Editing Linear combination (section)

=== Functions ===
Let ''K'' be the set '''C''' of all [[complex number]]s, and let ''V'' be the set C<sub>'''C'''</sub>(''R'') of all [[continuous function]]s from the [[real line]] '''R''' to the [[complex plane]] '''C'''.
Consider the vectors (functions) ''f'' and ''g'' defined by ''f''(''t'')&nbsp;:= ''e''<sup>''it''</sup> and ''g''(''t'')&nbsp;:= ''e''<sup>−''it''</sup>.
(Here, ''e'' is the [[e (mathematical constant)|base of the natural logarithm]], about 2.71828..., and ''i'' is the [[imaginary unit]], a square root of −1.)
Some linear combinations of ''f'' and ''g''&nbsp;are:
*<div style="vertical-align: 0%;display:inline;"><math> \cos t = \tfrac12 \, e^{i t} + \tfrac12 \, e^{-i t} </math></div> 
*<div style="vertical-align: 0%;display:inline;"><math> 2 \sin t = (-i) e^{i t} + (i) e^{-i t}. </math></div> 
On the other hand, the constant function 3 is ''not'' a linear combination of ''f'' and ''g''. To see this, suppose that 3 could be written as a linear combination of ''e''<sup>''it''</sup> and ''e''<sup>−''it''</sup>. This means that there would exist complex scalars ''a'' and ''b'' such that {{nowrap|1=''ae''<sup>''it''</sup> + ''be''<sup>−''it''</sup> = 3}} for all real numbers ''t''. Setting ''t'' = 0 and ''t'' = π gives the equations {{nowrap|1=''a'' + ''b'' = 3}} and {{nowrap|1=''a'' + ''b'' = −3}}, and clearly this cannot happen.  See [[Euler's identity]].