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Let $(e_1,...,e_n)$ be an orthonomal basis and $(,f_1,...,f_n)$ vectors such that $|f_k-e_k|_2<dfrac{1}{sqrt n},forall k,$.

I'd like to show that $(,f_1,...,f_n)$ are linearly independent.

I don't know where to start:

I tried to see what happens for $n=2$ but, because $(,f_1,f_2)$ are two different vectors, they are linearly independent, and it does not help to see what happens in higher dimension.

It's easy to see that the $f_i$ are distinct because they are in separated balls.

I also tried to elevate to the square the relation but I found nothing.

If the $f_k$ are not linearly independent, then they can not span the space, so

$M=$span$({{f_k}})^{perp}neq 0.$ Therefore, there is a non-zero vector $uin M$ such that for each integer $1le kle n, uperp f_k.$

Of course, $u=sum^n_{k=1}langle u,e_krangle e_k, $ so that $|u|^2=sum^n_{k=1}|langle u,e_k-f_krangle |^2. $ Now, apply Cauchy-Schwarz to this last item and then use the hypothesis:

$|u|^2le sum^n_{k=1}|u|^2cdot |e_k-f_k|^2<sum^n_{k=1}|u|^2cdot left ( frac{1}{sqrt n} right )^2=ncdot left ( frac{1}{sqrt n} right )^2cdot |u|^2=|u|^2$, and so we get that $|u|^2<|u|^2$, which is absurd.

The condition
$|f_k - e_k|_2 < frac{1}{sqrt n}$ can be written
$$ |(I-T)e_k|_2 < frac{1}{sqrt n}$$
Where $T:Xto X$ is the linear map on $X=operatorname{span}{e_1dots e_n}$ that maps $e_k mapsto f_k$, and $I:Xmapsto X$ is the identity map.

Suppose now $x=sum_i x_i e_i $ is an arbitrary vector in $X$ with $|x|^2_2 = sum_i x_i ^2 = 1$. Then
begin{align} &|(I-T)x|_2
\ le& sum_{i=1}^n | x_i|| (I-T) e_i|_2
\ le &sqrt{sum_i |x_i|^2} sqrt{sum_i | (I-T) e_i|^2_2} \ =& sqrt{sum_i | (I-T) e_i|^2_2}
end{align}

Thus the operator norm $|I-T|_{op} := sup_{x: |x|_2 = 1} |(I-T)x|_2 le sqrt{sum_i | (I-T) e_i|^2_2} < 1 $. By results on Neumann series (see Proof of Neumann Lemma ), $T$ is invertible, and the result follows.

Also, here's a picture showing why a strict inequality is important: