Why are $(\,f_1,...,f_n)$ linearly independent if $\|\,f_k-e_k\|_2<\dfrac{1}{\sqrt n}$, where $(e_k)$ is an orthonormal basis?

Question

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.

Answer 1

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 $u\in M$ such that for each integer $1\le k\le n,\ u\perp f_k.$

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

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

Answer 2

The condition $\|f_k - e_k\|_2 < \frac{1}{\sqrt n}$ can be written $$ \|(I-T)e_k\|_2 < \frac{1}{\sqrt n}$$ Where $T:X\to X$ is the linear map on $X=\operatorname{span}\{e_1\dots e_n\}$ that maps $e_k \mapsto f_k$, and $I:X\mapsto X$ 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: