The n-tuple space, Fn. Let F be any field, and let V be the set of all n-tuples α = (x1,x2,…,xn) of scalars xi in F. If β= (y1,y2,…,yn) with yi in F, the sum of α and β is defined by   α + β = (x1+y1,x2+y2,…,xn+yn). The product of a scalar c and vector α is defined by   cα = (cx1,cx2,…,cxn). The fact that this vector addition and scalar multiplication satisfy conditions (3) and (4) is easy to verify, using the similar properties of addition and multiplication of elements of F.

We only need to verify 3 and 4 of the definition of the vector space when we already have a field F and a set Fn of vector objects. Firstly, let γ = (z1,z2,⋯,zn). For addition, we have

  1. Since addition in F is commutative: $$\begin{aligned} \alpha+\beta &=\left(x_1+y_1, x_2+y_2, \ldots, x_n+y_n\right)\\ &=\left(y_1+x_1,y_2+x_2, \ldots, y_n+x_n\right)\\ &=\beta + \alpha. \end{aligned}$$

  2. Since addition in F is associative: $$\begin{aligned} \alpha+(\beta+\gamma) &=\left(x_1+(y_1+z_1), x_2+(y_2+z_2), \ldots, x_n+(y_n+z_n)\right)\\ &=\left((x_1+y_1)+z_1, (x_2+y_2)+z_2, \ldots, (x_n+y_n)+z_n\right)\\ &=(\alpha+\beta)+\gamma. \end{aligned}$$

  3. We must show there is a unique vector 0 in V such that α + 0 = α for all α in V. Consider (0,…,0) the vector of all 0 ’s of length n. Then we have (0,…,0) + (x1,…,xn) = (0F+x1,…,0F+xn) = (x1,…,xn) since 0 + x = x for all x ∈ F. Thus we can have 0 = (0,…,0) + (x1,…,xn). To show this vector is unique with respect to this property, suppose 0′ = (x1,…,xn) also satisfies the property that 0′ + α = α for all α in V. Notice (x1,…,xn) = (x1+0,…,xn+0) = (x1,…,xn) + (0,…,0) but by definition of 0′, if we let α = (0,…,0), this equals (0,…,0). Thus (x1,…,xn) = (0,…,0) = 0. Thus 0 = 0′ and the zero element is unique.