Prove that $forall t in mathbb R$ the set $f^{-1}({t})$ is a hyperplane of $X$
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Exercise :
Let $X$ be a vector space and $f:X to mathbb R$ be a linear functional. Show that for all $t in mathbb R$, the set $f^{-1}({t})$ is a hyperplane of $X$.
Attempt :
I have proved a slightly different example, showing that if $W$ is a hyperplane of $X$ then there exists a linear functional such that $W=f^{-1}({t})$. This was carried out by using the trick and setting $f(x) = f(lambda x + y)= lambda$, since we just needed to show that there exists some linear functional that would fullfill the given condition for some $t in mathbb R$
The case in this exercise though, is different, since we need to generally prove that for any linear functional and all $t in mathbb R$ the hyperplane condition holds.
Essentialy what I need to prove is that $f^{-1}({t})$ is a subspace of $X$ of $text{co}dim=1$ which then means that it is a hyperplane. Or, to prove that every element in this image can be written as $x = lambda x_0 + y$ with $x_0 notin Y$.
Question - Request : I can't see how to proceed proving the fact above though, as the only $text{co}dim$ statement that I recall is the kernel one. I would really appreciate any tips, hints or elaboration to help me work over this exercise and understand it.
real-analysis functional-analysis vector-spaces operator-theory geometric-functional-analysis
asked Nov 18 at 11:47
Rebellos
13k21142
add a comment |
up vote
2
down vote
favorite
Exercise :
Let $X$ be a vector space and $f:X to mathbb R$ be a linear functional. Show that for all $t in mathbb R$, the set $f^{-1}({t})$ is a hyperplane of $X$.
Attempt :
I have proved a slightly different example, showing that if $W$ is a hyperplane of $X$ then there exists a linear functional such that $W=f^{-1}({t})$. This was carried out by using the trick and setting $f(x) = f(lambda x + y)= lambda$, since we just needed to show that there exists some linear functional that would fullfill the given condition for some $t in mathbb R$
The case in this exercise though, is different, since we need to generally prove that for any linear functional and all $t in mathbb R$ the hyperplane condition holds.
Essentialy what I need to prove is that $f^{-1}({t})$ is a subspace of $X$ of $text{co}dim=1$ which then means that it is a hyperplane. Or, to prove that every element in this image can be written as $x = lambda x_0 + y$ with $x_0 notin Y$.
Question - Request : I can't see how to proceed proving the fact above though, as the only $text{co}dim$ statement that I recall is the kernel one. I would really appreciate any tips, hints or elaboration to help me work over this exercise and understand it.
real-analysis functional-analysis vector-spaces operator-theory geometric-functional-analysis
asked Nov 18 at 11:47
Rebellos
13k21142
Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
1
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
Please see my answer.
– asdq
Nov 18 at 13:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Exercise :
Let $X$ be a vector space and $f:X to mathbb R$ be a linear functional. Show that for all $t in mathbb R$, the set $f^{-1}({t})$ is a hyperplane of $X$.
Attempt :
I have proved a slightly different example, showing that if $W$ is a hyperplane of $X$ then there exists a linear functional such that $W=f^{-1}({t})$. This was carried out by using the trick and setting $f(x) = f(lambda x + y)= lambda$, since we just needed to show that there exists some linear functional that would fullfill the given condition for some $t in mathbb R$
The case in this exercise though, is different, since we need to generally prove that for any linear functional and all $t in mathbb R$ the hyperplane condition holds.
Essentialy what I need to prove is that $f^{-1}({t})$ is a subspace of $X$ of $text{co}dim=1$ which then means that it is a hyperplane. Or, to prove that every element in this image can be written as $x = lambda x_0 + y$ with $x_0 notin Y$.
Question - Request : I can't see how to proceed proving the fact above though, as the only $text{co}dim$ statement that I recall is the kernel one. I would really appreciate any tips, hints or elaboration to help me work over this exercise and understand it.
real-analysis functional-analysis vector-spaces operator-theory geometric-functional-analysis
asked Nov 18 at 11:47
Rebellos
13k21142
Exercise :
Let $X$ be a vector space and $f:X to mathbb R$ be a linear functional. Show that for all $t in mathbb R$, the set $f^{-1}({t})$ is a hyperplane of $X$.
Attempt :
I have proved a slightly different example, showing that if $W$ is a hyperplane of $X$ then there exists a linear functional such that $W=f^{-1}({t})$. This was carried out by using the trick and setting $f(x) = f(lambda x + y)= lambda$, since we just needed to show that there exists some linear functional that would fullfill the given condition for some $t in mathbb R$
The case in this exercise though, is different, since we need to generally prove that for any linear functional and all $t in mathbb R$ the hyperplane condition holds.
Essentialy what I need to prove is that $f^{-1}({t})$ is a subspace of $X$ of $text{co}dim=1$ which then means that it is a hyperplane. Or, to prove that every element in this image can be written as $x = lambda x_0 + y$ with $x_0 notin Y$.
Question - Request : I can't see how to proceed proving the fact above though, as the only $text{co}dim$ statement that I recall is the kernel one. I would really appreciate any tips, hints or elaboration to help me work over this exercise and understand it.
real-analysis functional-analysis vector-spaces operator-theory geometric-functional-analysis
real-analysis functional-analysis vector-spaces operator-theory geometric-functional-analysis
asked Nov 18 at 11:47
Rebellos
13k21142
asked Nov 18 at 11:47
Rebellos
13k21142
edited Nov 18 at 11:54
asked Nov 18 at 11:47
Rebellos
13k21142
asked Nov 18 at 11:47
Rebellos
13k21142
asked Nov 18 at 11:47
Rebellos
13k21142
13k21142
Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
1
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
Please see my answer.
– asdq
Nov 18 at 13:37
add a comment |
Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
1
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
Please see my answer.
– asdq
Nov 18 at 13:37
Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
1
1
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
Please see my answer.
– asdq
Nov 18 at 13:37
Please see my answer.
– asdq
Nov 18 at 13:37
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $vin f^{-1}(t)$. Then we have $f^{-1}(t)=v + ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $win f^{-1}(t)$ we have $v-w in ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
answered Nov 18 at 13:36
asdq
1,7211418
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $vin f^{-1}(t)$. Then we have $f^{-1}(t)=v + ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $win f^{-1}(t)$ we have $v-w in ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
answered Nov 18 at 13:36
asdq
1,7211418
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
add a comment |
up vote
2
down vote
accepted
Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $vin f^{-1}(t)$. Then we have $f^{-1}(t)=v + ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $win f^{-1}(t)$ we have $v-w in ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
answered Nov 18 at 13:36
asdq
1,7211418
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $vin f^{-1}(t)$. Then we have $f^{-1}(t)=v + ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $win f^{-1}(t)$ we have $v-w in ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
answered Nov 18 at 13:36
asdq
1,7211418
Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $vin f^{-1}(t)$. Then we have $f^{-1}(t)=v + ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $win f^{-1}(t)$ we have $v-w in ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
answered Nov 18 at 13:36
asdq
1,7211418
answered Nov 18 at 13:36
asdq
1,7211418
answered Nov 18 at 13:36
asdq
1,7211418
answered Nov 18 at 13:36
asdq
1,7211418
1,7211418
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
add a comment |
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
Could you clarify what you mean by the word "affine" ? I'm afraid we haven't been using that.
– Rebellos
Nov 18 at 13:41
1
1
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
An affine subspace of a vector space just a linear subspace that does not necessarily contain $0$. This means that such a subspace is obtained by translating a linear subspace with some element in he vector space, i.e. exactly in the way I described.
– asdq
Nov 18 at 14:02
add a comment |
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Hint: Since $f$ is linear, there is a relation between $f^{-1}(t)$ and $ker f$. Note that $f^{-1}(t)$ is not a linear subspace but only an affine one in general, since $0notin f^{-1}(t)$ for $tneq 0$.
– asdq
Nov 18 at 11:55
@asdq Hi, thanks for your input ! I am probably missing something, but what's the connection between $f^{-1}({t})$ and the kernel of $f$ ?
– Rebellos
Nov 18 at 12:01
1
If $f$ is nonzero then $f^{-1}(t)$ is obtained by translation of the kernel.
– asdq
Nov 18 at 13:11
@asdq I apologise but I cannot see a way to work around it. I would really appreciate an elaboration !
– Rebellos
Nov 18 at 13:20
Please see my answer.
– asdq
Nov 18 at 13:37