Kernel and image of matrix transformation

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Julia Goedecke 2 years, 10 months ago

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Removed the brackets round matrix entry, as the entry is not really a matrix but a collection of columns.

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Created this as a copy of Kernel and image randomised for practice.

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Question statement

Let $T_A\colon {\mathbb{R}^{\var{colnum}} \to \mathbb{R}^{\var{rownum}}}$ with $A = \var{D}$ . Find

a basis for the kernel of the matrix transformation $T_A$ ;
a basis for the image of the matrix transformation $T_A$ ;
the rank and nullity of $T_A$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

pivots3	list	[ 1, 3 ]
Aech	matrix	Matrix of size 4×4
A	matrix	Matrix of size 4×4
pivots4	list	[ 2, 3 ]
Bech	matrix	Matrix of size 3×4
Cech	matrix	Matrix of size 4×3
B	matrix	Matrix of size 3×4
C	matrix	Matrix of size 4×3
D	matrix	Matrix of size 3×4
colnum	number	4
rankD	number	2
kernel_sized_zero_matrix	matrix	Matrix of size 4×2
rownum	number	3
Dech	matrix	Matrix of size 3×4
pivots	list	[ 2, 3 ]
stuffcols	list	[ 1, 4 ]
nullityD	number	2
kernelbasismatrix	matrix	Matrix of size 4×2
kernel	matrix	Matrix of size 4×2
test	matrix	matrix([0,0],[0,0],[0,0])
image	matrix	Matrix of size 3×2
imagecheckmatrix	matrix	Matrix of size 1×3
imagecheck	matrix	matrix([0,0])

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random([[1,2,3],[1,3],[2,3],[2]])

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `list`

[ 1, 3 ]

→ Used by:

Cech
pivots

Parts

Kernel Gap-fill
1. Gap Matrix entry
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Image Gap-fill
1. Gap Matrix entry
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Rank and nullity Gap-fill
1. Gap rank Number entry
  1. Alt Unnamed alternative
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2. Gap nullity Number entry
  1. Alt Unnamed alternative
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Enter the basis for the kernel of $T_A$ as the columns of a matrix. E.g. if there is only one basis vector, select "1 column" and enter the vector. If there are two basis vectors, select "2 columns" and enter the first basis vector in the first column and the second basis vector in the second column, etc. This way you can determine how many basis vectors there are.

If the kernel is $0$, then just enter a zero vector of the correct size.

A basis of the kernel of $T_A$ is:

Unnamed gap

Advice

Kernel and nullity

The kernel of $T_A$ is the set of all vectors which gets mapped to zero: $\operatorname{Ker}(T_A)=\left\{v \middle| Av=0\right\}$. In this case, the kernel of $T_A$ is the same as the nullspace of $A$. So we solve the linear system $Av=0$.

The matrix $\var{D}$ has reduced row echelon form $\var[fractionNumbers]{Dech}$. So we see that there are no stuff columns, so the nullspace is $0$ So we see that there is are $\var{length(stuffcols)}$ stuff columns, and a basis of the kernel is given by $\var[fractionNumbers]{transpose(kernelbasismatrix)[0]}$, $\var[fractionNumbers]{transpose(kernelbasismatrix)[1]}$, $\var[fractionNumbers]{transpose(kernelbasismatrix)[2]}$.

Any correct basis will be accepted above.

The nullity of $T_A$ is the dimension of the kernel, so we see that it is $n(T_A)=\var{nullityD}$.

Image and rank

The image of $T_A$ is the set of all vectors which are reached: $\operatorname{Im}(T_A)=\left\{w\middle| \exists v \text{ s.t. } Av=w\right\}$. The image of the matrix transformation $T_A$ is the same as the column space of matrix $A$. We can find a basis for the column space by taking the columns in $A$ which turn into pivot columns in the RREF of $A$. We see from the above RREF that this is columns $\var{pivots[0]}$, $\var{pivots[1]}$, $\var{pivots[2]}$. (Why? See lecture notes, "how to find a basis for a column space".)

Therefore the rank of $T_A$, the dimension of the image, is $r(T_A)=\var{rankD}$. So actually $\operatorname{Im}(A)= \mathbb{R}^{\var{rownum}}$, so any basis of $\mathbb{R}^{\var{rownum}}$ is correct here, such as the standard basis.

So a basis for the image is given by $\var{transpose(image)[0]}$, $\var{transpose(image)[1]}$, $\var{transpose(image)[2]}$.

We see that the image is not all of the codomain $\mathbb{R}^{\var{rownum}}$.

Any correct basis will be accepted above.

Rank-Nullity Theorem

We can check that rank + nullity = dimension of domain, i.e. $r(T_A)+n(T_A)=\dim(\mathbb{R}^{\var{colnum}})$: $\var{rankD}+\var{nullityD}=\var{colnum}$.

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Part	Test	Passed?
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Matrix entry
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Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 11 months ago

Error in variable testing condition

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Generated value: `list`

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Kernel and image of matrix transformation

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 10 months ago

Julia Goedecke 2 years, 11 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

Generated value: `list`