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# 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

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### Euclidean vectors

Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R3. Consider the vectors e1 = (1,0,0), e2 = (0,1,0) and e3 = (0,0,1). Then any vector in R3 is a linear combination of e1e2 and e3.

To see that this is so, take an arbitrary vector (a1,a2,a3) in R3, 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

Let K be the set C of all complex numbers, and let V be the set CC(R) of all continuous functions from the real line R to the complex plane C. Consider the vectors (functions) f and gdefined by f(t) := eit and g(t) := eit. (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 eit and eit. This means that there would exist complex scalars a and b such that aeit + beit = 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.