Sum of Linear Subspaces
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Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that
a) $U+W = { u-w | u in U, w in W }$
b) $S cap (U + W) = (S cap U) + (S cap W)$?
My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.
linear-algebra vector-spaces
asked Jan 7 at 19:22
TimTim
246
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add a comment |
$begingroup$
Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that
a) $U+W = { u-w | u in U, w in W }$
b) $S cap (U + W) = (S cap U) + (S cap W)$?
My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.
linear-algebra vector-spaces
asked Jan 7 at 19:22
TimTim
246
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If $win W$ is $-win W$?
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– Doug M
Jan 7 at 19:27
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@DougM OP already mentioned that
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– Shubham Johri
Jan 7 at 19:28
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Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
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– bounceback
Jan 7 at 19:31
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@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
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– Tim
Jan 7 at 19:35
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No, your first equality is false - see Bernard's answer below for the counterexample
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– bounceback
Jan 7 at 19:38
add a comment |
$begingroup$
Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that
a) $U+W = { u-w | u in U, w in W }$
b) $S cap (U + W) = (S cap U) + (S cap W)$?
My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.
linear-algebra vector-spaces
asked Jan 7 at 19:22
TimTim
246
$endgroup$
Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that
a) $U+W = { u-w | u in U, w in W }$
b) $S cap (U + W) = (S cap U) + (S cap W)$?
My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 7 at 19:22
TimTim
246
asked Jan 7 at 19:22
TimTim
246
edited Jan 7 at 19:30
Tim
asked Jan 7 at 19:22
TimTim
246
asked Jan 7 at 19:22
TimTim
246
asked Jan 7 at 19:22
TimTim
246
246
$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27
$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28
$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
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– bounceback
Jan 7 at 19:31
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@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
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– Tim
Jan 7 at 19:35
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No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38
add a comment |
$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27
$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28
$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31
$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35
$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38
$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27
$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27
$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28
$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28
$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31
$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31
$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35
$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35
$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38
$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38
add a comment |
1 Answer
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For $a)$ it's fine, as $win Wiff -win W$.
For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.
Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.
answered Jan 7 at 19:34
BernardBernard
124k741117
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$begingroup$
For $a)$ it's fine, as $win Wiff -win W$.
For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.
Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.
answered Jan 7 at 19:34
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
For $a)$ it's fine, as $win Wiff -win W$.
For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.
Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.
answered Jan 7 at 19:34
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
For $a)$ it's fine, as $win Wiff -win W$.
For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.
Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.
answered Jan 7 at 19:34
BernardBernard
124k741117
$endgroup$
For $a)$ it's fine, as $win Wiff -win W$.
For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.
Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.
answered Jan 7 at 19:34
BernardBernard
124k741117
answered Jan 7 at 19:34
BernardBernard
124k741117
answered Jan 7 at 19:34
BernardBernard
124k741117
answered Jan 7 at 19:34
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
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$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27
$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28
$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31
$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35
$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38