# Schwarz inequality and linear dependence

Let $\{a_i\}$ and $\{b_i\}$ be families of complex numbers.

I know that if $\{a_i\}$ and $\{b_i\}$ are linear dependence, then Schwarz inequality becomes equality, but I cannot prove the converse. Please help.

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Look at the steps you follow to prove Scwarz inequality. Assume that you have equality. What do you get in the middle? –  leo Jul 13 '12 at 4:42
@leo i don't understand what do you mean by 'the middle'. I have that if equality holds, then $0=\sum c_i(a_i - (a_k/b_k)b_i)$ for every $k\in I$ such that $b_k≠0$. ($c_i$ denotes conjugate of $b_i$ since i dunno how to type conjugate in LaTeX) –  Katlus Jul 13 '12 at 5:07
BTW your families of complex numbers are finite, right? I.e., we have $a_i$, $b_i$ for $i=1,\dots,n$. –  Martin Sleziak Jul 13 '12 at 7:26
I've just edited the proof on Wikipedia to make the converse part stand out more. Do you have any questions about it? –  Marc van Leeuwen Jul 13 '12 at 8:24