Mixing
Question about vector negation
$begingroup$
I have to decide if the following statement is true or false:
The operation of vector negation is a bijection from the set of free vectors
to itself.
Obviously the domain and codomain are correct, but is it a bijection?
I want to say yes, because, although a free vector (and thus its negative) can be represented by an infinite amount of bound vectors, they all have to have the same length and direction, so they represent the same unique free vector. In other words, no matter their positions in 3-space, can all free vectors with the same length and direction be considered one and the same free vector, thus making vector negation a bijection?
Thanks in advance!
vectors
asked Jan 12 at 19:32
JBuckJBuck
667
$endgroup$
|
show 1 more comment
$begingroup$
I have to decide if the following statement is true or false:
The operation of vector negation is a bijection from the set of free vectors
to itself.
Obviously the domain and codomain are correct, but is it a bijection?
I want to say yes, because, although a free vector (and thus its negative) can be represented by an infinite amount of bound vectors, they all have to have the same length and direction, so they represent the same unique free vector. In other words, no matter their positions in 3-space, can all free vectors with the same length and direction be considered one and the same free vector, thus making vector negation a bijection?
Thanks in advance!
vectors
asked Jan 12 at 19:32
JBuckJBuck
667
$endgroup$
$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58
|
show 1 more comment
$begingroup$
I have to decide if the following statement is true or false:
The operation of vector negation is a bijection from the set of free vectors
to itself.
Obviously the domain and codomain are correct, but is it a bijection?
I want to say yes, because, although a free vector (and thus its negative) can be represented by an infinite amount of bound vectors, they all have to have the same length and direction, so they represent the same unique free vector. In other words, no matter their positions in 3-space, can all free vectors with the same length and direction be considered one and the same free vector, thus making vector negation a bijection?
Thanks in advance!
vectors
asked Jan 12 at 19:32
JBuckJBuck
667
$endgroup$
I have to decide if the following statement is true or false:
The operation of vector negation is a bijection from the set of free vectors
to itself.
Obviously the domain and codomain are correct, but is it a bijection?
I want to say yes, because, although a free vector (and thus its negative) can be represented by an infinite amount of bound vectors, they all have to have the same length and direction, so they represent the same unique free vector. In other words, no matter their positions in 3-space, can all free vectors with the same length and direction be considered one and the same free vector, thus making vector negation a bijection?
Thanks in advance!
vectors
vectors
asked Jan 12 at 19:32
JBuckJBuck
667
asked Jan 12 at 19:32
JBuckJBuck
667
asked Jan 12 at 19:32
JBuckJBuck
667
asked Jan 12 at 19:32
JBuckJBuck
667
asked Jan 12 at 19:32
JBuckJBuck
667
667
$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58
|
show 1 more comment
$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58
$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $f:Vrightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $yin V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $yin V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $f:Vrightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $yin V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $yin V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
add a comment |
$begingroup$
Let $f:Vrightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $yin V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $yin V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
add a comment |
$begingroup$
Let $f:Vrightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $yin V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $yin V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
$endgroup$
Let $f:Vrightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $yin V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $yin V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
answered Jan 13 at 1:06
NicNic8NicNic8
4,50531123
4,50531123
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
add a comment |
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
$begingroup$
Thank you! What I had in mind was along those lines.
$endgroup$
– JBuck
Jan 13 at 1:38
add a comment |
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$begingroup$
It always helps to come back to the definition. A bijection is a function that is one-one and onto. Can you prove that vector negation is one-one? Is it onto?
$endgroup$
– NicNic8
Jan 12 at 19:38
$begingroup$
Barring a terrible mistake on my part, I think that vector negation f(x), where x is a free vector is, indeed, onto, since f(x) can take all values in the codomain (free vectors).
$endgroup$
– JBuck
Jan 12 at 19:47
$begingroup$
However, when we consider its injectiveness, things get a little bit murky for me. If f is injective, then no two free vectors must have the same negation. As the time passes, I am more and more inclined to say that f is injective, because we can easily prove by the definition that if u and v have the same negation, then u=v.
$endgroup$
– JBuck
Jan 12 at 19:50
$begingroup$
This is also taking into account that 0 and -0 have the same negation (0), because we define 0=-0. Please feel free to point any mistakes in my explanation.
$endgroup$
– JBuck
Jan 12 at 19:51
$begingroup$
Another way to prove a map is bijective is to prove that it's invertible. Can you show that negation is an invertible function?
$endgroup$
– CyclotomicField
Jan 12 at 19:58