The Four Fundamental Subspaces

• Four subspaces and their basis and dimension
• [[#column-space-ca|Column space $C(A)$]]
• [[#nullspace-na|Nullspace $N(A)$]]
• [[#row-space-camathrmt|Row space $C(A^{\mathrm{T}})$]]
• [[#left-nullspace-namathrmt|Left Nullspace $N(A^{\mathrm{T}})$]] - [[#how-to-find-a-basis-for-namathrmt|how to find a basis for $N(A^{\mathrm{T}})$]]
• New vector space

Four subspaces and their basis and dimension §

Any $m$ by $n$ matrix $A$ determines four Subspaces (possibly containing only the zero vector):

Column space $C(A)$ §

All combinations of the columns of $A$

A subspace of ${\mathbb{R}}^m$

The $r$ pivot columns form a basis

$\text{dim }C(A)=r$

Nullspace $N(A)$ §

All solutions $\mathbf{x}$ of the equation $A\mathbf{x}=0$

A subspace of ${\mathbb{R}}^n$

The $n-r$ special solutions from a basis

$\text{dim }N(A)=n-r$

Row space $C(A^{\mathrm{T}})$ §

All combinations of the rows of $A$

A subspace of ${\mathbb{R}}^n$

The first $r$ rows of $R$ (the reduced row echelon form of $A$) form a basis

$\text{dim }C(A^{\mathrm{T}})=r$

Left Nullspace $N(A^{\mathrm{T}})$ §

All solutions $\mathbf{y}$ of the equation ${\mathbf{y}}^{\mathrm{T}}A=0$

A subspace of ${\mathbb{R}}^m$

The bottom $m-r$ of $E$ form a basis

$\text{dim }N(A^{\mathrm{T}})=m-r$

how to find a basis for $N(A^{\mathrm{T}})$ §

First we get $E$ through Gauss-Jordan elimination:

$$E_{m\times n}\left[\begin{array}{cc}{A}_{m\times n}& I_{m\times n}\end{array}\right]=\left[\begin{array}{cc}{R}_{m\times n}& E_{m\times n}\end{array}\right]$$

Then the bottom $m-r$ rows of $E$ describe linear dependencies of rows of $A$ since the bottom $m-r$ rows of $R$ are zero. For example:

$$EA=\left[\begin{array}{ccc}-1& 2& 0\\ 1& -1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{cccc}1& 2& 3& 1\\ 1& 1& 2& 1\\ 1& 2& 3& 1\end{array}\right]=\left[\begin{array}{cccc}1& 0& 1& 1\\ 0& 1& 1& 0\\ 0& 0& 0& 0\end{array}\right]=R$$

In this example, The last row of $E$ satisfy the equation ${\mathbf{y}}^{\mathrm{T}}A=0$ and thus form a basis for the left Nullspace of $A$.

New vector space §

We put all $3$ by $3$ matrices together and see the collection as a new vector space; we call it M.

Some subspace of $M$ include:

• all upper triangular matrices
• all symmetric matrices
• $D$, all diagonal matrices

$D$ is the intersection of the first two space. It also has a dimension of $3$ and a basis for $D$ is:

$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 7\end{array}\right].$$