Subspaces, Affine sets, Convex sets, Cones
John Mitchell

• Let v1,,vk be k vectors in IRn. Let λ 1,k be k scalars. The vector v := i=1kλ ivi is a linear combination of v1,,vk.
• Let S be a subset of IRn. S is a subspace if it is closed under linear combinations. Thus, for any k > 0, for any vectors v1,,vk S, and for any scalars λ 1,k, the linear combination v := i=1kλ ivi is also in S. Notice that the origin is in any nonempty subspace — just take all λi = 0.
• The row space, range, and null space of a matrix are all subspaces.
• Let v1,,vk be k vectors in IRn. Let λ 1,k be k scalars satisfying i=1kλ i = 1. (Note: some of the scalars may be negative.) The vector v := i=1kλ ivi is an affine combination of v1,,vk.
• Let S be a subset of IRn. S is an affine space if it is closed under affine combinations. Thus, for any k > 0, for any vectors v1,,vk S, and for any scalars λ 1,k satisfying i=1kλ i = 1, the affine combination v := i=1kλ ivi is also in S.
• The set of solutions to the system of equations Ax = b is an affine space.
• An affine space is a translation of a subspace.
• Any subspace is also an affine space.
• Let v1,,vk be k vectors in IRn. Let λ 1,k be k nonnegative scalars satisfying i=1kλ i = 1. The vector v := i=1kλ ivi is a convex combination of v1,,vk.
• Let S be a subset of IRn. S is an convex set if it is closed under convex combinations. Thus, for any k > 0, for any vectors v1,,vk S, and for any nonnegative scalars λ 1,k satisfying i=1kλ i = 1, the convex combination v := i=1kλ ivi is also in S.
• A hyperplane is the set of solutions to a linear equality aT x = b with a IRn \{0}.
• A half-space is the set of solutions to a linear inequality aT x b with a IRn \{0}.
• A polyhedron is the intersection of a finite number of half-spaces. Polyhedra are convex sets.
• A polytope is defined to be a bounded polyhedron. Note that every point in a polytope is a convex combination of the extreme points.
• Any subspace is a convex set. Any affine space is a convex set.
• Let S be a subset of IRn. S is a cone if it is closed under nonnegative scalar multiplication. Thus, for any vector v S and for any nonnegative scalar λ, the vector λv is also in S.
• Let S be a subset of IRn. S is a convex cone if it is a cone and it is convex. It can be shown that this is equivalent to saying that S is closed under nonnegative linear combinations. Thus, for any k > 0, for any vectors v1,,vk S, and for any nonnegative scalars λ 1,k, the linear combination v := i=1kλ ivi is also in S.
• The origin is in any nonempty cone — just take λ = 0.
• Any subspace is a convex cone.