Section05 Dimensions of the Four Subspaces

Four Fundamental Subspaces

Type	Represented by Mathical Symbol	Subspace of	Dimension
*row space*	$\boldsymbol{C}(\boldsymbol{A}^{T})$	$\mathbb{R}^{n}$	$r$ (the rank of matrix)
*column space*	$\boldsymbol{C}(\boldsymbol{A})$	$\mathbb{R}^{m}$	$r$ (the rank of matrix)
*nullspce*	$\boldsymbol{N}(\boldsymbol{A})$	$\mathbb{R}^{n}$	$n-r$
*left nullspace*	$\boldsymbol{N}(\boldsymbol{A}^{T})$	$\mathbb{R}^{m}$	$m-r$

Row Space and nullspace are perpendicular!
Column Space and left nullspace are perpendicular!

Type	Basis	Dimension
*row space*	The nonzero rows of $\boldsymbol{R}$ .	The number of nonzero rows of $\boldsymbol{R}$ . (Rank of Matrix)
*column space*	The pivot columns of $\boldsymbol{A}$ , which coresponded with the columns have pivot of $\boldsymbol{R}$ .	The number of pivots columns. (Rank of Matrix)
*nullspce*	Special solutions of $\boldsymbol{Ax} = \boldsymbol{0}$	The number of free columns of $\boldsymbol{A} \text{ or } \boldsymbol{R}$ . (free variables of solutions.) ( $n-r$ )
*left nullspace*	Special solutions of $\boldsymbol{A}^{T}\boldsymbol{y} = \boldsymbol{0}$ .	The number of free columns of $\boldsymbol{A}^{T}$ ( $m-r$ )

Type	Properties
*row space*	$\boldsymbol{A}$ has the same row space with $\boldsymbol{R}$
*column space*	$\boldsymbol{A}$ 's column space is different with $\boldsymbol{R}$ 's, but has the same dimension with $\boldsymbol{R}$
*nullspce*	$\boldsymbol{A}$ has the same nullspace with $\boldsymbol{R}$
*left nullspace*	$\boldsymbol{A}$ 's left nullspace has the dimension of $m-r$

Counting Theorem: $\dim(\text{column space}) + \dim(\text{nullspace}) = \dim(\mathbb{R}^{n})$

Every rank matrix is one column times one row. $\boldsymbol{A} = \boldsymbol{u}\boldsymbol{v}^{T}$

Proof $\begin{array}{ll} & \text{Suppose } \boldsymbol{A} \text{ is a rank 2 matrix.} \\ \because & \boldsymbol{A} = (\text{pivots columns of }\boldsymbol{A})(\text{nonzero rows of } \boldsymbol{R}) \\ \therefore & \boldsymbol{A} = \left[ \begin{matrix} \boldsymbol{u}_{1} & \boldsymbol{u}_{2} \\ \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{v}_{1}^{T} \\ \boldsymbol{v}_{2}^{T} \end{matrix} \right] = \boldsymbol{u}_{1}\boldsymbol{v}_{1}^{T} + \boldsymbol{u}_{2}\boldsymbol{v}_{2}^{T}\\ \because & \boldsymbol{u}\boldsymbol{v}^{T} \text{ is a rank 1 matrix.} \\ \therefore & \text{Rank 2 matrices = Rank 1 matrix + Rank 1 matrix} \\ \end{array}$

Subspace of
- Row Space
  - rows of $\boldsymbol{AB}$ is the combination of $\boldsymbol{B}$ 's rows, therefore
  - $\text{Row space of } \boldsymbol{AB} \subseteq \text{Row space of } \boldsymbol{B}; \text{Rank}(\boldsymbol{AB}) \le \text{Rank}(\boldsymbol{B})$
- Column Space
  - coluns of $\boldsymbol{AB}$ is the combination of $\boldsymbol{A}$ 's columns, therefore
  - $\text{Column space of } \boldsymbol{AB} \subseteq \text{Column space of } \boldsymbol{A}; \text{Rank}(\boldsymbol{AB}) \le \text{Rank}(\boldsymbol{A})$
Find a Matrix whose one basis of column space and row space is $\boldsymbol{u}, \boldsymbol{v}$ . $\boldsymbol{A} = \left[ \begin{matrix} \boldsymbol{u} & \boldsymbol{v} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{u}^{T} \\ \boldsymbol{v}^{T} \end{matrix} \right] \text{ is one matrix which satisfied with these requirements.}$
Suppose $\boldsymbol{A} = \sum_{i=1}^{n}\boldsymbol{u}_{i}\boldsymbol{v}_{i}^{T}$ The columns is the linearly combination of $\boldsymbol{u}_{i}$ , The rows are the linearly combination of $\boldsymbol{v}_{i}$ .