If $U$ is a subspace of $Bbb R^n$ and let $x,yin Bbb R^n$ with $(x+y)in U$ then is it true that $xin U$ and...
$begingroup$
If $U$ is a subspace of $Bbb R^n$ and let $x,yin Bbb R^n$ with $(x+y)in U$ then is it true that $xin U$ and $yin U$.
It looks like it is true since x+y follows under addition, under multiplication and the trivial solution zero vector but I would like a formal interpretation, thank you for your answers
linear-algebra
edited Dec 6 '18 at 19:42
UserS
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
$endgroup$
add a comment |
$begingroup$
If $U$ is a subspace of $Bbb R^n$ and let $x,yin Bbb R^n$ with $(x+y)in U$ then is it true that $xin U$ and $yin U$.
It looks like it is true since x+y follows under addition, under multiplication and the trivial solution zero vector but I would like a formal interpretation, thank you for your answers
linear-algebra
edited Dec 6 '18 at 19:42
UserS
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
$endgroup$
1
$begingroup$
Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50
add a comment |
$begingroup$
If $U$ is a subspace of $Bbb R^n$ and let $x,yin Bbb R^n$ with $(x+y)in U$ then is it true that $xin U$ and $yin U$.
It looks like it is true since x+y follows under addition, under multiplication and the trivial solution zero vector but I would like a formal interpretation, thank you for your answers
linear-algebra
edited Dec 6 '18 at 19:42
UserS
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
$endgroup$
If $U$ is a subspace of $Bbb R^n$ and let $x,yin Bbb R^n$ with $(x+y)in U$ then is it true that $xin U$ and $yin U$.
It looks like it is true since x+y follows under addition, under multiplication and the trivial solution zero vector but I would like a formal interpretation, thank you for your answers
linear-algebra
linear-algebra
edited Dec 6 '18 at 19:42
UserS
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
edited Dec 6 '18 at 19:42
UserS
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
edited Dec 6 '18 at 19:42
UserS
1,5391112
edited Dec 6 '18 at 19:42
UserS
1,5391112
edited Dec 6 '18 at 19:42
UserS
1,5391112
1,5391112
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
asked Dec 6 '18 at 19:23
Venik N.Venik N.
32
32
1
$begingroup$
Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50
add a comment |
1
$begingroup$
Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50
1
1
$begingroup$
Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50
$begingroup$
Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50
add a comment |
3 Answers
3
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No. Its not true. Here is a counterexample. Suppose we are in $mathbb{R} ^2$. Take $x=(1, 0),y=(0,1)$ and let $U=span(1,1)$. Then $x+yin U$ but neither $x$ nor $y$ is.
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
$endgroup$
add a comment |
$begingroup$
Note that every vector subspace contains zero vector. Now let $$U={lambda (1,0,0,....,0)in Bbb R^n :lambdain Bbb R}.$$ Then $(0,-1,0,0,....,0),(0,1,0,0,....,0)$ are not in $U$ but there sum is $(0,0,0,....,0)=0(1,0,0,....,0)in U$.
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Let $U$={$(a,b)|a=b$}, $x=(1,0)$, $y=(0,1)$.
answered Dec 6 '18 at 19:50
NedNed
1,993910
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
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votes
$begingroup$
No. Its not true. Here is a counterexample. Suppose we are in $mathbb{R} ^2$. Take $x=(1, 0),y=(0,1)$ and let $U=span(1,1)$. Then $x+yin U$ but neither $x$ nor $y$ is.
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
$endgroup$
add a comment |
$begingroup$
No. Its not true. Here is a counterexample. Suppose we are in $mathbb{R} ^2$. Take $x=(1, 0),y=(0,1)$ and let $U=span(1,1)$. Then $x+yin U$ but neither $x$ nor $y$ is.
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
$endgroup$
add a comment |
$begingroup$
No. Its not true. Here is a counterexample. Suppose we are in $mathbb{R} ^2$. Take $x=(1, 0),y=(0,1)$ and let $U=span(1,1)$. Then $x+yin U$ but neither $x$ nor $y$ is.
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
$endgroup$
No. Its not true. Here is a counterexample. Suppose we are in $mathbb{R} ^2$. Take $x=(1, 0),y=(0,1)$ and let $U=span(1,1)$. Then $x+yin U$ but neither $x$ nor $y$ is.
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
answered Dec 6 '18 at 19:52
Arpan DasArpan Das
837
837
add a comment |
add a comment |
$begingroup$
Note that every vector subspace contains zero vector. Now let $$U={lambda (1,0,0,....,0)in Bbb R^n :lambdain Bbb R}.$$ Then $(0,-1,0,0,....,0),(0,1,0,0,....,0)$ are not in $U$ but there sum is $(0,0,0,....,0)=0(1,0,0,....,0)in U$.
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that every vector subspace contains zero vector. Now let $$U={lambda (1,0,0,....,0)in Bbb R^n :lambdain Bbb R}.$$ Then $(0,-1,0,0,....,0),(0,1,0,0,....,0)$ are not in $U$ but there sum is $(0,0,0,....,0)=0(1,0,0,....,0)in U$.
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that every vector subspace contains zero vector. Now let $$U={lambda (1,0,0,....,0)in Bbb R^n :lambdain Bbb R}.$$ Then $(0,-1,0,0,....,0),(0,1,0,0,....,0)$ are not in $U$ but there sum is $(0,0,0,....,0)=0(1,0,0,....,0)in U$.
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
$endgroup$
Note that every vector subspace contains zero vector. Now let $$U={lambda (1,0,0,....,0)in Bbb R^n :lambdain Bbb R}.$$ Then $(0,-1,0,0,....,0),(0,1,0,0,....,0)$ are not in $U$ but there sum is $(0,0,0,....,0)=0(1,0,0,....,0)in U$.
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
answered Dec 6 '18 at 19:33
UserSUserS
1,5391112
1,5391112
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add a comment |
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Let $U$={$(a,b)|a=b$}, $x=(1,0)$, $y=(0,1)$.
answered Dec 6 '18 at 19:50
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
Let $U$={$(a,b)|a=b$}, $x=(1,0)$, $y=(0,1)$.
answered Dec 6 '18 at 19:50
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
Let $U$={$(a,b)|a=b$}, $x=(1,0)$, $y=(0,1)$.
answered Dec 6 '18 at 19:50
NedNed
1,993910
$endgroup$
Let $U$={$(a,b)|a=b$}, $x=(1,0)$, $y=(0,1)$.
answered Dec 6 '18 at 19:50
NedNed
1,993910
answered Dec 6 '18 at 19:50
NedNed
1,993910
answered Dec 6 '18 at 19:50
NedNed
1,993910
answered Dec 6 '18 at 19:50
NedNed
1,993910
1,993910
add a comment |
add a comment |
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Take $U$ to be the zero subspace. Take $x$ to be a non zero vector. Take $y$ to be $-x$.
$endgroup$
– user614671
Dec 6 '18 at 19:50