Editing Linear combination (section)

== Examples and counterexamples ==
{{inline citations|section|date=August 2013}}

=== Euclidean vectors ===
Let the field ''K'' be the set '''R''' of [[real number]]s, and let the vector space ''V'' be the [[Euclidean space]] '''R'''<sup>3</sup>.
Consider the vectors {{nowrap|1='''e'''<sub>1</sub> = (1,0,0)}}, {{nowrap|1='''e'''<sub>2</sub> = (0,1,0)}} and {{nowrap|1='''e'''<sub>3</sub> = (0,0,1)}}.
Then ''any'' [[Euclidean vector|vector]] in '''R'''<sup>3</sup> is a linear combination of '''e'''<sub>1</sub>, '''e'''<sub>2</sub>, and&nbsp;'''e'''<sub>3</sub>.

To see that this is so, take an arbitrary vector (''a''<sub>1</sub>,''a''<sub>2</sub>,''a''<sub>3</sub>) in '''R'''<sup>3</sup>, and write:
:<math>
\begin{align}
( a_1 , a_2 , a_3) & = ( a_1 ,0,0) + (0, a_2 ,0) + (0,0, a_3) \\[6pt]
& =  a_1 (1,0,0) + a_2 (0,1,0) + a_3 (0,0,1) \\[6pt]
& =  a_1 \mathbf e_1 +  a_2 \mathbf e_2 + a_3 \mathbf e_3.
\end{align}
</math>

=== 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]].

=== Polynomials ===
Let ''K'' be '''R''', '''C''', or any field, and let ''V'' be the set ''P'' of all [[polynomial]]s with coefficients taken from the field ''K''.
Consider the vectors (polynomials) ''p''<sub>1</sub>&nbsp;:=&nbsp;1, {{nowrap|1=''p''<sub>2</sub> := ''x'' + 1}}, and {{nowrap|1=''p''<sub>3</sub> := ''x''<sup>2</sup> + ''x'' + 1}}.

Is the polynomial ''x''<sup>2</sup>&nbsp;−&nbsp;1 a linear combination of ''p''<sub>1</sub>, ''p''<sub>2</sub>, and ''p''<sub>3</sub>?
To find out, consider an arbitrary linear combination of these vectors and try to see when it equals the desired vector ''x''<sup>2</sup>&nbsp;−&nbsp;1.
Picking arbitrary coefficients ''a''<sub>1</sub>, ''a''<sub>2</sub>, and ''a''<sub>3</sub>, we&nbsp;want
:<math> a_1 (1) + a_2 ( x + 1) +  a_3 ( x^2 + x + 1) =  x^2 - 1. </math>
Multiplying the polynomials out, this&nbsp;means
:<math> ( a_1 ) + ( a_2 x + a_2) + ( a_3 x^2 + a_3 x + a_3) =  x^2 - 1 </math> 
and collecting like powers of ''x'', we&nbsp;get
:<math> a_3 x^2 + ( a_2 + a_3 ) x + ( a_1 + a_2 + a_3 ) = 1 x^2 + 0 x + (-1). </math> 
Two polynomials are equal [[if and only if]] their corresponding coefficients are equal, so we can conclude
:<math> a_3 = 1, \quad a_2 + a_3 = 0, \quad a_1 + a_2 + a_3 = -1. </math> 
This [[system of linear equations]] can easily be solved.
First, the first equation simply says that ''a''<sub>3</sub> is 1.
Knowing that, we can solve the second equation for ''a''<sub>2</sub>, which comes out to −1.
Finally, the last equation tells us that ''a''<sub>1</sub> is also −1.
Therefore, the only possible way to get a linear combination is with these coefficients.
Indeed,
:<math> x^2 - 1 = -1 - ( x + 1) + ( x^2 + x + 1) = - p_1 -  p_2 +  p_3 </math>
so ''x''<sup>2</sup>&nbsp;−&nbsp;1 ''is'' a linear combination of ''p''<sub>1</sub>, ''p''<sub>2</sub>, and&nbsp;''p''<sub>3</sub>.

On the other hand, what about the polynomial ''x''<sup>3</sup>&nbsp;−&nbsp;1? If we try to make this vector a linear combination of ''p''<sub>1</sub>, ''p''<sub>2</sub>, and ''p''<sub>3</sub>, then following the same process as before, we get the equation

:<math>
\begin{align}
& 0 x^3 + a_3 x^2 + ( a_2 + a_3 ) x + ( a_1 + a_2 + a_3 ) \\[5pt]
= {} & 1 x^3 + 0 x^2 + 0 x + (-1).
\end{align}
</math>

However, when we set corresponding coefficients equal in this case, the equation for ''x''<sup>3</sup>&nbsp;is
:<math> 0 = 1 </math> 
which is always false.
Therefore, there is no way for this to work, and ''x''<sup>3</sup>&nbsp;−&nbsp;1 is ''not'' a linear combination of ''p''<sub>1</sub>, ''p''<sub>2</sub>, and&nbsp;''p''<sub>3</sub>.