1.12 Image and Kernel

The image and kernel of a linear mapping are vector subspaces with certain important properties

Definition:  Image and Kernel

For $\Phi: V \to W$, we define the kernel or null space

$ker(\Phi)=\Phi^{-1}(0_w)=\{v \in V: \Phi(v)=0_w\}$

and image or range

$Im(\Phi) =\Phi(V)=\{w \in W| \exists v \in V: \Phi(v)=w\}$

We also call $V$ and $W$ as domain and codomain of $\Phi$, respectively.

Intuitively, the kernel is the set of vectors in $v \in V$ that $\Phi$ maps onto the neutral element $0_w \in W$. The image is the set of vectors $w \in W$  that can be “reached” by $\Phi$ from any vector in $V$ . An illustration is given in figure.

Remark: Consider a linear mapping $\Phi: V \to W$, where $V,W$ are vector spaces.

• It always holds that $\Phi(0_v)=0_w$ and therefore $0_v \in ker(\Phi)$.In particular the null space is never empty.
• $Im(\Phi) \subseteq W$ is a subspace of $W$, and $ker(\Phi) \subseteq V$ is a subspace of $V$.

Remark(Null space and Column space)

Let us consider $A \in \mathbb{R}^{m \times n}$ and a linear mapping $\Phi: \mathbb{R}^n \to \mathbb{R}^m, x \to Ax$

• for $A=[a_1,\ldots,a_n]$, where $a_i$ are the columns of $A_i$, we obtain$Im(\Phi)={Ax:x \in \mathbb{R}^n}=\sum_{i=1}^n x_ia_i : x_1,\ldots,x_n \in \mathbb{R}$             $=span[a_1,\ldots,a_n] \subseteq \mathbb{R}^m$
• i.e., the image is the span of the columns of $A$, also called the column space. Therefore, the column space(image) is a subspace of $\mathbb{R}^m$ ,where $m$ is the height of the matrix.
• $rk(A)=dim(Im(\Phi))$
• The kernel/null space $ker(\Phi)$ is the general solution to the homogenious system of linear equations $Ax=0$ and captures all possible linear combination of elements in $\mathbb{R}^n$ that produces $0 \in \mathbb{R}^m$
• The kernel is a subspace of $\mathbb{R}^n$, where $n$ is the width of the matrix.
• The kernel focuses on the relationship among the columns, and we can use it to determine whether/how we can express a column as a linear combination of other columns

Rank Nullity Theorem

For vector spaces $V,W$ and a linear mapping $\Phi:V \to W$ it holds that

$dim(ker(\Phi))+ dim(Im(\Phi))= dim(V)$

The rank nullity theorem is also referred to as the fundamental theorem of linear mapping.

If  $dim(Im(\Phi)) < dim(V)$, then $ker(\Phi)$ is non trivial, i.e, the kernel contains more than $0_v$ and $dim(ker(\Phi)) \ge 1$.

If $A_\phi$ is the transformation matrix of $\Phi$ with respect to an ordered basis and $dim(Im(\Phi)) < dim(V)$, then the system of linear equations $A_\phi x=0$ has infinitely many solutions.

If $dim(V)=dim(W)$, then the following three way equivalence holds

$\Phi$ is injective

$\Phi$ is surjective

$\Phi$ is bijective

since $Im(\Phi) \subseteq W$

import sympy as sp

import numpy as np

row1=[1,2,-1,0]

row2=[1,0,0,1]

M = sp.Matrix((row1,row2))

print("Transformation Matrix")

display(M)

print("Reduced Echelon Form")

display(M.rref())

O/P

Transformation Matrix
$\displaystyle \left[\begin{matrix}1 & 2 & -1 & 0\\1 & 0 & 0 & 1\end{matrix}\right]$

Reduced Echelon Form
(Matrix([ [1, 0, 0, 1], 

             [0, 1, -1/2, -1/2]]), (0, 1))

Note: from the reduced echelon form we can find out the null space.

We can also sympy package nullspace method to find null space

# import sympy

from sympy import *

M = Matrix([[1, 2, -1, 0], [1, 0, 0, 1]])

display(M)

# Use sympy.nullspace() method

M_nullspace = M.nullspace()

print("Nullspace of a matrix : ")

display(M_nullspace)

O/P

$\displaystyle \left[\begin{matrix}1 & 2 & -1 & 0\\1 & 0 & 0 & 1\end{matrix}\right]$
Nullspace of a matrix :
[Matrix([ [ 0], [1/2], [ 1], [ 0]]), 

 Matrix([ [ -1], [1/2], [ 0], [ 1]])]

Watch the video https://www.youtube.com/watch?v=4SlMWZxgZFE

Example:

Find the Image and Kernel of a linear mapping $A=\begin{bmatrix} 2 &3 \\ 6 & 9 \end{bmatrix}$

After row reduction the echelon form is

$A=\begin{bmatrix} 2 &3 \\ 0 & 0 \end{bmatrix}$

$A=\begin{bmatrix} 1 &3/2 \\ 0 & 0 \end{bmatrix}$

So the pivot column indicate linear independent vector

$Im(A)=span(\begin{bmatrix} 2 \\ 6 \\ \end{bmatrix})$

The null space is spanned by

$Null(A)=span(\begin{bmatrix} 3/2 \\ -1 \\ \end{bmatrix})$

Example ( university question)

Consider the transformation $T (x, y) = (x + y, x + 2y, 2x + 3y)$. Obtain ker $T$ and use this to calculate the nullity. Also find the transformation matrix for $T$.

The transformation matrix is

$T=\begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 2 & 3 \end{bmatrix}$

The null space consist of  the trivial vector 

$\begin{bmatrix}0 \\ 0\\ \end{bmatrix}$

Given the linear transformation $T(x,y,z,w)=(x+2y-z,x+w)$, find the basis each for Ker(T) and Range(T) ( University Question)

The transformation matrix is

$\begin{bmatrix} 1 & 2 & -1 & 0 \\ 1 & 0 & 0 & 1 \\ \end{bmatrix}$

After row reduction, the reduced echelon form is
$\begin{bmatrix} 1 & 2 & -1 & 0 \\ 0 & -2 & 1 & 1 \\ \end{bmatrix} R2= R2-R1$

$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -2 & 1 & 1 \\ \end{bmatrix}  R1=R1+R2$

$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -1/2 & -1/2 \\ \end{bmatrix}  R2=R2/-2$

Basis for Range(T) is $\begin{bmatrix} 1 \\ 1\\ \end{bmatrix}\begin{bmatrix} 2 \\ 0 \\ \end{bmatrix}$
Basis for Ker(T) is $\begin{bmatrix} 0 \\ -1/2 \\ -1\\ 0 \end{bmatrix}\begin{bmatrix} 1 \\ -1/2 \\ 0\\ -1 \end{bmatrix}$

 

 Let $T$ be the linear transformation from $R^2$ to $R^3$ defined by $T(v)=Av$, with standard matrix $A=\begin{bmatrix} 2& -6 & 4 \\ 1 & -1 & 2 \end{bmatrix}$. Find the kernel and image of $T$.

After row reduction, the reduced echelon form is

 

$A=\begin{bmatrix} 2& -6 & 4 \\ 1 & -1 & 2 \end{bmatrix}$

 

$R1=R1/2$

 $A=\begin{bmatrix} 1& -3& 2 \\ 1 & -1 & 2 \end{bmatrix}$

 

 

$R2=R2-R1$

 $A=\begin{bmatrix} 1& -3& 2 \\ 0 & 2 & 0 \end{bmatrix}$

 

$R2=R2/2$

 $A=\begin{bmatrix} 1& -3& 2 \\ 0 & 1 & 0 \end{bmatrix}$

 

 

$R1=R2*3 + R1$

 $A=\begin{bmatrix} 1& 0& 2 \\ 0 & 1 & 0 \end{bmatrix}$

 

Basis for Range(T) is $\begin{bmatrix} 2 \\ 1\\ \end{bmatrix}\begin{bmatrix} -6 \\ -1 \\ \end{bmatrix}$
Basis for Ker(T) is 

$\begin{bmatrix} 2 \\ 0 \\ -1\\ \end{bmatrix}$