We consider the two subsets of the R−vector space R 3 V = (x, y, z) ∈ R 3 : x + y − z = 0 and F = (x, y, z) ∈ R 3 : x = y = z 1) Show that V and F are two vector subspaces of R 3 . 2) Determine dimR V and dimR F. 3) Deduce that V and F are two vector subspaces supplementary in R 3

To show that V and F are two vector subspaces of R^3, we need to verify three properties:

a) The zero vector of R^3 is in V and F. For V, if we let x = y = z = 0, we get 0 + 0 - 0 = 0, which is true. So, the zero vector is in V. For F, if we let x = y = z = 0, we get 0 = 0 = 0, which is also true. So, the zero vector is in F.

b) V and F are closed under addition. For V, if we take two vectors (x1, y1, z1) and (x2, y2, z2) in V, their sum (x1 + x2, y1 + y2, z1 + z2) is also in V because (x1 + x2) + (y1 + y2) - (z1 + z2) = (x1 + y1 - z1) + (x2 + y2 - z2) = 0 + 0 = 0. For F, if we take two vectors (x1, x1, x1) and (x2, x2, x2) in F, their sum (x1 + x2, x1 + x2, x1 + x2) is also in F because x1 + x2 = x1 + x2 = x1 + x2.

c) V and F are closed under scalar multiplication. For V, if we take a vector (x, y, z) in V and a scalar k in R, the product k(x, y, z) = (kx, ky, kz) is in V because kx + ky - kz = k(x + y - z) = k*0 = 0. For F, if we take a vector (x, x, x) in F and a scalar k in R, the product k(x, x, x) = (kx, kx, kx) is in F because kx = kx = kx.

The dimension of a vector space is the number of vectors in its basis. For V, a basis can be {(1, 1, 2), (0, 1, 1)}. So, dimR V = 2. For F, a basis can be {(1, 1, 1)}. So, dimR F = 1.

Two vector subspaces V and F of a vector space R^3 are supplementary if their direct sum is R^3 and their intersection is {0}. The direct sum of V and F is R^3 because every vector in R^3 can be written as a sum of a vector in V and a vector in F. The intersection of V and F is {0} because the only vector that is in both V and F is the zero vector. Therefore, V and F are two vector subspaces supplementary in R^3.