subspace of $\Re^n$ is any set H in $\Re^n$ that has properties: a) The zero vector is in H b) For each u and v in H, the sum $u + v$ is in H c) For each u in H and each scalar c, the vector $cu$ is in H
the zero subspace is the set that only contains zero vector in $\Re^n$
column space: set of all linear combinations of the columns of a matrix. it’s the span of the columns of the matrix.
column space of m × n matrix is subspace of $\Re^m$
null space: set of all solutions of equation $Ax = 0$.
null space of an m × n matrix is subspace of $\Re^n$.
to determine if p is in Nul A, check if Ap = 0. if so, p is in Nul A.
basis for subspace H of $\Re^n$ is linearly independent set in H spanning H
the pivot columns of a matrix form the basis for its column space.
let $H \in \Re^n$ be subspace with $B = { b_1, \dots, b_p}$. then for all x ∈ H, there are unique $c_1, \dots, c_p$ such that $x = c_1 b_2 + \dots + c_p b_p$. (to prove this theorem, use a contradiction on uniqueness)
the coordinates of x w.r.t. B are $c_1, \dots, c_p$.
the coordinate system of x w.r.t. B is $[x]_B = \begin{bmatrix}c_1\\ \dots\\ c_p\end{bmatrix}$
let $H \in \Re^n$ be a subspace with basis $B={ b_1, \dots, b_p }$. then every basis for H comprises p vectors.
the dimension of H is the number of basis vectors in any basis for H.
dim Col A = #pivot columns (rank A)
dim Nul A = #free variables in Ax = 0
Rank theorem: dim Col A + dim Nul A = #columns