Isomorphic Vector Spaces Proof

In summary, the problem is asking you to prove that in a vector space V over a field F, if there is a linear map f from V to F that is not the zero map, then the quotient space V/Ker(f) is isomorphic to F. This can be proven by using the first isomorphism theorem, which states that if f is a linear map from V to W, then V/Ker(f) is isomorphic to Im(f), the image of f. In this case, the image of f is just F, since f is a map from V to F. Therefore, V/Ker(f) is isomorphic to F.
  • #1
Ninty64
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Homework Statement


Let [itex]V[/itex] be a vector space over the field [itex]F[/itex] and consider [itex]F[/itex] to be a vector space over [itex]F[/itex] in dimension one. Let [itex]f \in L(V,F), f \neq \vec{0}_{V\rightarrow F}[/itex]. Prove that [itex]V/Ker(f)[/itex] is isomorphic to [itex]F[/itex] as a vector space.


Homework Equations


[itex]L(V,F)[/itex] is the set of all linear maps from [itex]V[/itex] to [itex]F[/itex]
Dim(V) = Rank(f) = Nullity(f)

The Attempt at a Solution


I tried to come up with a proof. There is no solution to the problem and I'm unsure of my answer. I also want to get better at proving things, so any responses are appreciated!

I know that two vector spaces are isomorphic if there exists a linear transformation between them that is injective and surjective. But I also know that they are isomorphic if [itex]Dim(V/Ker(f))[/itex] = [itex]Dim(F)[/itex], which I tried to prove.

By definition, [itex]Dim(F)=1[/itex]
Assume [itex]Dim(V) = n[/itex]
Let [itex]\left\{\vec{v}_1,\vec{v}_2,...,\vec{v}_n\right\}[/itex] be a basis for V
Let [itex]\left\{\vec{x}\right\}[/itex] be a basis for [itex]F[/itex]
Then there exist unique [itex]f(\vec{v}_i)=\vec{w}_i[/itex] such that
[itex]f(c_1\vec{v}_1+c_2\vec{v}_2 + ... + c_n\vec{v}_n) = c_1\vec{w}_1 + ... + c_n\vec{w}_n[/itex] with some [itex]c_i \neq 0[/itex] since [itex]f \neq \vec{0}_{V\rightarrow F}[/itex], where [itex]c_1,c_2,...,c_n \in F[/itex]
By definition, [itex]c_1\vec{w}_1+...+c_n\vec{w}_n \in Span(\vec{x})[/itex]
[itex]\Rightarrow ax = c_1\vec{w}_1+...+c_n\vec{w}_n \neq \vec{0}, a \in F[/itex]
[itex]\Rightarrow x = a^{-1}c_1\vec{w}_1 + ... + a^{-1}c_n\vec{w}_n[/itex] since [itex]a \neq 0[/itex]
Therefore, [itex]Span({\vec{w}_1,\vec{w}_2,...,\vec{w}_n}) = F[/itex]
By the Rank Nullity Theorem,
[itex]Dim(V) = Rank(f) + Nullity(f)[/itex]
[itex]\Rightarrow n = 1 + Nullity(f) \Rightarrow Nullity(f)=n-1[/itex]
Let [itex]\left\{\vec{y}_1,\vec{y}_2,...,\vec{y}_{n-1}\right\}[/itex] be a basis for [itex]Ker(f)[/itex]
Since ker(f) is a subspace of V, by definition, then
[itex]Span(\left\{ \vec{y}_1,\vec{y}_2,...,\vec{y}_{n-1}\right\} ) \subseteq Span(\left\{ \vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\} )[/itex]
[itex]\Rightarrow \vec{y}_1 \in Span(\left\{ \vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\} )[/itex]
[itex]\Rightarrow \vec{y}_1 \cup \left\{\vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\}[/itex] is linearly dependent
[itex]\Rightarrow[/itex] there exists some [itex]\vec{v}_i[/itex] such that it is a linear combination of the preceding vectors (let [itex]\vec{v}_i = \vec{v}_n[/itex] by reordering)
Thus, [itex]\vec{v}_n \in Span(\left\{ \vec{y}_1, \vec{v}_1,\vec{v}_2,...,\vec{v}_{n-1} \right\} )[/itex]
Since [itex]\left\{ \vec{y}_1, ..., \vec{y}_{n-1} \right\}[/itex] is linearly independent, we can continue with this method until we get
[itex]Span(\left\{ \vec{y}_1, \vec{y}_2, ..., \vec{y}_{n-1}, \vec{v} \right\} ) = V[/itex] where [itex]\vec{v} \in \left\{ \vec{v}_1, \vec{v}_2, ... , \vec{v}_n \right\} [/itex]
Let [itex]\vec{z} \in V/Ker(f)[/itex]
[itex]\Rightarrow \vec{z} = \left[ b_1\vec{y}_1 + b_2\vec{y}_2 + ... + b_{n-1}\vec{y}_{n-1} + b_n\vec{v} \right]_{Ker(f)} = [b_n\vec{v}]_{Ker(f)}[/itex]
Thus, [itex]\vec{v}[/itex] spans [itex] V/Ker(f)[/itex]
[itex]\Rightarrow Dim(V/Ker(f))=1[/itex]
Thus, [itex]V/Ker(f)[/itex] and [itex]F[/itex] are isomorphic.
 
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  • #2
This is a special case of a more general result, usually called the first isomorphism theorem. I don't suppose you have learned it? If so, it's a simple application.

There's a simple proof that doesn't require working with a basis. However, I don't know if it is elementary enough. It depends on what facts you know already. What is the definition of V/ker(f)? Do you know about quotient spaces in general? And cosets?
 
  • #3
jbunniii said:
This is a special case of a more general result, usually called the first isomorphism theorem. I don't suppose you have learned it? If so, it's a simple application.

There's a simple proof that doesn't require working with a basis. However, I don't know if it is elementary enough. It depends on what facts you know already.

We went over all three of those. I don't fully understand them, though. I think that's my main problem. I went back to basis because I was familiar with that.

jbunniii said:
What is the definition of V/ker(f)?
It is "the collection of cosets of [itex]V[/itex] modulo [itex]Ker(f)[/itex]".
I quoted that from the book because I would have gotten that wrong. I understand that the coset is [itex]\vec{v} + Ker(f)[/itex] where [itex]\vec{v} \in V[/itex]. I also understand that it relates to the modulus because if
[itex]\vec{v} = \vec{x} + \vec{y}, \vec{x} \in V-Ker(f), \vec{y} \in Ker(f)[/itex]
then [itex]\vec{v} + Ker(f) = \vec{x} + \vec{y} + Ker(f) = \vec{x} + Ker(f)[/itex]
since the kernel is a vector space and any element in the vector space added to every element in the vector space is just every element in the vector space.
jbunniii said:
Do you know about quotient spaces in general and cosets?
I know that every element in [itex]Ker(f)[/itex] is [itex][ \vec{0} ]_{V/Ker(f)}[/itex] and that all the other elements in [itex]V-Ker(f)[/itex] are grouped into congruence classes.

It's all somewhat of a gray area though. I should familiarize myself with these concepts before I go back to the drawing board.

Edit: And we did cover the isomorphism theorem (somewhat briefly). This question is in the section that that was covered. I spent a couple of hours reading/trying to understand the 3 isomorphism theorems in my book (Advanced Linear Algebra by Bruce Cooperstein). The fact I'm having so much trouble from it probably stems from the fact that I'm lacking full understanding of some fundamental concepts.
 
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  • #4
Ninty64 said:
Edit: And we did cover the isomorphism theorem (somewhat briefly). This question is in the section that that was covered. I spent a couple of hours reading/trying to understand the 3 isomorphism theorems in my book (Advanced Linear Algebra by Bruce Cooperstein). The fact I'm having so much trouble from it probably stems from the fact that I'm lacking full understanding of some fundamental concepts.
OK, here's what I recommend. Focus for now on the first isomorphism theorem - that's all you need for this problem. Don't even worry about its proof, just try to understand what the theorem is saying, and how it could be applied to this problem. Applying the theorem like this will help you understand it better, and once you understand it better, its proof will eventually seem almost obvious. I have to take off now but I'll check in later this evening to see how it goes.
 
  • #5
What I take away from the First Isomorphism Theorem is that if two vectors [itex]\vec{v}, \vec{u} \in V[/itex], then for any Linear transformation [itex]T:V\rightarrow W[/itex],
if [itex]T(\vec{u}) = T(\vec{v})[/itex] then [itex] \vec{u} \equiv \vec{v}[/itex] modulus [itex]Ker(T)[/itex]
So if you only take the cosets of [itex]V[/itex] mod [itex]Ker(T)[/itex], then it follows that
[itex]\overline{T}:V/Ker(T) \rightarrow W[/itex] is 1-to-1 since any elements that mapped to the same element in [itex]W[/itex] were in the same coset.
Furthermore, if you restrict [itex]W[/itex] to [itex]Range(T)[/itex], then the new transformation,
[itex]\hat{T}:V/Ker(T) \rightarrow Range(T)[/itex] is surjective since only the elements that mapped to the same point were "grouped together", so to speak.

So back to the original problem.
Let [itex]f \in L(V,F)[/itex]
Since [itex]f \neq \vec{0}_{V \rightarrow F}[/itex], then [itex]Dim(Range(f)) \geq 1[/itex]
Since [itex]Dim(F) = 1[/itex], then it follows that [itex]Rank(f) = 1[/itex] and, furthermore, [itex]Range(f) = F[/itex]
By the first isomorphism theorem, the linear transformation [itex]\hat{T}:V/Ker(f) \rightarrow Range(f)[/itex] is isomorphic.
Since [itex]Range(f) = F[/itex], then [itex]V/Ker(f)[/itex] and [itex]F[/itex] are isomorphic.

Haha! I feel good about this now! So I hope it's right.
 
  • #6
Ninty64 said:
What I take away from the First Isomorphism Theorem is that if two vectors [itex]\vec{v}, \vec{u} \in V[/itex], then for any Linear transformation [itex]T:V\rightarrow W[/itex],
if [itex]T(\vec{u}) = T(\vec{v})[/itex] then [itex] \vec{u} \equiv \vec{v}[/itex] modulus [itex]Ker(T)[/itex]
So if you only take the cosets of [itex]V[/itex] mod [itex]Ker(T)[/itex], then it follows that
[itex]\overline{T}:V/Ker(T) \rightarrow W[/itex] is 1-to-1 since any elements that mapped to the same element in [itex]W[/itex] were in the same coset.
Furthermore, if you restrict [itex]W[/itex] to [itex]Range(T)[/itex], then the new transformation,
[itex]\hat{T}:V/Ker(T) \rightarrow Range(T)[/itex] is surjective since only the elements that mapped to the same point were "grouped together", so to speak.

So back to the original problem.
Let [itex]f \in L(V,F)[/itex]
Since [itex]f \neq \vec{0}_{V \rightarrow F}[/itex], then [itex]Dim(Range(f)) \geq 1[/itex]
Since [itex]Dim(F) = 1[/itex], then it follows that [itex]Rank(f) = 1[/itex] and, furthermore, [itex]Range(f) = F[/itex]
By the first isomorphism theorem, the linear transformation [itex]\hat{T}:V/Ker(f) \rightarrow Range(f)[/itex] is isomorphic.
Since [itex]Range(f) = F[/itex], then [itex]V/Ker(f)[/itex] and [itex]F[/itex] are isomorphic.

Haha! I feel good about this now! So I hope it's right.

Yes, it looks right to me.
 
  • #7
Thank you so much for your help!
 

FAQ: Isomorphic Vector Spaces Proof

1. What is an isomorphic vector space?

An isomorphic vector space is a mathematical concept in linear algebra where two vector spaces have the same structure and can be mapped onto each other in a way that preserves the algebraic operations and properties of vectors.

2. How can you prove that two vector spaces are isomorphic?

To prove that two vector spaces are isomorphic, you need to show that there exists a bijective linear transformation between the two spaces. This means that the transformation must be one-to-one and onto, and must preserve the vector space operations such as addition and scalar multiplication.

3. What are some common techniques used in isomorphic vector space proofs?

Some common techniques used in isomorphic vector space proofs include showing that the transformation is both one-to-one and onto, using the invertibility of matrices, and using familiar vector space properties such as associativity and distributivity to show that the transformation preserves the structure of the vector space.

4. Can a vector space be isomorphic to itself?

Yes, a vector space can be isomorphic to itself. This is known as an identity transformation, where the transformation maps each element of the vector space to itself. In this case, the vector space has the same structure and properties as itself, making it isomorphic.

5. Are all vector spaces isomorphic to each other?

No, not all vector spaces are isomorphic to each other. For two vector spaces to be isomorphic, they must have the same number of dimensions. Additionally, the transformation between the two spaces must preserve the vector space operations and properties. If these conditions are not met, then the vector spaces are not isomorphic.

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