Mixing
Proof $dim(text{im}(f)capker(g)) = dim(text{im}(f))-dim(text{im}(gf))$
$begingroup$
In one of the exercises in my textbook we had to prove the following:
Consider $3$ $K$-vectorspaces $U$, $V$ and $W$ and $2$ maps:
- $fintext{Hom}_K(U,V)$
- $gintext{Hom}_K(V,W)$
then :
begin{equation}
dim(text{im}(f)capker(g)) = dim(text{im}(f))-dim(text{im}(gf))
end{equation}
This proof wasn't addressed during the lectures so I tried to understand it at home but I don't understand most of the steps. The proof is as follows:
We consider the map $g|_{f(U)}:f(U)rightarrow W$ which is the restriction of $g$ to the image $text{im}(f)leqslant V$.
begin{equation}
begin{split}
dim(text{im}(f)) & = dim(ker(g|_{f(U)})) + dim(text{im}(g|_{f(U)})) \
& = dim({vintext{im}(f);|;g(v) = 0}) + dim({g(v);|; vintext{im}(f)}) \
& = dim(text{im}(f)capker(g)) + dim({(gf)(w);|;win U}) \
& = dim(text{im}(f)capker(g)) + dim(text{im}(gf))
end{split}
end{equation}
And then the formula follows from this.
My guess is that in the firs step they use the following:
begin{equation}
dim(V) = dim(ker(f))+dim(text{im}(f))
end{equation}
But I don't quite get why they use $g|_{f(U)}$ as the function in this equation and where it comes from. I also need some help understanding the transition between step 1 and 2 and between 2 and 3. The last step I do understand. Would someone be able to explain this?
linear-algebra functions vector-spaces
asked Jan 12 at 15:11
ViktorViktor
1389
$endgroup$
add a comment |
$begingroup$
In one of the exercises in my textbook we had to prove the following:
Consider $3$ $K$-vectorspaces $U$, $V$ and $W$ and $2$ maps:
- $fintext{Hom}_K(U,V)$
- $gintext{Hom}_K(V,W)$
then :
begin{equation}
dim(text{im}(f)capker(g)) = dim(text{im}(f))-dim(text{im}(gf))
end{equation}
This proof wasn't addressed during the lectures so I tried to understand it at home but I don't understand most of the steps. The proof is as follows:
We consider the map $g|_{f(U)}:f(U)rightarrow W$ which is the restriction of $g$ to the image $text{im}(f)leqslant V$.
begin{equation}
begin{split}
dim(text{im}(f)) & = dim(ker(g|_{f(U)})) + dim(text{im}(g|_{f(U)})) \
& = dim({vintext{im}(f);|;g(v) = 0}) + dim({g(v);|; vintext{im}(f)}) \
& = dim(text{im}(f)capker(g)) + dim({(gf)(w);|;win U}) \
& = dim(text{im}(f)capker(g)) + dim(text{im}(gf))
end{split}
end{equation}
And then the formula follows from this.
My guess is that in the firs step they use the following:
begin{equation}
dim(V) = dim(ker(f))+dim(text{im}(f))
end{equation}
But I don't quite get why they use $g|_{f(U)}$ as the function in this equation and where it comes from. I also need some help understanding the transition between step 1 and 2 and between 2 and 3. The last step I do understand. Would someone be able to explain this?
linear-algebra functions vector-spaces
asked Jan 12 at 15:11
ViktorViktor
1389
$endgroup$
$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31
add a comment |
$begingroup$
In one of the exercises in my textbook we had to prove the following:
Consider $3$ $K$-vectorspaces $U$, $V$ and $W$ and $2$ maps:
- $fintext{Hom}_K(U,V)$
- $gintext{Hom}_K(V,W)$
then :
begin{equation}
dim(text{im}(f)capker(g)) = dim(text{im}(f))-dim(text{im}(gf))
end{equation}
This proof wasn't addressed during the lectures so I tried to understand it at home but I don't understand most of the steps. The proof is as follows:
We consider the map $g|_{f(U)}:f(U)rightarrow W$ which is the restriction of $g$ to the image $text{im}(f)leqslant V$.
begin{equation}
begin{split}
dim(text{im}(f)) & = dim(ker(g|_{f(U)})) + dim(text{im}(g|_{f(U)})) \
& = dim({vintext{im}(f);|;g(v) = 0}) + dim({g(v);|; vintext{im}(f)}) \
& = dim(text{im}(f)capker(g)) + dim({(gf)(w);|;win U}) \
& = dim(text{im}(f)capker(g)) + dim(text{im}(gf))
end{split}
end{equation}
And then the formula follows from this.
My guess is that in the firs step they use the following:
begin{equation}
dim(V) = dim(ker(f))+dim(text{im}(f))
end{equation}
But I don't quite get why they use $g|_{f(U)}$ as the function in this equation and where it comes from. I also need some help understanding the transition between step 1 and 2 and between 2 and 3. The last step I do understand. Would someone be able to explain this?
linear-algebra functions vector-spaces
asked Jan 12 at 15:11
ViktorViktor
1389
$endgroup$
In one of the exercises in my textbook we had to prove the following:
Consider $3$ $K$-vectorspaces $U$, $V$ and $W$ and $2$ maps:
- $fintext{Hom}_K(U,V)$
- $gintext{Hom}_K(V,W)$
then :
begin{equation}
dim(text{im}(f)capker(g)) = dim(text{im}(f))-dim(text{im}(gf))
end{equation}
This proof wasn't addressed during the lectures so I tried to understand it at home but I don't understand most of the steps. The proof is as follows:
We consider the map $g|_{f(U)}:f(U)rightarrow W$ which is the restriction of $g$ to the image $text{im}(f)leqslant V$.
begin{equation}
begin{split}
dim(text{im}(f)) & = dim(ker(g|_{f(U)})) + dim(text{im}(g|_{f(U)})) \
& = dim({vintext{im}(f);|;g(v) = 0}) + dim({g(v);|; vintext{im}(f)}) \
& = dim(text{im}(f)capker(g)) + dim({(gf)(w);|;win U}) \
& = dim(text{im}(f)capker(g)) + dim(text{im}(gf))
end{split}
end{equation}
And then the formula follows from this.
My guess is that in the firs step they use the following:
begin{equation}
dim(V) = dim(ker(f))+dim(text{im}(f))
end{equation}
But I don't quite get why they use $g|_{f(U)}$ as the function in this equation and where it comes from. I also need some help understanding the transition between step 1 and 2 and between 2 and 3. The last step I do understand. Would someone be able to explain this?
linear-algebra functions vector-spaces
linear-algebra functions vector-spaces
asked Jan 12 at 15:11
ViktorViktor
1389
asked Jan 12 at 15:11
ViktorViktor
1389
asked Jan 12 at 15:11
ViktorViktor
1389
asked Jan 12 at 15:11
ViktorViktor
1389
asked Jan 12 at 15:11
ViktorViktor
1389
1389
$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31
add a comment |
$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31
$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31
$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31
add a comment |
1 Answer
1
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votes
$begingroup$
Before the first step in your proof, the well known consequence is used:
$$ (*) space dim(V) = dim(Ker(g)) + dim(Im(g)) $$
Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) subseteq Dom(g) = V$ therefore, if we study only in the restriction $gmid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
$endgroup$
add a comment |
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$begingroup$
Before the first step in your proof, the well known consequence is used:
$$ (*) space dim(V) = dim(Ker(g)) + dim(Im(g)) $$
Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) subseteq Dom(g) = V$ therefore, if we study only in the restriction $gmid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
Before the first step in your proof, the well known consequence is used:
$$ (*) space dim(V) = dim(Ker(g)) + dim(Im(g)) $$
Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) subseteq Dom(g) = V$ therefore, if we study only in the restriction $gmid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
Before the first step in your proof, the well known consequence is used:
$$ (*) space dim(V) = dim(Ker(g)) + dim(Im(g)) $$
Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) subseteq Dom(g) = V$ therefore, if we study only in the restriction $gmid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
$endgroup$
Before the first step in your proof, the well known consequence is used:
$$ (*) space dim(V) = dim(Ker(g)) + dim(Im(g)) $$
Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) subseteq Dom(g) = V$ therefore, if we study only in the restriction $gmid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
answered Jan 29 at 19:05
freehumoristfreehumorist
351214
351214
add a comment |
add a comment |
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$begingroup$
yes, the definition of kernel is: $ker(f) := {vin V;|;f(v) = 0}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $text{im}(f) := {f(v);|; vin V}$ if $f:Vrightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa.
$endgroup$
– Viktor
Jan 12 at 15:31