# Polynomials

Note that by definition, a linear combination involves only finitely many vectors (except as described in **Generalizations** below). However, the set *S* that the vectors are taken from (if one is mentioned) can still be infinite; each individual linear combination will only involve finitely many vectors. Also, there is no reason that *n* cannot be zero; in that case, we declare by convention that the result of the linear combination is the zero vector in *V*.

## Examples and counterexamples[edit]

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### Euclidean vectors[edit]

Let the field *K* be the set **R** of real numbers, and let the vector space *V* be the Euclidean space **R**^{3}. Consider the vectors *e*_{1} = (1,0,0), *e*_{2} = (0,1,0) and *e*_{3} = (0,0,1). Then *any* vector in **R**^{3} is a linear combination of *e*_{1}, *e*_{2} and *e*_{3}.

To see that this is so, take an arbitrary vector (*a*_{1},*a*_{2},*a*_{3}) in **R**^{3}, and write:{\displaystyle (a_{1},a_{2},a_{3})=(a_{1},0,0)+(0,a_{2},0)+(0,0,a_{3})\,}

{\displaystyle =a_{1}(1,0,0)+a_{2}(0,1,0)+a_{3}(0,0,1)\,}

{\displaystyle =a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}.\,}

### Functions[edit]

Let *K* be the set **C** of all complex numbers, and let *V* be the set C_{C}(*R*) of all continuous functions from the real line **R** to the complex plane **C**. Consider the vectors (functions) *f* and *g*defined by *f*(*t*) := *e*^{it} and *g*(*t*) := *e*^{−it}. (Here, *e* is the 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* are:

- {\displaystyle \cos t={\begin{matrix}{\frac {1}{2}}\end{matrix}}e^{it}+{\begin{matrix}{\frac {1}{2}}\end{matrix}}e^{-it}\,}
- {\displaystyle 2\sin t=(-i)e^{it}+(i)e^{-it}.\,}

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*^{it} and *e*^{−it}. This means that there would exist complex scalars *a* and *b* such that *ae*^{it} + *be*^{−it} = 3 for all real numbers *t*. Setting *t* = 0 and *t* = π gives the equations *a* + *b* = 3 and *a* + *b* = −3, and clearly this cannot happen. See Euler’s identity.