Mixing
Find a basis of $ker f$ and $im f$ for linear map $f: mathbb R^{3} rightarrow mathbb R^{3}$
$begingroup$
My $f$ is: $f(x_{1},x_{2},x_{3})=(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3})$
1. Basis of $ker f$: $$f(x_{1},x_{2},x_{3})=(0,0,0)$$ So I have a matrix: $$begin{bmatrix} 1 & 2 & -1 \ 1 & 1 & 2 \ 2 & 3 & 1end{bmatrix}$$ Then I have $$x_{1}=-5x_{3}, x_{2}=3x_{3}, x_{3} in mathbb R$$ Finally basic of $ker f$ is $(-5,3,1)$
It is my solution but I'm not sure if this is the good job, so please rate.
2. Basis of $im f$: $$im f={(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3}): x_{1}, x_{2}, x_{3} in mathbb R }$$ So I have a matrix: $$begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 3 \ -1 & 2 & 1end{bmatrix}$$ However I do not understand what can I do after elementary matrix because in books are only some steps and I do not know where the solution comes from.
linear-algebra
asked Jan 11 at 1:14
MP3129MP3129
3049
$endgroup$
add a comment |
$begingroup$
My $f$ is: $f(x_{1},x_{2},x_{3})=(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3})$
1. Basis of $ker f$: $$f(x_{1},x_{2},x_{3})=(0,0,0)$$ So I have a matrix: $$begin{bmatrix} 1 & 2 & -1 \ 1 & 1 & 2 \ 2 & 3 & 1end{bmatrix}$$ Then I have $$x_{1}=-5x_{3}, x_{2}=3x_{3}, x_{3} in mathbb R$$ Finally basic of $ker f$ is $(-5,3,1)$
It is my solution but I'm not sure if this is the good job, so please rate.
2. Basis of $im f$: $$im f={(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3}): x_{1}, x_{2}, x_{3} in mathbb R }$$ So I have a matrix: $$begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 3 \ -1 & 2 & 1end{bmatrix}$$ However I do not understand what can I do after elementary matrix because in books are only some steps and I do not know where the solution comes from.
linear-algebra
asked Jan 11 at 1:14
MP3129MP3129
3049
$endgroup$
$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33
add a comment |
$begingroup$
My $f$ is: $f(x_{1},x_{2},x_{3})=(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3})$
1. Basis of $ker f$: $$f(x_{1},x_{2},x_{3})=(0,0,0)$$ So I have a matrix: $$begin{bmatrix} 1 & 2 & -1 \ 1 & 1 & 2 \ 2 & 3 & 1end{bmatrix}$$ Then I have $$x_{1}=-5x_{3}, x_{2}=3x_{3}, x_{3} in mathbb R$$ Finally basic of $ker f$ is $(-5,3,1)$
It is my solution but I'm not sure if this is the good job, so please rate.
2. Basis of $im f$: $$im f={(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3}): x_{1}, x_{2}, x_{3} in mathbb R }$$ So I have a matrix: $$begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 3 \ -1 & 2 & 1end{bmatrix}$$ However I do not understand what can I do after elementary matrix because in books are only some steps and I do not know where the solution comes from.
linear-algebra
asked Jan 11 at 1:14
MP3129MP3129
3049
$endgroup$
My $f$ is: $f(x_{1},x_{2},x_{3})=(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3})$
1. Basis of $ker f$: $$f(x_{1},x_{2},x_{3})=(0,0,0)$$ So I have a matrix: $$begin{bmatrix} 1 & 2 & -1 \ 1 & 1 & 2 \ 2 & 3 & 1end{bmatrix}$$ Then I have $$x_{1}=-5x_{3}, x_{2}=3x_{3}, x_{3} in mathbb R$$ Finally basic of $ker f$ is $(-5,3,1)$
It is my solution but I'm not sure if this is the good job, so please rate.
2. Basis of $im f$: $$im f={(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3}): x_{1}, x_{2}, x_{3} in mathbb R }$$ So I have a matrix: $$begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 3 \ -1 & 2 & 1end{bmatrix}$$ However I do not understand what can I do after elementary matrix because in books are only some steps and I do not know where the solution comes from.
linear-algebra
linear-algebra
asked Jan 11 at 1:14
MP3129MP3129
3049
asked Jan 11 at 1:14
MP3129MP3129
3049
edited Jan 11 at 1:30
MP3129
asked Jan 11 at 1:14
MP3129MP3129
3049
asked Jan 11 at 1:14
MP3129MP3129
3049
asked Jan 11 at 1:14
MP3129MP3129
3049
3049
$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33
add a comment |
$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33
$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33
add a comment |
1 Answer
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$begingroup$
You have the correct kernel.
You can should verify this by multiplying
$begin{bmatrix} 1&2&-1\1&1&2\2&3&1end{bmatrix}begin{bmatrix}-5\3\1end{bmatrix} = begin{bmatrix}0\0\0end{bmatrix}$
As for the image, since the degree of the Kernel is 1.
Choose 2 independent columns, any two will do. They will span the image.
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
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add a comment |
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1 Answer
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$begingroup$
You have the correct kernel.
You can should verify this by multiplying
$begin{bmatrix} 1&2&-1\1&1&2\2&3&1end{bmatrix}begin{bmatrix}-5\3\1end{bmatrix} = begin{bmatrix}0\0\0end{bmatrix}$
As for the image, since the degree of the Kernel is 1.
Choose 2 independent columns, any two will do. They will span the image.
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
You have the correct kernel.
You can should verify this by multiplying
$begin{bmatrix} 1&2&-1\1&1&2\2&3&1end{bmatrix}begin{bmatrix}-5\3\1end{bmatrix} = begin{bmatrix}0\0\0end{bmatrix}$
As for the image, since the degree of the Kernel is 1.
Choose 2 independent columns, any two will do. They will span the image.
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
You have the correct kernel.
You can should verify this by multiplying
$begin{bmatrix} 1&2&-1\1&1&2\2&3&1end{bmatrix}begin{bmatrix}-5\3\1end{bmatrix} = begin{bmatrix}0\0\0end{bmatrix}$
As for the image, since the degree of the Kernel is 1.
Choose 2 independent columns, any two will do. They will span the image.
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
$endgroup$
You have the correct kernel.
You can should verify this by multiplying
$begin{bmatrix} 1&2&-1\1&1&2\2&3&1end{bmatrix}begin{bmatrix}-5\3\1end{bmatrix} = begin{bmatrix}0\0\0end{bmatrix}$
As for the image, since the degree of the Kernel is 1.
Choose 2 independent columns, any two will do. They will span the image.
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
answered Jan 11 at 1:28
Doug MDoug M
44.8k31854
44.8k31854
add a comment |
add a comment |
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$begingroup$
Just to clarify, by "basic" I think you mean "basis".
$endgroup$
– Dave
Jan 11 at 1:21
$begingroup$
"math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$
$endgroup$
– Mohamad Misto
Jan 11 at 1:24
$begingroup$
Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be.
$endgroup$
– amd
Jan 11 at 2:33