A list is a basis if norm of difference with orthonormal basis is bounded by given constant less than unity

Question

This is a problem from Linear Algebra Done Right, 3rd Edition, problem 14 in 6.B.

Given $\{e_1, \dotsc, e_n\}$ is an orthonormal basis of $V$, and $||v_j -e_j|| < \frac{1}{\sqrt{n}} \; \forall \; 1\leq j\leq n$, show that $\{v_1, \dotsc, v_n\}$ is a basis of $V$.

I have tried to start with a linear combination of the $\{v_1, \dotsc, v_n\}$ that is zero but am at a loss to use the information given. I also tried using the fact that each $v_j$ is in the span of the given orthonormal basis. Any hints will be appreciated.

Answer 1

Assume that $x\in V$, $\|x\|=1$, such that $\langle x,v_k\rangle = 0$ for all $k$. Then $|\langle x,e_k\rangle| < 1/\sqrt n$. Now use Parseval's equality to obtain a contradiction.