Subspaces, Affine sets, Convex sets, Cones
John Mitchell

This document describes various subsets of $I\!\!R^n$.

• Let $v_1,\ldots,v_k$ be k vectors in $I\!\!R^n$. Let $\lambda_1,\ldots,\lambda_k$ be k scalars. The vector $v:=\sum_{i=1}^k\lambda_iv_i$ is a linear combination of $v_1,\ldots,v_k$.
• Let S be a subset of $I\!\!R^n$. S is a subspace if it is closed under linear combinations. Thus, for any k>0, for any vectors $v_1,\ldots,v_k \in S$, and for any scalars $\lambda_1,\ldots,\lambda_k$, the linear combination $v:=\sum_{i=1}^k\lambda_iv_i$ is also in S. Notice that the origin is in any nonempty subspace -- just take all $\lambda_i=0$.
• The row space, range, and null space of a matrix are all subspaces.
• Let $v_1,\ldots,v_k$ be k vectors in $I\!\!R^n$. Let $\lambda_1,\ldots,\lambda_k$ be k scalars satisfying $\sum_{i=1}^k\lambda_i=1$. (Note: some of the scalars may be negative.) The vector $v:=\sum_{i=1}^k\lambda_iv_i$ is an affine combination of $v_1,\ldots,v_k$.
• Let S be a subset of $I\!\!R^n$. S is an affine space if it is closed under affine combinations. Thus, for any k>0, for any vectors $v_1,\ldots,v_k \in S$, and for any scalars $\lambda_1,\ldots,\lambda_k$ satisfying $\sum_{i=1}^k\lambda_i=1$, the affine combination $v:=\sum_{i=1}^k\lambda_iv_i$ is also in S.
• The set of solutions to the system of equations Ax=b is an affine space. This is why we talk about affine spaces in this course!
• An affine space is a translation of a subspace.
• Any subspace is also an affine space.
• Let $v_1,\ldots,v_k$ be k vectors in $I\!\!R^n$. Let $\lambda_1,\ldots,\lambda_k$ be k nonnegative scalars satisfying $\sum_{i=1}^k\lambda_i=1$. The vector $v:=\sum_{i=1}^k\lambda_iv_i$ is a convex combination of $v_1,\ldots,v_k$.
• Let S be a subset of $I\!\!R^n$. S is an convex set if it is closed under convex combinations. Thus, for any k>0, for any vectors $v_1,\ldots,v_k \in S$, and for any nonnegative scalars $\lambda_1,\ldots,\lambda_k$ satisfying $\sum_{i=1}^k\lambda_i=1$, the convex combination $v:=\sum_{i=1}^k\lambda_iv_i$ is also in S.
• 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 $I\!\!R^n$. S is a cone if it is closed under nonnegative scalar multiplication. Thus, for any vector $v\in S$ and for any nonnegative scalar $\lambda$, the vector $\lambda v$ is also in S.
• Let S be a subset of $I\!\!R^n$. 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 $v_1,\ldots,v_k \in S$, and for any nonnegative scalars $\lambda_1,\ldots,\lambda_k$, the linear combination $v:=\sum_{i=1}^k\lambda_iv_i$ is also in S.
• The origin is in any nonempty cone -- just take $\lambda=0$.
• Any subspace is a convex cone.