Mixing
Finding kernel and range of a linear transformation
$begingroup$
We are given:
Find $ker(T)$, and $textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:mathbb{R^3} rightarrow mathbb{R^3}$$
with standard matrix
$$ A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]textrm{.}
$$
The kernel can be found in a $2 times 2$ matrix as follows:
$$ L = left[begin{array}{rrr}
a & b\
c & d\
end{array}right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = left[begin{array}{rrr}
a & b\
-b & -a\
end{array}right]
$$
but I do not know how to apply that to this problem.
linear-algebra linear-transformations
edited May 31 '15 at 1:07
Ken
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
$endgroup$
add a comment |
$begingroup$
We are given:
Find $ker(T)$, and $textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:mathbb{R^3} rightarrow mathbb{R^3}$$
with standard matrix
$$ A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]textrm{.}
$$
The kernel can be found in a $2 times 2$ matrix as follows:
$$ L = left[begin{array}{rrr}
a & b\
c & d\
end{array}right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = left[begin{array}{rrr}
a & b\
-b & -a\
end{array}right]
$$
but I do not know how to apply that to this problem.
linear-algebra linear-transformations
edited May 31 '15 at 1:07
Ken
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
$endgroup$
3
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56
add a comment |
$begingroup$
We are given:
Find $ker(T)$, and $textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:mathbb{R^3} rightarrow mathbb{R^3}$$
with standard matrix
$$ A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]textrm{.}
$$
The kernel can be found in a $2 times 2$ matrix as follows:
$$ L = left[begin{array}{rrr}
a & b\
c & d\
end{array}right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = left[begin{array}{rrr}
a & b\
-b & -a\
end{array}right]
$$
but I do not know how to apply that to this problem.
linear-algebra linear-transformations
edited May 31 '15 at 1:07
Ken
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
$endgroup$
We are given:
Find $ker(T)$, and $textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:mathbb{R^3} rightarrow mathbb{R^3}$$
with standard matrix
$$ A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]textrm{.}
$$
The kernel can be found in a $2 times 2$ matrix as follows:
$$ L = left[begin{array}{rrr}
a & b\
c & d\
end{array}right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = left[begin{array}{rrr}
a & b\
-b & -a\
end{array}right]
$$
but I do not know how to apply that to this problem.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited May 31 '15 at 1:07
Ken
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
edited May 31 '15 at 1:07
Ken
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
edited May 31 '15 at 1:07
Ken
3,63151728
edited May 31 '15 at 1:07
Ken
3,63151728
edited May 31 '15 at 1:07
Ken
3,63151728
3,63151728
asked May 31 '15 at 0:37
user4640007user4640007
1711213
asked May 31 '15 at 0:37
user4640007user4640007
1711213
asked May 31 '15 at 0:37
user4640007user4640007
1711213
1711213
3
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56
add a comment |
3
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56
3
3
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$
A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)={x in R^n|Ax=0}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $xne0$ and $0$ is a vector here.
$$
left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
0\0\0
end{array}right]
$$
In row-reduced form,
$$
A = left[begin{array}{rrr}
1 & 0 & frac{14}{11}\
0 & 1 & frac{-19}{11}\
0 & 0 & 0\
end{array}right]
$$
$$x=frac{-14}{11}z$$
$$y=frac{19}{11}z$$
$$
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
-14\19\11
end{array}right]z
$$
Similarly for $2×2$ matrix .
answered May 31 '15 at 1:23
vidhanvidhan
853315
$endgroup$
add a comment |
$begingroup$
For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
$endgroup$
add a comment |
$begingroup$
To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
$endgroup$
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)={x in R^n|Ax=0}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $xne0$ and $0$ is a vector here.
$$
left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
0\0\0
end{array}right]
$$
In row-reduced form,
$$
A = left[begin{array}{rrr}
1 & 0 & frac{14}{11}\
0 & 1 & frac{-19}{11}\
0 & 0 & 0\
end{array}right]
$$
$$x=frac{-14}{11}z$$
$$y=frac{19}{11}z$$
$$
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
-14\19\11
end{array}right]z
$$
Similarly for $2×2$ matrix .
answered May 31 '15 at 1:23
vidhanvidhan
853315
$endgroup$
add a comment |
$begingroup$
$$
A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)={x in R^n|Ax=0}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $xne0$ and $0$ is a vector here.
$$
left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
0\0\0
end{array}right]
$$
In row-reduced form,
$$
A = left[begin{array}{rrr}
1 & 0 & frac{14}{11}\
0 & 1 & frac{-19}{11}\
0 & 0 & 0\
end{array}right]
$$
$$x=frac{-14}{11}z$$
$$y=frac{19}{11}z$$
$$
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
-14\19\11
end{array}right]z
$$
Similarly for $2×2$ matrix .
answered May 31 '15 at 1:23
vidhanvidhan
853315
$endgroup$
add a comment |
$begingroup$
$$
A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)={x in R^n|Ax=0}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $xne0$ and $0$ is a vector here.
$$
left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
0\0\0
end{array}right]
$$
In row-reduced form,
$$
A = left[begin{array}{rrr}
1 & 0 & frac{14}{11}\
0 & 1 & frac{-19}{11}\
0 & 0 & 0\
end{array}right]
$$
$$x=frac{-14}{11}z$$
$$y=frac{19}{11}z$$
$$
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
-14\19\11
end{array}right]z
$$
Similarly for $2×2$ matrix .
answered May 31 '15 at 1:23
vidhanvidhan
853315
$endgroup$
$$
A = left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)={x in R^n|Ax=0}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $xne0$ and $0$ is a vector here.
$$
left[begin{array}{rrr}
1 & -1 & 3\
5 & 6 & -4\
7 & 4 & 2\
end{array}right]
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
0\0\0
end{array}right]
$$
In row-reduced form,
$$
A = left[begin{array}{rrr}
1 & 0 & frac{14}{11}\
0 & 1 & frac{-19}{11}\
0 & 0 & 0\
end{array}right]
$$
$$x=frac{-14}{11}z$$
$$y=frac{19}{11}z$$
$$
left[begin{array}{r}
a\b\c
end{array}right]
=left[begin{array}{r}
-14\19\11
end{array}right]z
$$
Similarly for $2×2$ matrix .
answered May 31 '15 at 1:23
vidhanvidhan
853315
edited May 31 '15 at 1:52
answered May 31 '15 at 1:23
vidhanvidhan
853315
answered May 31 '15 at 1:23
vidhanvidhan
853315
answered May 31 '15 at 1:23
vidhanvidhan
853315
853315
add a comment |
add a comment |
$begingroup$
For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
$endgroup$
add a comment |
$begingroup$
For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
$endgroup$
add a comment |
$begingroup$
For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
$endgroup$
For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
answered May 31 '15 at 15:22
Sam ChristopherSam Christopher
585323
585323
add a comment |
add a comment |
$begingroup$
To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
$endgroup$
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
add a comment |
$begingroup$
To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
$endgroup$
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
add a comment |
$begingroup$
To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
$endgroup$
To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
answered Jan 23 at 13:01
sesan Amusasesan Amusa
1
1
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
add a comment |
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
2
2
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 23 at 13:22
2
2
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
$begingroup$
Giving a hurried and partial (you do not even mention the kernel of $T$) Answer after so much time has passed is of negligible value. When an older Question already has an Accepted and/or upvoted Answer, it is expedient to carefully highlight what new information is being added (thus demonstrating that you've considered the existing Answers and are not simply repeating the work of others).
$endgroup$
– hardmath
Jan 23 at 16:25
add a comment |
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3
$begingroup$
Kernels are defined for linear transformations, not matrices. Usually when we say the "kernel of a matrix $A$", what we really mean is the kernel of the linear transformation $x mapsto Ax$ for a column matrix $x$. The kernel in that case will be a set of column matrices. So I don't understand what you mean when you say that the kernel of $L$ is the set of matrices $begin{bmatrix} a & b \ -b & -aend{bmatrix}$. Can you expand on what exactly you mean and where this comes from?
$endgroup$
– user137731
May 31 '15 at 0:56