Show that S x 1 x is a real number is not a subspace of R2

Show that S = {(x, 1) x is a real number} is not a subspace of R^2 24) Suppose that the vectors V_1 = (-2,1,0, 0, 0), V_2 = (-4,0, -3, -2,1) are a basis for nullspace of a 4X5 matrix A. Find a vector X such that X notequalto 0, X notequalto V_1, X notequalto V_2 and AX = 0.

Solution

Consider any two vectors in S

as (X1 1) and (x2 1) where x1 and x2 are real

(x1 1)+(x2 1) = (x1+x2 2) not in the form (x 1)

Hence addition is not closed

So does not form a subspace

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