Prove or disprove If $u_1, u_2, u_3,u_4,u_5 $ form basis in $mathbbR^5 $ and $V$ is 2-d subspace then show that V has basis from $mathbbR^5$
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I have a doubt. If $u_1, u_2, u_3,u_4,u_5 $ form basis in $mathbbR^5 $ and $V$ is 2-dimensional subspace of $mathbbR^5$ then show that V has basis of 2 elements from given basis of $mathbbR^5$
Is this correct?
linear-algebra vector-spaces
asked Sep 7 at 3:22
Magneto
874213
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up vote
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I have a doubt. If $u_1, u_2, u_3,u_4,u_5 $ form basis in $mathbbR^5 $ and $V$ is 2-dimensional subspace of $mathbbR^5$ then show that V has basis of 2 elements from given basis of $mathbbR^5$
Is this correct?
linear-algebra vector-spaces
asked Sep 7 at 3:22
Magneto
874213
1
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
2
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
2
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
thank you sir, got it..........
– Magneto
Sep 7 at 3:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a doubt. If $u_1, u_2, u_3,u_4,u_5 $ form basis in $mathbbR^5 $ and $V$ is 2-dimensional subspace of $mathbbR^5$ then show that V has basis of 2 elements from given basis of $mathbbR^5$
Is this correct?
linear-algebra vector-spaces
asked Sep 7 at 3:22
Magneto
874213
I have a doubt. If $u_1, u_2, u_3,u_4,u_5 $ form basis in $mathbbR^5 $ and $V$ is 2-dimensional subspace of $mathbbR^5$ then show that V has basis of 2 elements from given basis of $mathbbR^5$
Is this correct?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Sep 7 at 3:22
Magneto
874213
asked Sep 7 at 3:22
Magneto
874213
edited Sep 7 at 3:47
asked Sep 7 at 3:22
Magneto
874213
asked Sep 7 at 3:22
Magneto
874213
asked Sep 7 at 3:22
Magneto
874213
874213
1
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
2
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
2
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
thank you sir, got it..........
– Magneto
Sep 7 at 3:47
add a comment |Â
1
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
2
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
2
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
thank you sir, got it..........
– Magneto
Sep 7 at 3:47
1
1
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
2
2
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
2
2
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
thank you sir, got it..........
– Magneto
Sep 7 at 3:47
thank you sir, got it..........
– Magneto
Sep 7 at 3:47
add a comment |Â
1 Answer
1
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up vote
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As others have said, this statement is false in general. I will give you this counter-example:
If $u_1=(1,0,0,0,0),; u_2=(0,1,0,0,0),; u_3=(0,0,1,0,0),;ldots$ and $V=(a,a,b,b,b):a,binmathbbR $. It is clear that for every member of the basis: $u_i notin V$. Therefore, it is impossible for any pair of the original basis vectors to span $V$.
answered Sep 7 at 3:53
olaphus
563
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As others have said, this statement is false in general. I will give you this counter-example:
If $u_1=(1,0,0,0,0),; u_2=(0,1,0,0,0),; u_3=(0,0,1,0,0),;ldots$ and $V=(a,a,b,b,b):a,binmathbbR $. It is clear that for every member of the basis: $u_i notin V$. Therefore, it is impossible for any pair of the original basis vectors to span $V$.
answered Sep 7 at 3:53
olaphus
563
add a comment |Â
up vote
2
down vote
accepted
As others have said, this statement is false in general. I will give you this counter-example:
If $u_1=(1,0,0,0,0),; u_2=(0,1,0,0,0),; u_3=(0,0,1,0,0),;ldots$ and $V=(a,a,b,b,b):a,binmathbbR $. It is clear that for every member of the basis: $u_i notin V$. Therefore, it is impossible for any pair of the original basis vectors to span $V$.
answered Sep 7 at 3:53
olaphus
563
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As others have said, this statement is false in general. I will give you this counter-example:
If $u_1=(1,0,0,0,0),; u_2=(0,1,0,0,0),; u_3=(0,0,1,0,0),;ldots$ and $V=(a,a,b,b,b):a,binmathbbR $. It is clear that for every member of the basis: $u_i notin V$. Therefore, it is impossible for any pair of the original basis vectors to span $V$.
answered Sep 7 at 3:53
olaphus
563
As others have said, this statement is false in general. I will give you this counter-example:
If $u_1=(1,0,0,0,0),; u_2=(0,1,0,0,0),; u_3=(0,0,1,0,0),;ldots$ and $V=(a,a,b,b,b):a,binmathbbR $. It is clear that for every member of the basis: $u_i notin V$. Therefore, it is impossible for any pair of the original basis vectors to span $V$.
answered Sep 7 at 3:53
olaphus
563
answered Sep 7 at 3:53
olaphus
563
answered Sep 7 at 3:53
olaphus
563
answered Sep 7 at 3:53
olaphus
563
563
add a comment |Â
add a comment |Â
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1
What is the source of this question ? Have you translated from another language ?
– Ahmad Bazzi
Sep 7 at 3:34
2
I don't think this is even true. For example, in $mathbb R^2$, consider the ordinary basis $(0,1), (1,0)$, and the subspace $V = (x,y) : x = y$ of $mathbb R^2$. Then, none of the basis elements even belong to $V$, so they cannot form a basis of $V$. Every basis can be extended to a basis of the overlying space, but every basis of the overlying space may not necessarily be contractible to span a given space.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 3:34
2
Do you mean that $V$ is spanned solely by only two vectors of the set $A=u_i, i=1...5$ ? If it is, then I think this is not correct.
– dmtri
Sep 7 at 3:34
thank you sir, got it..........
– Magneto
Sep 7 at 3:47