: Start with $\sum_i \lambda_iv_i = 0$ and assume one $\lambda_i\neq 0$ and denote $d_i = e_i -v_i$
$\Rightarrow \sum_i \lambda_i(e_i +d_i) =0$
$\Rightarrow \sum_i \lambda_ie_i =- \sum_i \lambda_id_i$
$\Rightarrow ||\sum_i\lambda_i e_i ||= ||\sum_i \lambda_id_i||$
(Take $d=\max\{||d_i||\}$) 
$\Rightarrow ||\sum_i\lambda_i e_i ||^2= ||\sum_i \lambda_id_i||^2
\leq (\sum_i |\lambda_i|\cdot ||d_i||)^2  
\leq (\sum_i |\lambda_i|\cdot d)^2 
< (\sum_i |\lambda_i|\cdot \frac{1}{\sqrt{n}})^2 
\leq \frac{1}{n} (\sum_i |\lambda_i| )^2
\leq \frac{1}{n}\cdot n (\sum_i |\lambda_i|^2) 
$
The last inequality uses Cauchy–Schwarz inequality.
Now, since {e_i} is orthonormal basis we know that $||\sum_i\lambda_i e_i ||^2= \sum_i|\lambda_i|^2$ and thus:
$\sum_i|\lambda_i|^2= ||\sum_i\lambda_i e_i ||^2< \frac{1}{n}\cdot n (\sum_i |\lambda_i|^2) = \sum_i|\lambda_i|^2$.
Contradiction.
