TITLE: Prove that $v_1, \dots v_n$ is a basis of V.
QUESTION [8 upvotes]: Prove that if $e_1, \dots e_n$ is an orthonormal basis of $ V$ and $v_1, \dots , v_n$ are vectors in $ V$ such that
$$\|e_j - v_j\| < \frac{1}{\sqrt{n}}$$ for each j, then $v_1, \dots v_n$ is a basis of $V$.
I have fiddled around with this for a while now unfortunately nothing came close, so I'll just say I have no idea how to do this. Help :(?

REPLY [7 votes]: WLOG suppose $V=\mathbb C^n$ with standard inner product and $(e_1,\ldots,e_n)$ is the standard basis.  Let $A$ be the $n$-by-$n$ matrix with columns $v_1,\ldots,v_n$.  Then the Frobenius norm of $I-A$ is the square root of $\sum\limits_{k=1}^n\|e_k-v_k\|^2<1$.  Consequently, $A$ is invertible (with inverse $\sum\limits_{k=0}^\infty (I-A)^k$), and therefore it has linearly independent columns.