Mixing
How to show that $dimker(AB) le dim ker A + dim ker B $?
$begingroup$
I want to show that
$$ dim ker(AB) le dim ker A + dim ker B. $$
My problem
I thought that I can do that in this way:
Let consider $x inker B$
$$Bx = 0$$
Let multiplicate this from left side by A and we get:
$$ABx = 0$$
so $$ker B subsetker AB $$
so $$dim ker(B) le dimker AB$$
We can do the same thing with $ker A$
let consider $ vec{y} in operatorname{im}(AB) $
so $$ y = (AB)x $$
what is equivalent to $$ vec{y} = A(Bvec{x}) = Avec{w} $$
So
$$ vec{y} in operatorname{im}(AB) rightarrow vec{y} in operatorname{im}(A)$$
so
$$ operatorname{rank} AB le operatorname{rank} A leftrightarrow dim ker A le dim ker AB $$
But I am not sure what I should do later...
edited
I have seen this post $A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ but I haven't got nothing like $operatorname{im}(A|_{operatorname{im}(B)})$ on my algebra lecture and I can't use that so I search for another proof (or similar without this trick)
linear-algebra vector-spaces linear-transformations function-and-relation-composition
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
$endgroup$
add a comment |
$begingroup$
I want to show that
$$ dim ker(AB) le dim ker A + dim ker B. $$
My problem
I thought that I can do that in this way:
Let consider $x inker B$
$$Bx = 0$$
Let multiplicate this from left side by A and we get:
$$ABx = 0$$
so $$ker B subsetker AB $$
so $$dim ker(B) le dimker AB$$
We can do the same thing with $ker A$
let consider $ vec{y} in operatorname{im}(AB) $
so $$ y = (AB)x $$
what is equivalent to $$ vec{y} = A(Bvec{x}) = Avec{w} $$
So
$$ vec{y} in operatorname{im}(AB) rightarrow vec{y} in operatorname{im}(A)$$
so
$$ operatorname{rank} AB le operatorname{rank} A leftrightarrow dim ker A le dim ker AB $$
But I am not sure what I should do later...
edited
I have seen this post $A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ but I haven't got nothing like $operatorname{im}(A|_{operatorname{im}(B)})$ on my algebra lecture and I can't use that so I search for another proof (or similar without this trick)
linear-algebra vector-spaces linear-transformations function-and-relation-composition
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
$endgroup$
$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20
add a comment |
$begingroup$
I want to show that
$$ dim ker(AB) le dim ker A + dim ker B. $$
My problem
I thought that I can do that in this way:
Let consider $x inker B$
$$Bx = 0$$
Let multiplicate this from left side by A and we get:
$$ABx = 0$$
so $$ker B subsetker AB $$
so $$dim ker(B) le dimker AB$$
We can do the same thing with $ker A$
let consider $ vec{y} in operatorname{im}(AB) $
so $$ y = (AB)x $$
what is equivalent to $$ vec{y} = A(Bvec{x}) = Avec{w} $$
So
$$ vec{y} in operatorname{im}(AB) rightarrow vec{y} in operatorname{im}(A)$$
so
$$ operatorname{rank} AB le operatorname{rank} A leftrightarrow dim ker A le dim ker AB $$
But I am not sure what I should do later...
edited
I have seen this post $A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ but I haven't got nothing like $operatorname{im}(A|_{operatorname{im}(B)})$ on my algebra lecture and I can't use that so I search for another proof (or similar without this trick)
linear-algebra vector-spaces linear-transformations function-and-relation-composition
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
$endgroup$
I want to show that
$$ dim ker(AB) le dim ker A + dim ker B. $$
My problem
I thought that I can do that in this way:
Let consider $x inker B$
$$Bx = 0$$
Let multiplicate this from left side by A and we get:
$$ABx = 0$$
so $$ker B subsetker AB $$
so $$dim ker(B) le dimker AB$$
We can do the same thing with $ker A$
let consider $ vec{y} in operatorname{im}(AB) $
so $$ y = (AB)x $$
what is equivalent to $$ vec{y} = A(Bvec{x}) = Avec{w} $$
So
$$ vec{y} in operatorname{im}(AB) rightarrow vec{y} in operatorname{im}(A)$$
so
$$ operatorname{rank} AB le operatorname{rank} A leftrightarrow dim ker A le dim ker AB $$
But I am not sure what I should do later...
edited
I have seen this post $A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ but I haven't got nothing like $operatorname{im}(A|_{operatorname{im}(B)})$ on my algebra lecture and I can't use that so I search for another proof (or similar without this trick)
linear-algebra vector-spaces linear-transformations function-and-relation-composition
linear-algebra vector-spaces linear-transformations function-and-relation-composition
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
edited Jan 8 at 20:23
VirtualUser
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
asked Jan 8 at 18:30
VirtualUserVirtualUser
69814
69814
$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20
add a comment |
$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20
$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20
add a comment |
1 Answer
1
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$begingroup$
This is a proof in general where $A:Vto W$ and $B:Uto V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is,
$$dim ker (AB) leq dim ker A+dimker B$$
is true whether or not the relevant dimensions are finite cardinals.
Note that $xin ker(AB)$ iff $Bxin ker A$, which is the same as saying $$xin B^{-1}(ker Acap operatorname{im}B).$$ Recall from the isomorphism theorems that $operatorname{im} Bcong U/ker B$ so there exists an isomorphism $$varphi: Uoverset{cong}{longrightarrow} ker Boplus operatorname{im}B.$$ In other words,
$$varphibig(B^{-1}(ker Acap operatorname{im}B)big)=ker Boplus (ker Acap operatorname{im}B).$$
Consequently,
begin{align}dimker(AB)&=dimbig(ker Boplus (ker Acap operatorname{im}B)big)\&=dimker B+dim(ker Acap operatorname{im}B).end{align}
Since $ker Acap operatorname{im}Bsubseteq ker A$, we obtain the desired inequality.
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
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$begingroup$
This is a proof in general where $A:Vto W$ and $B:Uto V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is,
$$dim ker (AB) leq dim ker A+dimker B$$
is true whether or not the relevant dimensions are finite cardinals.
Note that $xin ker(AB)$ iff $Bxin ker A$, which is the same as saying $$xin B^{-1}(ker Acap operatorname{im}B).$$ Recall from the isomorphism theorems that $operatorname{im} Bcong U/ker B$ so there exists an isomorphism $$varphi: Uoverset{cong}{longrightarrow} ker Boplus operatorname{im}B.$$ In other words,
$$varphibig(B^{-1}(ker Acap operatorname{im}B)big)=ker Boplus (ker Acap operatorname{im}B).$$
Consequently,
begin{align}dimker(AB)&=dimbig(ker Boplus (ker Acap operatorname{im}B)big)\&=dimker B+dim(ker Acap operatorname{im}B).end{align}
Since $ker Acap operatorname{im}Bsubseteq ker A$, we obtain the desired inequality.
$endgroup$
add a comment |
$begingroup$
This is a proof in general where $A:Vto W$ and $B:Uto V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is,
$$dim ker (AB) leq dim ker A+dimker B$$
is true whether or not the relevant dimensions are finite cardinals.
Note that $xin ker(AB)$ iff $Bxin ker A$, which is the same as saying $$xin B^{-1}(ker Acap operatorname{im}B).$$ Recall from the isomorphism theorems that $operatorname{im} Bcong U/ker B$ so there exists an isomorphism $$varphi: Uoverset{cong}{longrightarrow} ker Boplus operatorname{im}B.$$ In other words,
$$varphibig(B^{-1}(ker Acap operatorname{im}B)big)=ker Boplus (ker Acap operatorname{im}B).$$
Consequently,
begin{align}dimker(AB)&=dimbig(ker Boplus (ker Acap operatorname{im}B)big)\&=dimker B+dim(ker Acap operatorname{im}B).end{align}
Since $ker Acap operatorname{im}Bsubseteq ker A$, we obtain the desired inequality.
$endgroup$
add a comment |
$begingroup$
This is a proof in general where $A:Vto W$ and $B:Uto V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is,
$$dim ker (AB) leq dim ker A+dimker B$$
is true whether or not the relevant dimensions are finite cardinals.
Note that $xin ker(AB)$ iff $Bxin ker A$, which is the same as saying $$xin B^{-1}(ker Acap operatorname{im}B).$$ Recall from the isomorphism theorems that $operatorname{im} Bcong U/ker B$ so there exists an isomorphism $$varphi: Uoverset{cong}{longrightarrow} ker Boplus operatorname{im}B.$$ In other words,
$$varphibig(B^{-1}(ker Acap operatorname{im}B)big)=ker Boplus (ker Acap operatorname{im}B).$$
Consequently,
begin{align}dimker(AB)&=dimbig(ker Boplus (ker Acap operatorname{im}B)big)\&=dimker B+dim(ker Acap operatorname{im}B).end{align}
Since $ker Acap operatorname{im}Bsubseteq ker A$, we obtain the desired inequality.
$endgroup$
This is a proof in general where $A:Vto W$ and $B:Uto V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is,
$$dim ker (AB) leq dim ker A+dimker B$$
is true whether or not the relevant dimensions are finite cardinals.
Note that $xin ker(AB)$ iff $Bxin ker A$, which is the same as saying $$xin B^{-1}(ker Acap operatorname{im}B).$$ Recall from the isomorphism theorems that $operatorname{im} Bcong U/ker B$ so there exists an isomorphism $$varphi: Uoverset{cong}{longrightarrow} ker Boplus operatorname{im}B.$$ In other words,
$$varphibig(B^{-1}(ker Acap operatorname{im}B)big)=ker Boplus (ker Acap operatorname{im}B).$$
Consequently,
begin{align}dimker(AB)&=dimbig(ker Boplus (ker Acap operatorname{im}B)big)\&=dimker B+dim(ker Acap operatorname{im}B).end{align}
Since $ker Acap operatorname{im}Bsubseteq ker A$, we obtain the desired inequality.
edited Jan 8 at 19:54
answered Jan 8 at 19:48
user593746
add a comment |
add a comment |
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$begingroup$
Perhaps see here? math.stackexchange.com/questions/269474/…
$endgroup$
– T. Fo
Jan 8 at 18:59
$begingroup$
@T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post
$endgroup$
– VirtualUser
Jan 8 at 19:01
$begingroup$
You have this backwards. Since $ker Bsubset ker(AB)$, we have $dimker B le dimker(AB)$, not the other way around.
$endgroup$
– Ted Shifrin
Jan 8 at 20:20