I 'm trying to solve an axler text book question (6B 14) which is
Suppose $e_1,\dots,e_n$ is an orthonomral 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 $j$. Prove that $v_1,\dots,v_n$ is a basis of $V$.
Here is a solution. But the mistake I have in this is that I've used Cauchy Shwartz for something which does not have the absolute value. (I've highlighted where in my proof below)
Suppose $dim V=n$, then any list of $n$ linearly independent vectors form a basis for $V$. Now suppose $v_1..., v_n$ is a list of linearly dependant vectors such that for some $a_1..., a_n \in \mathbb{F}$ for atleast one $a_i \neq 0$, we have $$\sum^n_{i=1}a_i v_i =0$$ We can have two possibilities. One is $$||\sum^n_{i=1}a_i(e_i - v_i)||^2= ||\sum^n_{i=1}a_i e_i||^2 = \sum^n_{i=1}|a_i|^2$$ Cross products cancel and it may be a basis. Another possibility is $$||\sum^n_{i=1}a_i(e_i - v_i)||^2=\langle\sum^n_{i=1}a_i(e_i - v_i), \sum^n_{i=1}a_i(e_i - v_i)\rangle$$ $$=\sum^n_{i=1} \sum^n_{j=1}\langle a_i(e_i - v_i), a_j(e_j - v_j)\rangle$$ $$\leq \sum^n_{i=1} \sum^n_{j=1} ||a_i(e_i - v_i)|| ||a_j(e_j - v_j)||$$ ^^^Here is where the mistake is used Cauchy Schwartz without absolute value.. $$=\leq \sum^n_{i=1} \sum^n_{j=1} |a_i| |a_j| ||a_i(e_i - v_i)|| ||a_i(e_j - v_j)||$$ $$< =\leq \sum^n_{i=1} \sum^n_{j=1} \frac{1}{n} |a_i| |a_j|$$ $$\frac{1}{n} (\sum^n_{i=1} |a_i|)^2$$ $$\leq \sum^n_{i=1} |a_i|^2$$ But then this means that $$\sum^n_{i=1} |a_i|^2 < \sum^n_{i=1} |a_i|^2$$ Which is not possible. This is a contradiction. So $v_1..., v_n$ is a linearly independent list of vectors of size $n$ and thus is a basis of $V$.
Would really appreciate help on this!
