Let $X$ be a infinite dimesional normed space, $M$, $N$ be subspaces with $ M subseteq N$. Show $dim X/N leq dim X/M$ The Next CEO of Stack OverflowFrom $dim Aleq dim B$, can we conclude that $Asubseteq B $?Given two subspaces $U,W$ of vector space $V$, how to show that $dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$Is the intersection of finite codimensional subspaces of finite codimension in an infinite dim vector space?Vector space of dim n and its subspacesInfinite dimensional normed space$U,W$ are subspaces. show $dim(U+W) = 1+dim(U cap W)$, then $U+W,Ucap W=U,W$Subspaces $X$ and $Y$ of a Hilbert Space with $dim X<infty$ and $dim X<dim Y$.Uniqueness of annihilator subspace in infinite dimensional normed spaceNormed space in finite dimensional subspacesIf $X$ is infinite dim'l and subspaces $N subseteq M$ satisfy $dim(X/M)=dim(X/N) lt infty$, then $N=M$ holds.
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Let $X$ be a infinite dimesional normed space, $M$, $N$ be subspaces with $ M subseteq N$. Show $dim X/N leq dim X/M$
The Next CEO of Stack OverflowFrom $dim Aleq dim B$, can we conclude that $Asubseteq B $?Given two subspaces $U,W$ of vector space $V$, how to show that $dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$Is the intersection of finite codimensional subspaces of finite codimension in an infinite dim vector space?Vector space of dim n and its subspacesInfinite dimensional normed space$U,W$ are subspaces. show $dim(U+W) = 1+dim(U cap W)$, then $U+W,Ucap W=U,W$Subspaces $X$ and $Y$ of a Hilbert Space with $dim X<infty$ and $dim X<dim Y$.Uniqueness of annihilator subspace in infinite dimensional normed spaceNormed space in finite dimensional subspacesIf $X$ is infinite dim'l and subspaces $N subseteq M$ satisfy $dim(X/M)=dim(X/N) lt infty$, then $N=M$ holds.
$begingroup$
Let $X$ be a infinite dimesional normed space. Let $M$, $N$ be subspaces such that $ M subseteq N$. How can we show $$dim X/N leq dim X/M$$?
I think this is intuitively clear and in the case $X$ is a finite dim'l, we can actually prove by noting $dim M leq dim N$.
Any help is appreciated.
linear-algebra functional-analysis
asked Mar 28 at 3:39
izimathizimath
421210
$endgroup$
add a comment |
$begingroup$
Let $X$ be a infinite dimesional normed space. Let $M$, $N$ be subspaces such that $ M subseteq N$. How can we show $$dim X/N leq dim X/M$$?
I think this is intuitively clear and in the case $X$ is a finite dim'l, we can actually prove by noting $dim M leq dim N$.
Any help is appreciated.
linear-algebra functional-analysis
asked Mar 28 at 3:39
izimathizimath
421210
$endgroup$
add a comment |
$begingroup$
Let $X$ be a infinite dimesional normed space. Let $M$, $N$ be subspaces such that $ M subseteq N$. How can we show $$dim X/N leq dim X/M$$?
I think this is intuitively clear and in the case $X$ is a finite dim'l, we can actually prove by noting $dim M leq dim N$.
Any help is appreciated.
linear-algebra functional-analysis
asked Mar 28 at 3:39
izimathizimath
421210
$endgroup$
Let $X$ be a infinite dimesional normed space. Let $M$, $N$ be subspaces such that $ M subseteq N$. How can we show $$dim X/N leq dim X/M$$?
I think this is intuitively clear and in the case $X$ is a finite dim'l, we can actually prove by noting $dim M leq dim N$.
Any help is appreciated.
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Mar 28 at 3:39
izimathizimath
421210
asked Mar 28 at 3:39
izimathizimath
421210
asked Mar 28 at 3:39
izimathizimath
421210
asked Mar 28 at 3:39
izimathizimath
421210
asked Mar 28 at 3:39
izimathizimath
421210
421210
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Consider the map $Phi : X/Mto X/N$, defined by $Phi(x+M) := x+N$. This definition makes sense since if $x+M = y+M$, then $x-yin Msubset N$ and thus $x+N = y+N$. The map $Phi$ is obviously linear and surjective. Hence, if $X/M$ is finite-dimensional, then so is $X/N$ and, in this case, $dim(X/M) = dimoperatornameranPhi + dimkerPhigedimoperatornameranPhi = dim(X/N)$.
Since you asked for cardinalities: The space $N/M$ is a subspace of $X/M$, so we can consider the space $tfracX/MN/M$ and the map $psi : tfracX/MN/Mto X/N$, defined by $psi((x+M) + N/M) := x+N$. This map is easily seen to be bijective. So, if $x_i + M: iin I$ is a basis for $X/M$, then $(x_i+M) + N/M : iin I$ spans $tfracX/MN/M$. Therefore, there exists $Jsubset I$ such that $(x_i+M) + N/M : iin J$ is a basis for $tfracX/MN/M$. Applying $psi$ we find that $x_i+N : iin J$ is a basis for $X/N$. Hence, $dim X/N = |J|le |I| = dim X/M$.
answered Mar 28 at 4:53
amsmathamsmath
3,288420
$endgroup$
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the map $Phi : X/Mto X/N$, defined by $Phi(x+M) := x+N$. This definition makes sense since if $x+M = y+M$, then $x-yin Msubset N$ and thus $x+N = y+N$. The map $Phi$ is obviously linear and surjective. Hence, if $X/M$ is finite-dimensional, then so is $X/N$ and, in this case, $dim(X/M) = dimoperatornameranPhi + dimkerPhigedimoperatornameranPhi = dim(X/N)$.
Since you asked for cardinalities: The space $N/M$ is a subspace of $X/M$, so we can consider the space $tfracX/MN/M$ and the map $psi : tfracX/MN/Mto X/N$, defined by $psi((x+M) + N/M) := x+N$. This map is easily seen to be bijective. So, if $x_i + M: iin I$ is a basis for $X/M$, then $(x_i+M) + N/M : iin I$ spans $tfracX/MN/M$. Therefore, there exists $Jsubset I$ such that $(x_i+M) + N/M : iin J$ is a basis for $tfracX/MN/M$. Applying $psi$ we find that $x_i+N : iin J$ is a basis for $X/N$. Hence, $dim X/N = |J|le |I| = dim X/M$.
answered Mar 28 at 4:53
amsmathamsmath
3,288420
$endgroup$
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
add a comment |
$begingroup$
Consider the map $Phi : X/Mto X/N$, defined by $Phi(x+M) := x+N$. This definition makes sense since if $x+M = y+M$, then $x-yin Msubset N$ and thus $x+N = y+N$. The map $Phi$ is obviously linear and surjective. Hence, if $X/M$ is finite-dimensional, then so is $X/N$ and, in this case, $dim(X/M) = dimoperatornameranPhi + dimkerPhigedimoperatornameranPhi = dim(X/N)$.
Since you asked for cardinalities: The space $N/M$ is a subspace of $X/M$, so we can consider the space $tfracX/MN/M$ and the map $psi : tfracX/MN/Mto X/N$, defined by $psi((x+M) + N/M) := x+N$. This map is easily seen to be bijective. So, if $x_i + M: iin I$ is a basis for $X/M$, then $(x_i+M) + N/M : iin I$ spans $tfracX/MN/M$. Therefore, there exists $Jsubset I$ such that $(x_i+M) + N/M : iin J$ is a basis for $tfracX/MN/M$. Applying $psi$ we find that $x_i+N : iin J$ is a basis for $X/N$. Hence, $dim X/N = |J|le |I| = dim X/M$.
answered Mar 28 at 4:53
amsmathamsmath
3,288420
$endgroup$
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
add a comment |
$begingroup$
Consider the map $Phi : X/Mto X/N$, defined by $Phi(x+M) := x+N$. This definition makes sense since if $x+M = y+M$, then $x-yin Msubset N$ and thus $x+N = y+N$. The map $Phi$ is obviously linear and surjective. Hence, if $X/M$ is finite-dimensional, then so is $X/N$ and, in this case, $dim(X/M) = dimoperatornameranPhi + dimkerPhigedimoperatornameranPhi = dim(X/N)$.
Since you asked for cardinalities: The space $N/M$ is a subspace of $X/M$, so we can consider the space $tfracX/MN/M$ and the map $psi : tfracX/MN/Mto X/N$, defined by $psi((x+M) + N/M) := x+N$. This map is easily seen to be bijective. So, if $x_i + M: iin I$ is a basis for $X/M$, then $(x_i+M) + N/M : iin I$ spans $tfracX/MN/M$. Therefore, there exists $Jsubset I$ such that $(x_i+M) + N/M : iin J$ is a basis for $tfracX/MN/M$. Applying $psi$ we find that $x_i+N : iin J$ is a basis for $X/N$. Hence, $dim X/N = |J|le |I| = dim X/M$.
answered Mar 28 at 4:53
amsmathamsmath
3,288420
$endgroup$
Consider the map $Phi : X/Mto X/N$, defined by $Phi(x+M) := x+N$. This definition makes sense since if $x+M = y+M$, then $x-yin Msubset N$ and thus $x+N = y+N$. The map $Phi$ is obviously linear and surjective. Hence, if $X/M$ is finite-dimensional, then so is $X/N$ and, in this case, $dim(X/M) = dimoperatornameranPhi + dimkerPhigedimoperatornameranPhi = dim(X/N)$.
Since you asked for cardinalities: The space $N/M$ is a subspace of $X/M$, so we can consider the space $tfracX/MN/M$ and the map $psi : tfracX/MN/Mto X/N$, defined by $psi((x+M) + N/M) := x+N$. This map is easily seen to be bijective. So, if $x_i + M: iin I$ is a basis for $X/M$, then $(x_i+M) + N/M : iin I$ spans $tfracX/MN/M$. Therefore, there exists $Jsubset I$ such that $(x_i+M) + N/M : iin J$ is a basis for $tfracX/MN/M$. Applying $psi$ we find that $x_i+N : iin J$ is a basis for $X/N$. Hence, $dim X/N = |J|le |I| = dim X/M$.
answered Mar 28 at 4:53
amsmathamsmath
3,288420
edited Mar 28 at 6:16
answered Mar 28 at 4:53
amsmathamsmath
3,288420
answered Mar 28 at 4:53
amsmathamsmath
3,288420
answered Mar 28 at 4:53
amsmathamsmath
3,288420
3,288420
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
add a comment |
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
Is there any result when $X/M$ is not finite dimensional? I guess there should be some cardinality involved..
$endgroup$
– izimath
Mar 28 at 4:58
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
I added a paragraph concerning this question.
$endgroup$
– amsmath
Mar 28 at 6:16
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
$begingroup$
So Third isomorphism theorem gives the answer. Thank you.
$endgroup$
– izimath
Mar 28 at 11:47
add a comment |
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