Computing the three fundamental subspaces

Example First, compute the RREF of $A$ (RREF)

Column space

$\subseteq {\mathbb{R}}^m$

NOTE

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

$R$ is in $RREF(1,3)$ so $B=\{\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\}$ given by the two independent columns 1 and 3

NOTE

row operation does not change the position of the linear independent colomn vectors

Row space

$\subseteq {\mathbb{R}}^n$

NOTE

$\text{dim}(\mathbf{R}(A))=r$

$B=\{\left(\begin{array}{c}1\\ 2\\ 0\\ 3\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ -2\end{array}\right)\}$ given by $R$

after row operation the linear independent row vectors is still linear independent.

Conclusion (Theorem 4.33)

Let $A$ be an $m\times n$ matrix. Then $\mathbf{rank}(A)=\mathbf{rank}(A^{\top })$

note that:

• $\mathbf{rank}(A^{\top })=\text{dim}(\mathbf{C}(A^{\top }))=\text{dim}(\mathbf{R}(A))=r$
• $\mathbf{rank}(A)=\text{dim}(\mathbf{C}(A))=r$

NOTE

The rank is at most the smaller of the two matrix dimensions. $r\le \text{min}(n,m)$

Null space

$\subseteq {\mathbb{R}}^n$

$\mathbf{N}(A)=\mathbf{N}(R)=\mathbf{N}(R^{\prime })$ by definition $\mathbf{x}\in \mathbf{N}(R^{\prime })⟺R^{\prime }\mathbf{x}=0$

example

we separate the linear independent and dependent columns

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\underset{\in {\mathbb{R}}^r}{\underbrace{\left(\begin{array}{c}{x}_1\\ x_3\end{array}\right)}}+\left[\begin{array}{cc}2& 3\\ 0& -2\end{array}\right]\underset{\in {\mathbb{R}}^{n-r}}{\underbrace{\left(\begin{array}{c}{x}_2\\ x_4\end{array}\right)}}=0$$

note that the first matrix is always $I$. Thus we have

$$\left(\begin{array}{c}{x}_1\\ x_3\end{array}\right)=-\left[\begin{array}{cc}2& 3\\ 0& -2\end{array}\right]\left(\begin{array}{c}{x}_2\\ x_4\end{array}\right)$$

$⟹\mathbf{N}(R^{\prime })$ is isomorphic to ${\mathbb{R}}^{n-r}$ and the dimension is therefore $n-r$

to get $\mathbf{N}(A)$ we first write the solution in the form of the elements corresponding to the dependent columns $\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)=\left(\begin{array}{c}-2x_2-3x_4\\ x_2\\ 2x_4\\ x_4\end{array}\right),x_2,x_4\in \mathbb{R}$

To get the basis of $\mathbf{N}(A)$ we give $x_2,x_4$ the values $\left(\begin{array}{c}0\\ 1\end{array}\right)$ and $\left(\begin{array}{c}1\\ 0\end{array}\right)$ So in this example, we get $\{\left(\begin{array}{c}-3\\ 0\\ 2\\ 1\end{array}\right),\left(\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right)\}$

Null space isomorphism

(still using the previous example)

$$T:\underset{\in \mathbf{N}(R^{\prime })}{\underbrace{\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)}}\to \underset{\in {\mathbb{R}}^{n-r}}{\underbrace{\left(\begin{array}{c}{x}_2\\ x_4\end{array}\right)}}$$

This is a bijective linear transformation. For every $\left(\begin{array}{c}{x}_2\\ x_4\end{array}\right)$ we can compute the input using the equation we got previously

Summary