Mixing
Prove $W cap [(W+V) cap U + (U+V) cap W]=(U+V)cap W$
$begingroup$
$W,U,V$ are all vector spaces.
Show that
$W cap [(W+V) cap U + (U+V) cap W]=(U+V)cap W$
I tried to show it this way, hoping for some feedback:
- Showing $(U+V) cap W) subseteq W cap [(W+V) cap U + (U+V) cap W] $:
Let $v in (U+V) cap W)$. It immediately follows that $v in W$.
$v = 0 + v$ where $0in (W+V)cap U$, $vin (U+V)cap W$
Then by defintion, $vin (W+V) cap U + (U+V) cap W$
- Showing $ W cap [(W+V) cap U + (U+V) cap W] subseteq (U+V) cap W) $:
Let $v in W cap [(W+V) cap U + (U+V) cap W]$ . It immediately follows that $ v in W$ .
There are $uin U,w in W$ such that $v = u + w$, where $u in (W+V) cap U)$ and $w in (U+V) cap W$
Then $v = (0 + u) + w$ where $(0+u) in U+V$ and $win W$ therefore $v in (U+V) cap W)$
linear-algebra
asked Jan 13 at 21:55
BBLNBBLN
1194
$endgroup$
add a comment |
$begingroup$
$W,U,V$ are all vector spaces.
Show that
$W cap [(W+V) cap U + (U+V) cap W]=(U+V)cap W$
I tried to show it this way, hoping for some feedback:
- Showing $(U+V) cap W) subseteq W cap [(W+V) cap U + (U+V) cap W] $:
Let $v in (U+V) cap W)$. It immediately follows that $v in W$.
$v = 0 + v$ where $0in (W+V)cap U$, $vin (U+V)cap W$
Then by defintion, $vin (W+V) cap U + (U+V) cap W$
- Showing $ W cap [(W+V) cap U + (U+V) cap W] subseteq (U+V) cap W) $:
Let $v in W cap [(W+V) cap U + (U+V) cap W]$ . It immediately follows that $ v in W$ .
There are $uin U,w in W$ such that $v = u + w$, where $u in (W+V) cap U)$ and $w in (U+V) cap W$
Then $v = (0 + u) + w$ where $(0+u) in U+V$ and $win W$ therefore $v in (U+V) cap W)$
linear-algebra
asked Jan 13 at 21:55
BBLNBBLN
1194
$endgroup$
1
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04
add a comment |
$begingroup$
$W,U,V$ are all vector spaces.
Show that
$W cap [(W+V) cap U + (U+V) cap W]=(U+V)cap W$
I tried to show it this way, hoping for some feedback:
- Showing $(U+V) cap W) subseteq W cap [(W+V) cap U + (U+V) cap W] $:
Let $v in (U+V) cap W)$. It immediately follows that $v in W$.
$v = 0 + v$ where $0in (W+V)cap U$, $vin (U+V)cap W$
Then by defintion, $vin (W+V) cap U + (U+V) cap W$
- Showing $ W cap [(W+V) cap U + (U+V) cap W] subseteq (U+V) cap W) $:
Let $v in W cap [(W+V) cap U + (U+V) cap W]$ . It immediately follows that $ v in W$ .
There are $uin U,w in W$ such that $v = u + w$, where $u in (W+V) cap U)$ and $w in (U+V) cap W$
Then $v = (0 + u) + w$ where $(0+u) in U+V$ and $win W$ therefore $v in (U+V) cap W)$
linear-algebra
asked Jan 13 at 21:55
BBLNBBLN
1194
$endgroup$
$W,U,V$ are all vector spaces.
Show that
$W cap [(W+V) cap U + (U+V) cap W]=(U+V)cap W$
I tried to show it this way, hoping for some feedback:
- Showing $(U+V) cap W) subseteq W cap [(W+V) cap U + (U+V) cap W] $:
Let $v in (U+V) cap W)$. It immediately follows that $v in W$.
$v = 0 + v$ where $0in (W+V)cap U$, $vin (U+V)cap W$
Then by defintion, $vin (W+V) cap U + (U+V) cap W$
- Showing $ W cap [(W+V) cap U + (U+V) cap W] subseteq (U+V) cap W) $:
Let $v in W cap [(W+V) cap U + (U+V) cap W]$ . It immediately follows that $ v in W$ .
There are $uin U,w in W$ such that $v = u + w$, where $u in (W+V) cap U)$ and $w in (U+V) cap W$
Then $v = (0 + u) + w$ where $(0+u) in U+V$ and $win W$ therefore $v in (U+V) cap W)$
linear-algebra
linear-algebra
asked Jan 13 at 21:55
BBLNBBLN
1194
asked Jan 13 at 21:55
BBLNBBLN
1194
asked Jan 13 at 21:55
BBLNBBLN
1194
asked Jan 13 at 21:55
BBLNBBLN
1194
asked Jan 13 at 21:55
BBLNBBLN
1194
1194
1
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04
add a comment |
1
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04
1
1
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04
add a comment |
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1
$begingroup$
Part 1 looks good to me. In part 2 you show that $vin W$ so it remains to show that $vin U+V$. I would suggest writing $v=u+w$ as you do, then you know that $uin U$ and $win U+V$, so $u+win U+V$.
$endgroup$
– Dave
Jan 13 at 22:04