// Numbas version: finer_feedback_settings {"name": "Kernel and image of matrix transformation", "extensions": ["linear-algebra", "linalg2"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Kernel and image of matrix transformation", "tags": ["image", "kernel", "Linear algebra", "Linear Algebra", "linear algebra", "matrix transformation"], "metadata": {"description": "

Student finds a basis for kernel and image of a matrix transformation. Any basis can be entered; there is a custom marking algorithm which checks if it is a correct basis.

There are options to adjust this question fairly easily, for example to get different variants for practice and for a test, by changing the options in the \"pivot columns\" in the variables. You should be careful to think about and test your pivot options, as some are easier or harder than others, and some don't work very well.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "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\\).

", "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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transpose(matrix(map(map(-Dech[l][k-1],l,0..(k-2))+[1]+repeat(0,colnum-k),k,stuffcols)))

transpose(matrix(map(map(-Dech[l][k-1],l,0..(k-2) except stuffcols)+[1]+repeat(0,colnum-k),k,stuffcols)))

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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:

[[0]]

Enter the basis for the image 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 image is \\(0\\), then just enter a zero vector of the correct size.

A basis of the image of \\(T_A\\) is:

[[0]]

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The rank and nullity of \\(T_A\\) are:

\\(r(T_A)= \\) [[0]]

\\(n(T_A)= \\) [[1]]

Check for yourself if they satisfy the Rank-Nullity Theorem.

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Check whether you have entered the rank and nullity the wrong way round.

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Check if you have entered the rank and nullity the wrong way round.

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