Mixing
Why does $dim W = dim V$ only if $W = V$ for a vector space $V$ and its subspace $W$? Removing one element...
$begingroup$
Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $dim W = dim V$ if and only if $W = V$.
Why is this? I understand the proof I was given (which uses the fact that any set of exactly $n$ linearly independent vectors, where $n = dim V$, in $V$ must be a basis of $V$).
But what about this reasoning: remove one element from $W$, so long as that element doesn't make up a basis of $W$ (though if there is more than one basis then even this condition isn't necessary). Now $W neq V$, but $dim W = dim V$, no?
I am guessing that the reasoning is in some way flawed. But why is it flawed/in what way?
linear-algebra vector-spaces
edited Jan 19 at 22:33
egreg
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
$endgroup$
add a comment |
$begingroup$
Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $dim W = dim V$ if and only if $W = V$.
Why is this? I understand the proof I was given (which uses the fact that any set of exactly $n$ linearly independent vectors, where $n = dim V$, in $V$ must be a basis of $V$).
But what about this reasoning: remove one element from $W$, so long as that element doesn't make up a basis of $W$ (though if there is more than one basis then even this condition isn't necessary). Now $W neq V$, but $dim W = dim V$, no?
I am guessing that the reasoning is in some way flawed. But why is it flawed/in what way?
linear-algebra vector-spaces
edited Jan 19 at 22:33
egreg
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
$endgroup$
$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
3
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22
add a comment |
$begingroup$
Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $dim W = dim V$ if and only if $W = V$.
Why is this? I understand the proof I was given (which uses the fact that any set of exactly $n$ linearly independent vectors, where $n = dim V$, in $V$ must be a basis of $V$).
But what about this reasoning: remove one element from $W$, so long as that element doesn't make up a basis of $W$ (though if there is more than one basis then even this condition isn't necessary). Now $W neq V$, but $dim W = dim V$, no?
I am guessing that the reasoning is in some way flawed. But why is it flawed/in what way?
linear-algebra vector-spaces
edited Jan 19 at 22:33
egreg
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
$endgroup$
Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $dim W = dim V$ if and only if $W = V$.
Why is this? I understand the proof I was given (which uses the fact that any set of exactly $n$ linearly independent vectors, where $n = dim V$, in $V$ must be a basis of $V$).
But what about this reasoning: remove one element from $W$, so long as that element doesn't make up a basis of $W$ (though if there is more than one basis then even this condition isn't necessary). Now $W neq V$, but $dim W = dim V$, no?
I am guessing that the reasoning is in some way flawed. But why is it flawed/in what way?
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 19 at 22:33
egreg
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
edited Jan 19 at 22:33
egreg
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
edited Jan 19 at 22:33
egreg
183k1486205
edited Jan 19 at 22:33
egreg
183k1486205
edited Jan 19 at 22:33
egreg
183k1486205
183k1486205
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
asked Jan 19 at 22:17
James RonaldJames Ronald
1297
1297
$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
3
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22
add a comment |
$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
3
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22
$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
3
3
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument fails because after removing one element from $W$, what you get is not a vector space anymore (unless you are working over the field $mathbb{F}_2$ and $dim W=1$).
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your argument fails because after removing one element from $W$, what you get is not a vector space anymore (unless you are working over the field $mathbb{F}_2$ and $dim W=1$).
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
add a comment |
$begingroup$
Your argument fails because after removing one element from $W$, what you get is not a vector space anymore (unless you are working over the field $mathbb{F}_2$ and $dim W=1$).
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
add a comment |
$begingroup$
Your argument fails because after removing one element from $W$, what you get is not a vector space anymore (unless you are working over the field $mathbb{F}_2$ and $dim W=1$).
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Your argument fails because after removing one element from $W$, what you get is not a vector space anymore (unless you are working over the field $mathbb{F}_2$ and $dim W=1$).
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
edited Jan 19 at 22:34
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 19 at 22:22
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
add a comment |
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
2
2
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
Or if the underlying field is $Bbb Z_2$ and $dim(W) = 1$
$endgroup$
– Omnomnomnom
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
$begingroup$
That makes sense, don't know how I didn't realize that. Thank you for the help!
$endgroup$
– James Ronald
Jan 19 at 22:23
3
3
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
Removing an element from $W={0}$ will not give you a vector space. The onyl working case is indeed a one-dimensional over $Bbb F_2$.
$endgroup$
– Hagen von Eitzen
Jan 19 at 22:27
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
$begingroup$
@HagenvonEitzen I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 19 at 22:34
add a comment |
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$begingroup$
Your reasoning is not clear. Can you spell it out?
$endgroup$
– Berci
Jan 19 at 22:19
3
$begingroup$
Removing one element from $W$ will (in general) not result in a vector space.
$endgroup$
– user1892304
Jan 19 at 22:21
$begingroup$
@user1892304 Ohhh I see, that makes sense thank you! And that would be because it would result in a linear sum or scalar multiple of one of the other items in the space to no longer be in it?
$endgroup$
– James Ronald
Jan 19 at 22:22