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How to find a basis for the kernel and image of a linear transformation matrix
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1
down vote
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Let $$A=
begin{bmatrix}
0 & 0 & 6 & -18 \
0 & 0 & -1 & 3 \
0 & 0 & -2 & 6 \
end{bmatrix}
$$
Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.
My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.
linear-algebra linear-transformations
edited Oct 9 '17 at 16:21
Jack
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
1
down vote
favorite
Let $$A=
begin{bmatrix}
0 & 0 & 6 & -18 \
0 & 0 & -1 & 3 \
0 & 0 & -2 & 6 \
end{bmatrix}
$$
Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.
My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.
linear-algebra linear-transformations
edited Oct 9 '17 at 16:21
Jack
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $$A=
begin{bmatrix}
0 & 0 & 6 & -18 \
0 & 0 & -1 & 3 \
0 & 0 & -2 & 6 \
end{bmatrix}
$$
Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.
My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.
linear-algebra linear-transformations
edited Oct 9 '17 at 16:21
Jack
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
Let $$A=
begin{bmatrix}
0 & 0 & 6 & -18 \
0 & 0 & -1 & 3 \
0 & 0 & -2 & 6 \
end{bmatrix}
$$
Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.
My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Oct 9 '17 at 16:21
Jack
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
edited Oct 9 '17 at 16:21
Jack
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
edited Oct 9 '17 at 16:21
Jack
27k1681196
edited Oct 9 '17 at 16:21
Jack
27k1681196
edited Oct 9 '17 at 16:21
Jack
27k1681196
27k1681196
asked Oct 10 '16 at 2:57
Aaron
24125
asked Oct 10 '16 at 2:57
Aaron
24125
asked Oct 10 '16 at 2:57
Aaron
24125
24125
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
3 Answers
3
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oldest
votes
up vote
0
down vote
By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.
Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.
answered Oct 10 '16 at 3:23
Jack
27k1681196
add a comment |
up vote
0
down vote
Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
add a comment |
up vote
-1
down vote
The kernel basis is:
$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$
The image basis is:
$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$
answered Oct 10 '16 at 3:32
Aaron
24125
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.
Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.
answered Oct 10 '16 at 3:23
Jack
27k1681196
add a comment |
up vote
0
down vote
By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.
Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.
answered Oct 10 '16 at 3:23
Jack
27k1681196
add a comment |
up vote
0
down vote
up vote
0
down vote
By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.
Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.
answered Oct 10 '16 at 3:23
Jack
27k1681196
By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.
Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.
answered Oct 10 '16 at 3:23
Jack
27k1681196
answered Oct 10 '16 at 3:23
Jack
27k1681196
answered Oct 10 '16 at 3:23
Jack
27k1681196
answered Oct 10 '16 at 3:23
Jack
27k1681196
27k1681196
add a comment |
add a comment |
up vote
0
down vote
Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
add a comment |
up vote
0
down vote
Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
add a comment |
up vote
0
down vote
up vote
0
down vote
Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
answered Apr 19 '17 at 17:58
Tomasz Tarka
424211
424211
add a comment |
add a comment |
up vote
-1
down vote
The kernel basis is:
$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$
The image basis is:
$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$
answered Oct 10 '16 at 3:32
Aaron
24125
add a comment |
up vote
-1
down vote
The kernel basis is:
$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$
The image basis is:
$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$
answered Oct 10 '16 at 3:32
Aaron
24125
add a comment |
up vote
-1
down vote
up vote
-1
down vote
The kernel basis is:
$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$
The image basis is:
$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$
answered Oct 10 '16 at 3:32
Aaron
24125
The kernel basis is:
$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$
The image basis is:
$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$
answered Oct 10 '16 at 3:32
Aaron
24125
answered Oct 10 '16 at 3:32
Aaron
24125
answered Oct 10 '16 at 3:32
Aaron
24125
answered Oct 10 '16 at 3:32
Aaron
24125
24125
add a comment |
add a comment |
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