Question

Suppose (e₁,e₂,...,e_n) is an orthonormal basis of the inner product space V and v₁,v₂,...,v_n are vectors of V such that ||e_j−v_j||<1/√n. Prove that (v₁,v₂,...,v_n) is a basis of V.

Answer 1

$\text{Note: It is enough to show that \((v_1 \ldots, v_n)\) are linearly independent.}\\ \text{Suppose \(v_i\) are linearly dependent then there are scalars \((\lambda_1 ,\ldots ,\lambda_n)\) whic are not all zero such that } \\ \sum_{i=1}^{n} \lambda_i v_i = 0 \\ \text{Then }\\ 0 = \sum_{i=1}^{n} \lambda_i v_i = \sum_{i=1}^{n} \lambda_i (v_i - e_i) + \sum_{i=1}^{n} \lambda_i e_i \\ \implies \| \sum_{i=1}^{n} \lambda_i (v_i - e_i ) \|= \| \sum_{i=1}^{n} \lambda_i e_i \| \\ \text{However, } \\ \|\sum_{i=1}^{n} \lambda_i e_i\| =\|\sum_{i=1}^{n} \lambda_i( v_i-e_i) \| \le \sum_{i=1}^{n} |\lambda_i| \|v_i - e_i\| \\ \qquad\qquad\quad\quad < \frac{1}{\sqrt{n}} \sum_{i=1}^{n} |\lambda_i| \le \left(\sum_{i=1}^{n} |\lambda_i|^2 \right)^{\frac{1}{2}} \,\,\,\,\text{(Cauchy-Schwarz)} \\ \implies \|\sum_{i=1}^{n} \lambda_i e_i\| \le \left(\sum_{i=1}^{n} |\lambda_i|^2 \right)^{\frac{1}{2}} \,\,\,\, (1)\\ \text{Since \(e_i\) are orthonormal we have }\\ \|\sum_{i=1}^{n} \lambda_i e_i\|^2 = \sum_{i=1}^{n} |\lambda_i|^2 \\ \text{So we have: }\\ \|\sum_{i=1}^{n} \lambda_i e_i\| = \left(\sum_{i=1}^{n} |\lambda_i|^2 \right)^{\frac{1}{2}}\\ \text{which is a contradiction to what we have above} \\\text{ (i.e inequality (1)) hence showing that \((v_1, \ldots,v_n)\) are linearly indepemdent and we have our result.}$