Editing Linear combination (section)

=== 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>.