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I need to show that $v_1,...,v_n$ is basis for $V$ whenever $e_1,...,e_n$ is an orthonormal basis for V and $v_1,...,v_n$ are vectors in $V$ such that $$\left\Vert e_i-v_i\right\Vert < \frac{1}{\sqrt{n}}.$$ I know already that I only need to prove that they are lineary independent. Also I tried to prove it by contradiction, but I got stuck fairly fast. Any help, or hint how to prove it (with ot without contradiction) will be appriciated.

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    $\begingroup$ Try the triangle inequality on an alleged non-trivial relation. $\endgroup$ – paul garrett Jun 12 '16 at 21:37
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Assume that $\sum_{i=1}^n a_i v_i = 0$. Then

$$ 0 = \sum_{i=1}^n a_i (v_i - e_i + e_i) \implies \sum_{i=1}^n a_i e_i =\sum_{i=1}^n a_i (e_i - v_i) $$

but then

$$ \sum_{i=1}^n |a_i|^2 = \left| \left| \sum_{i=1}^n a_i e_i \right| \right|^2 = \left| \left| \sum_{i=1}^n a_i(e_i - v_i) \right| \right|^2 \leq \left( \sum_{i=1}^n |a_i| \cdot ||e_i - v_i || \right)^2 \leq \left( \sum_{i=1}^n |a_i|^2 \right) \left( \sum_{i=1}^n ||e_i - v_i||^2 \right) $$

where we used both the triangle inequality for $(V, ||\cdot||)$ and the Cauchy-Schwartz inequality for $\mathbb{R}^n$. This implies that

$$ \left( \sum_{i=1}^n |a_i|^2 \right) \left( \sum_{i=1}^n ||e_i - v_i||^2 - 1 \right) \geq 0 $$

but since $\sum_{i=1}^n||e_i - v_i||^2 < \sum_{i=1}^n \frac{1}{n} = 1$, we must have $\sum_{i=1}^n |a_i|^2 = 0$ as required.

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