Some counterexamplesVector subspace decomposition problem (Linear algebra)$Sp(A)bigoplus Sp(B) Leftrightarrow Acup B$ is linearly independentLinear Algebra- Independence [Probably a Stupid Question]Finding some isomorphismsProve that $V=U bigoplus W approx U times W$some questions about vector spaceCharacterizing direct sumsCould two complement spaces of two isomorphic subspace be non-isomorphic?Prove that exist some base of V such that does not have any vector from subspaceDefining isomorphism between dual spaces
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Some counterexamples
Vector subspace decomposition problem (Linear algebra)$Sp(A)bigoplus Sp(B) Leftrightarrow Acup B$ is linearly independentLinear Algebra- Independence [Probably a Stupid Question]Finding some isomorphismsProve that $V=U bigoplus W approx U times W$some questions about vector spaceCharacterizing direct sumsCould two complement spaces of two isomorphic subspace be non-isomorphic?Prove that exist some base of V such that does not have any vector from subspaceDefining isomorphism between dual spaces
$begingroup$
I want to know examples of the following statements.
If $V=Sbigoplus T=S'bigoplus T'$, then $Sapprox S'$ does not imply $Tapprox T'$.
If $S$ is a subspace of both of the vector spaces $V$ and $W$, then $Vapprox W$ does not imply $frac VSapprox frac WS$.
linear-algebra
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
$endgroup$
add a comment |
$begingroup$
I want to know examples of the following statements.
If $V=Sbigoplus T=S'bigoplus T'$, then $Sapprox S'$ does not imply $Tapprox T'$.
If $S$ is a subspace of both of the vector spaces $V$ and $W$, then $Vapprox W$ does not imply $frac VSapprox frac WS$.
linear-algebra
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
$endgroup$
add a comment |
$begingroup$
I want to know examples of the following statements.
If $V=Sbigoplus T=S'bigoplus T'$, then $Sapprox S'$ does not imply $Tapprox T'$.
If $S$ is a subspace of both of the vector spaces $V$ and $W$, then $Vapprox W$ does not imply $frac VSapprox frac WS$.
linear-algebra
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
$endgroup$
I want to know examples of the following statements.
If $V=Sbigoplus T=S'bigoplus T'$, then $Sapprox S'$ does not imply $Tapprox T'$.
If $S$ is a subspace of both of the vector spaces $V$ and $W$, then $Vapprox W$ does not imply $frac VSapprox frac WS$.
linear-algebra
linear-algebra
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
asked Mar 17 at 23:38
Jiexiong687691Jiexiong687691
865
865
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take $S$ be infinite dimensional and $T,T'$ finite dimensional vector spaces of different dimension.
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
add a comment |
$begingroup$
Hint If $V$ is a finite-dimensional vector space over $Bbb K$, then $V cong Bbb K^n$ for some unique $n$, and if $S$ and $T$ are finite-dimensional vector spaces over $Bbb K$, then $dim (S oplus T) = dim S + dim T$. So, if $V$ is finite-dimensional, $V = S oplus T = S' oplus T'$ and $S cong S'$, we conclude that $T cong T'$. In particular, any counterexample must have $V$ infinite-dimensional and hence at least one of $S$ and $T$ infinite-dimensional.
A similar hint applies to the question about quotients.
answered Mar 17 at 23:51
TravisTravis
63.8k769151
$endgroup$
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $S$ be infinite dimensional and $T,T'$ finite dimensional vector spaces of different dimension.
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
add a comment |
$begingroup$
Take $S$ be infinite dimensional and $T,T'$ finite dimensional vector spaces of different dimension.
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
add a comment |
$begingroup$
Take $S$ be infinite dimensional and $T,T'$ finite dimensional vector spaces of different dimension.
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
Take $S$ be infinite dimensional and $T,T'$ finite dimensional vector spaces of different dimension.
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
answered Mar 17 at 23:41
Tsemo AristideTsemo Aristide
59.9k11446
59.9k11446
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
add a comment |
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
$begingroup$
For a direct sum to make sense, isn't the intersection of $S$ and $T$ should be $0$?
$endgroup$
– Jiexiong687691
Mar 18 at 2:12
add a comment |
$begingroup$
Hint If $V$ is a finite-dimensional vector space over $Bbb K$, then $V cong Bbb K^n$ for some unique $n$, and if $S$ and $T$ are finite-dimensional vector spaces over $Bbb K$, then $dim (S oplus T) = dim S + dim T$. So, if $V$ is finite-dimensional, $V = S oplus T = S' oplus T'$ and $S cong S'$, we conclude that $T cong T'$. In particular, any counterexample must have $V$ infinite-dimensional and hence at least one of $S$ and $T$ infinite-dimensional.
A similar hint applies to the question about quotients.
answered Mar 17 at 23:51
TravisTravis
63.8k769151
$endgroup$
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
add a comment |
$begingroup$
Hint If $V$ is a finite-dimensional vector space over $Bbb K$, then $V cong Bbb K^n$ for some unique $n$, and if $S$ and $T$ are finite-dimensional vector spaces over $Bbb K$, then $dim (S oplus T) = dim S + dim T$. So, if $V$ is finite-dimensional, $V = S oplus T = S' oplus T'$ and $S cong S'$, we conclude that $T cong T'$. In particular, any counterexample must have $V$ infinite-dimensional and hence at least one of $S$ and $T$ infinite-dimensional.
A similar hint applies to the question about quotients.
answered Mar 17 at 23:51
TravisTravis
63.8k769151
$endgroup$
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
add a comment |
$begingroup$
Hint If $V$ is a finite-dimensional vector space over $Bbb K$, then $V cong Bbb K^n$ for some unique $n$, and if $S$ and $T$ are finite-dimensional vector spaces over $Bbb K$, then $dim (S oplus T) = dim S + dim T$. So, if $V$ is finite-dimensional, $V = S oplus T = S' oplus T'$ and $S cong S'$, we conclude that $T cong T'$. In particular, any counterexample must have $V$ infinite-dimensional and hence at least one of $S$ and $T$ infinite-dimensional.
A similar hint applies to the question about quotients.
answered Mar 17 at 23:51
TravisTravis
63.8k769151
$endgroup$
Hint If $V$ is a finite-dimensional vector space over $Bbb K$, then $V cong Bbb K^n$ for some unique $n$, and if $S$ and $T$ are finite-dimensional vector spaces over $Bbb K$, then $dim (S oplus T) = dim S + dim T$. So, if $V$ is finite-dimensional, $V = S oplus T = S' oplus T'$ and $S cong S'$, we conclude that $T cong T'$. In particular, any counterexample must have $V$ infinite-dimensional and hence at least one of $S$ and $T$ infinite-dimensional.
A similar hint applies to the question about quotients.
answered Mar 17 at 23:51
TravisTravis
63.8k769151
answered Mar 17 at 23:51
TravisTravis
63.8k769151
answered Mar 17 at 23:51
TravisTravis
63.8k769151
answered Mar 17 at 23:51
TravisTravis
63.8k769151
63.8k769151
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
add a comment |
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
For the second one, we know that either both $V$ and $W$ are infinite dimensional or both are finite dimensional since $Vapprox W$. Then in either case, actually four cases since $S$ can be either finite dimensional or infinite dimensional, why $V/S notapprox W/S$?
$endgroup$
– Jiexiong687691
Mar 18 at 0:00
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
$begingroup$
In fact, if $V$ and $W$ are both finite-dimensional, then, $V / S$ and $W / S$ both have dimension $dim V - dim S$, so $dim V$ must be infinite.
$endgroup$
– Travis
Mar 18 at 1:40
add a comment |
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