Nathan S. answered • 03/13/20

Research Mathematician, Game Designer

All of your sets are *subsets* of R^{2} or R^{3} , so they are *in* a subspace. That doesn't seem to be the question though. I think what you want to know is which sets are themselves subspaces.

Lets remember the criteria. A subset U of a vector space V will be a linear subspace if for every pair of vectors u and u' in U, their linear combination is also in U. so in particular 0*u = 0 is in U, and u + u' is in U.

A. y = 4x - 2 doesn't contain the origin, so it isn't a linear subspace of R^{2 }

B. y = cos(x) - 1 does contains the vector (2π, 0), but it doesn't contain the scalar multiple 1/2π(2π,0), or (1,0). so it isnt a linear subspace either

C. 3x - 4y + 2z = 0 contains the origin (0, 0, 0), and if (a, b, c) solves 3x - 4y + 2z = 0, then so does (t*a, t*b, t*c). This is because 3ta - 4tb + 2tc = t(3a - 4b + 2c) = t*0 = 0. Similarly, if (a', b', c') also solves the equation, we have 3(a+a') - 4(b+b') + 2(c+c') = 3a + 3a' - 4b - 4b' + 2c + 2c' = (3a - 4b + 2c) + (3a' - 4b' + 2c') = 0 + 0 = 0. So this IS a linear subspace

D. this set contains 0, and for any two vectors in the set, their sum is also in the set. However, if we scale a vector in the set by a negative number, it escapes. For example (1, 1) is in the set, but -1(1, 1) = (-1, -1) is not in the set. This isnt a linear subspace.

In general with linear algebra, you want to think of linear subspaces as "flat" things which pass through the origin. For R^{n} , these will always be lines, planes, or plane-like things of a higher dimension. If it looks curvy, doesnt contain the origin, or has an edge, it isnt a linear subspace