Mixing
The relation between the intersection of two subspaces and the sum of two subspaces
$begingroup$
Prove or give a counter example: Let $W$ and $U$ are two sub spaces of $V$ and $x in V$. If $x notin W$ and $x notin U$, then $x notin W+U$.
So far I’m trying to disprove this problem by using the fact that the sum of two subsets is bigger than the intersection of U and W, but I’m not completely confident in this approach. Any advice?
linear-algebra
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
$endgroup$
add a comment |
$begingroup$
Prove or give a counter example: Let $W$ and $U$ are two sub spaces of $V$ and $x in V$. If $x notin W$ and $x notin U$, then $x notin W+U$.
So far I’m trying to disprove this problem by using the fact that the sum of two subsets is bigger than the intersection of U and W, but I’m not completely confident in this approach. Any advice?
linear-algebra
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
$endgroup$
1
$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58
add a comment |
$begingroup$
Prove or give a counter example: Let $W$ and $U$ are two sub spaces of $V$ and $x in V$. If $x notin W$ and $x notin U$, then $x notin W+U$.
So far I’m trying to disprove this problem by using the fact that the sum of two subsets is bigger than the intersection of U and W, but I’m not completely confident in this approach. Any advice?
linear-algebra
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
$endgroup$
Prove or give a counter example: Let $W$ and $U$ are two sub spaces of $V$ and $x in V$. If $x notin W$ and $x notin U$, then $x notin W+U$.
So far I’m trying to disprove this problem by using the fact that the sum of two subsets is bigger than the intersection of U and W, but I’m not completely confident in this approach. Any advice?
linear-algebra
linear-algebra
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
edited Jan 29 at 5:36
YuiTo Cheng
2,1862937
2,1862937
asked Jan 28 at 23:21
noobiskonoobisko
765
asked Jan 28 at 23:21
noobiskonoobisko
765
asked Jan 28 at 23:21
noobiskonoobisko
765
765
1
$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58
add a comment |
1
$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58
1
1
$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58
add a comment |
1 Answer
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$begingroup$
Hint: In $Bbb R^2$. How about $W$ is the $x$-axis, $U$ the $y$-axis? Consider $x=(1,1)$.
(Ahh, I see you came up with this example. You just hadn't specified $W$ and $U$.)
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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$begingroup$
Hint: In $Bbb R^2$. How about $W$ is the $x$-axis, $U$ the $y$-axis? Consider $x=(1,1)$.
(Ahh, I see you came up with this example. You just hadn't specified $W$ and $U$.)
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: In $Bbb R^2$. How about $W$ is the $x$-axis, $U$ the $y$-axis? Consider $x=(1,1)$.
(Ahh, I see you came up with this example. You just hadn't specified $W$ and $U$.)
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: In $Bbb R^2$. How about $W$ is the $x$-axis, $U$ the $y$-axis? Consider $x=(1,1)$.
(Ahh, I see you came up with this example. You just hadn't specified $W$ and $U$.)
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
$endgroup$
Hint: In $Bbb R^2$. How about $W$ is the $x$-axis, $U$ the $y$-axis? Consider $x=(1,1)$.
(Ahh, I see you came up with this example. You just hadn't specified $W$ and $U$.)
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
edited Jan 29 at 0:59
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 0:51
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
Think about the logic first. To disprove the statement you do not need to use any general facts ("the sum of two subsets is bigger than the intersection"), you just need to give one example where the result is not true. Hint: there are easy examples with $V=Bbb R^2$.
$endgroup$
– David
Jan 28 at 23:28
$begingroup$
I believe I made some progress, so when V is a vector in the field $R^2$. Then $(0,1) in W$ and $(1,0) in U$. So, $ (0,1) + (1,0) = (1,1) in W+U$,
$endgroup$
– noobisko
Jan 28 at 23:42
$begingroup$
Looks like a good set-up. Now you need to be clear about what is your vector $x$, what are your subspaces $U$ and $W$, and how this shows the statement is false. Keep on thinking about the logic: to show "if $A$ then $B$" is false, you need an example where $A$ is true and $B$ is false.
$endgroup$
– David
Jan 28 at 23:58