Matrix

Fundamental Subspace of a Matrix

Nullspace

Let $A\in M_{m\times n}(\mathbb{R})$. The nullspace (also called kernel) of $A$ is the subset of ${\mathbb{R}}^n$ defined by $\text{Null}(A)=\{\vec{x}\in {\mathbb{R}}^n\mid A\vec{x}=\vec{0}\}$

• the nullspace of A is simply the solution space of the homogeneous system of equations $A\vec{x}=\vec{0}$ and is hence a subspace of ${\mathbb{R}}^n$

"Find a basis for Null(A)"

Given the matrix $A$ in RREF form $\left[\begin{array}{cccc}1& 0& 3& 0\\ 0& 1& 2& 1\\ 0& 0& 0& 0\end{array}\right]$. The solution to the homogeneous system $A\vec{x}=\vec{0}$ is $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=s\left[\begin{array}{c}-3\\ -2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ -1\\ 0\\ 1\end{array}\right]$ So we have that $B_1=\{\left[\begin{array}{c}u\\ v\\ w\end{array}\right],\left[\begin{array}{c}u\\ v\\ w\end{array}\right]\}$ is a basis for Null(A) and dim(Null(A)) = 2.

Column Space

Let $A=\left[\begin{array}{ccc}{\vec{a}}_1& \dots & {\vec{a}}_n\end{array}\right]\in M_{m\times n}(\mathbb{R})$. The column space of $A$ is the subset of ${\mathbb{R}}^m$ defined by $\text{Col}(A)=\{A\vec{x}\mid \vec{x}\in {\mathbb{R}}^n\}=\text{span}\{{\vec{a}}_1,\dots ,{\vec{a}}_n\}$

Find a basis for Col(A)

To find a basis. to complete: page 167 of notes

Row Space

Let $A=\left[\begin{array}{c}{\vec{r}}_1^T\\ ⋮\\ {\vec{r}}_m^T\end{array}\right]\in M_{m\times n}(\mathbb{R})$

The row space of $A$ is the subset of ${\mathbb{R}}^n$ defined by $\text{Row}(A)=\{A^T\vec{x}\mid \vec{x}\in {\mathbb{R}}^m\}=\text{span}\{{\vec{r}}_1,\dots ,{\vec{r}}_m\}$

System-Rank Theorem

$\text{dim(Null(A))}=n-\text{rank(A)}$