# 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?

#### Solutions Collecting From Web of "Proving a basis for inner product space V when $||e_j-v_j||< \frac{1}{\sqrt{n}}$."

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.

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 $\lambda_1 e_1 + \cdots + \lambda_n e_n = 0$.

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:

$\sum_i|\lambda_i|^2= ||\sum_i\lambda_i e_i ||^2< \frac{1}{n}\cdot n (\sum_i |\lambda_i|^2) = \sum_i|\lambda_i|^2$.