Calculation of Cokernel? Is my solution correct?
$begingroup$
I was trying to find the following:
$$ M= begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}\
coker(M) = (x,y,z)*
begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}= 0\
2x+3y+z = 0\
4x+2y+z = 0\
z = -x+2y\
2x+3y+(-4x-2y)=-2x+y Rightarrow y = 2x Rightarrow z = -8x\
Rightarrow coker(M) = begin{bmatrix}x\2x\-8x end{bmatrix}$$
Is that correct?
linear-algebra
edited Jan 5 at 12:02
Shubham Johri
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
$endgroup$
add a comment |
$begingroup$
I was trying to find the following:
$$ M= begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}\
coker(M) = (x,y,z)*
begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}= 0\
2x+3y+z = 0\
4x+2y+z = 0\
z = -x+2y\
2x+3y+(-4x-2y)=-2x+y Rightarrow y = 2x Rightarrow z = -8x\
Rightarrow coker(M) = begin{bmatrix}x\2x\-8x end{bmatrix}$$
Is that correct?
linear-algebra
edited Jan 5 at 12:02
Shubham Johri
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
$endgroup$
$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01
add a comment |
$begingroup$
I was trying to find the following:
$$ M= begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}\
coker(M) = (x,y,z)*
begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}= 0\
2x+3y+z = 0\
4x+2y+z = 0\
z = -x+2y\
2x+3y+(-4x-2y)=-2x+y Rightarrow y = 2x Rightarrow z = -8x\
Rightarrow coker(M) = begin{bmatrix}x\2x\-8x end{bmatrix}$$
Is that correct?
linear-algebra
edited Jan 5 at 12:02
Shubham Johri
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
$endgroup$
I was trying to find the following:
$$ M= begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}\
coker(M) = (x,y,z)*
begin{bmatrix}
2 & 4\
3 & 2\
1 & 1
end{bmatrix}= 0\
2x+3y+z = 0\
4x+2y+z = 0\
z = -x+2y\
2x+3y+(-4x-2y)=-2x+y Rightarrow y = 2x Rightarrow z = -8x\
Rightarrow coker(M) = begin{bmatrix}x\2x\-8x end{bmatrix}$$
Is that correct?
linear-algebra
linear-algebra
edited Jan 5 at 12:02
Shubham Johri
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
edited Jan 5 at 12:02
Shubham Johri
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
edited Jan 5 at 12:02
Shubham Johri
5,550818
edited Jan 5 at 12:02
Shubham Johri
5,550818
edited Jan 5 at 12:02
Shubham Johri
5,550818
5,550818
asked Jan 5 at 11:41
thebillythebilly
566
asked Jan 5 at 11:41
thebillythebilly
566
asked Jan 5 at 11:41
thebillythebilly
566
566
$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01
add a comment |
$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01
$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01
$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = mathbb{R}$. Then I would conclude that $V = mathbb{R}^3, W = mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : mathbb{R}^3 to mathbb{R}^2$ with respect to the standard bases of $mathbb{R}^3$ and $mathbb{R}^2$. Writing the elements of $mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / text{im}(f)$. It seems you consider the kernel of $f$ which defined as $text{ker}(f) = { x in V mid f(x) = 0 }$. In our case $text{ker}(f) = { x in V mid x * M = 0 }$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
$endgroup$
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = mathbb{R}$. Then I would conclude that $V = mathbb{R}^3, W = mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : mathbb{R}^3 to mathbb{R}^2$ with respect to the standard bases of $mathbb{R}^3$ and $mathbb{R}^2$. Writing the elements of $mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / text{im}(f)$. It seems you consider the kernel of $f$ which defined as $text{ker}(f) = { x in V mid f(x) = 0 }$. In our case $text{ker}(f) = { x in V mid x * M = 0 }$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
$endgroup$
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
add a comment |
$begingroup$
Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = mathbb{R}$. Then I would conclude that $V = mathbb{R}^3, W = mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : mathbb{R}^3 to mathbb{R}^2$ with respect to the standard bases of $mathbb{R}^3$ and $mathbb{R}^2$. Writing the elements of $mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / text{im}(f)$. It seems you consider the kernel of $f$ which defined as $text{ker}(f) = { x in V mid f(x) = 0 }$. In our case $text{ker}(f) = { x in V mid x * M = 0 }$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
$endgroup$
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
add a comment |
$begingroup$
Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = mathbb{R}$. Then I would conclude that $V = mathbb{R}^3, W = mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : mathbb{R}^3 to mathbb{R}^2$ with respect to the standard bases of $mathbb{R}^3$ and $mathbb{R}^2$. Writing the elements of $mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / text{im}(f)$. It seems you consider the kernel of $f$ which defined as $text{ker}(f) = { x in V mid f(x) = 0 }$. In our case $text{ker}(f) = { x in V mid x * M = 0 }$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
$endgroup$
Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = mathbb{R}$. Then I would conclude that $V = mathbb{R}^3, W = mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : mathbb{R}^3 to mathbb{R}^2$ with respect to the standard bases of $mathbb{R}^3$ and $mathbb{R}^2$. Writing the elements of $mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / text{im}(f)$. It seems you consider the kernel of $f$ which defined as $text{ker}(f) = { x in V mid f(x) = 0 }$. In our case $text{ker}(f) = { x in V mid x * M = 0 }$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
answered Jan 5 at 12:33
Paul FrostPaul Frost
12.4k31035
12.4k31035
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
add a comment |
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
$ker(f)={xin V|Mx=0}$
$endgroup$
– Shubham Johri
Jan 5 at 13:05
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
I see. Thank you. @PaulFrost
$endgroup$
– thebilly
Jan 5 at 13:21
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
Paul, I understood the question in the following sense: $Vmapsto V^*$ is exact contravariant autoequivalence of category of fd $K$-vector spaces and thus $operatorname{coker} Mcong ker M^*$. Thus, it makes sense to define cokernel as subspace. It shouldn't work for infinite dimensional vector spaces, yet alone modules in general.
$endgroup$
– Ennar
Jan 5 at 13:35
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
$begingroup$
@Ennar So we see that the question needs clarification ...
$endgroup$
– Paul Frost
Jan 5 at 13:38
add a comment |
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$begingroup$
Yes, that is correct. You can complete the answer by stating $xinBbb R$ or $text{coker}(M)=text{span}{(1,2,-8)^T}$
$endgroup$
– Shubham Johri
Jan 5 at 12:01