# Isomorphic Vector Spaces Proof

1. Sep 26, 2012

### Ninty64

1. The problem statement, all variables and given/known data
Let $V$ be a vector space over the field $F$ and consider $F$ to be a vector space over $F$ in dimension one. Let $f \in L(V,F), f \neq \vec{0}_{V\rightarrow F}$. Prove that $V/Ker(f)$ is isomorphic to $F$ as a vector space.

2. Relevant equations
$L(V,F)$ is the set of all linear maps from $V$ to $F$
Dim(V) = Rank(f) = Nullity(f)

3. 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 $Dim(V/Ker(f))$ = $Dim(F)$, which I tried to prove.

By definition, $Dim(F)=1$
Assume $Dim(V) = n$
Let $\left\{\vec{v}_1,\vec{v}_2,...,\vec{v}_n\right\}$ be a basis for V
Let $\left\{\vec{x}\right\}$ be a basis for $F$
Then there exist unique $f(\vec{v}_i)=\vec{w}_i$ such that
$f(c_1\vec{v}_1+c_2\vec{v}_2 + ... + c_n\vec{v}_n) = c_1\vec{w}_1 + ... + c_n\vec{w}_n$ with some $c_i \neq 0$ since $f \neq \vec{0}_{V\rightarrow F}$, where $c_1,c_2,...,c_n \in F$
By definition, $c_1\vec{w}_1+...+c_n\vec{w}_n \in Span(\vec{x})$
$\Rightarrow ax = c_1\vec{w}_1+...+c_n\vec{w}_n \neq \vec{0}, a \in F$
$\Rightarrow x = a^{-1}c_1\vec{w}_1 + ... + a^{-1}c_n\vec{w}_n$ since $a \neq 0$
Therefore, $Span({\vec{w}_1,\vec{w}_2,...,\vec{w}_n}) = F$
By the Rank Nullity Theorem,
$Dim(V) = Rank(f) + Nullity(f)$
$\Rightarrow n = 1 + Nullity(f) \Rightarrow Nullity(f)=n-1$
Let $\left\{\vec{y}_1,\vec{y}_2,...,\vec{y}_{n-1}\right\}$ be a basis for $Ker(f)$
Since ker(f) is a subspace of V, by definition, then
$Span(\left\{ \vec{y}_1,\vec{y}_2,...,\vec{y}_{n-1}\right\} ) \subseteq Span(\left\{ \vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\} )$
$\Rightarrow \vec{y}_1 \in Span(\left\{ \vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\} )$
$\Rightarrow \vec{y}_1 \cup \left\{\vec{v}_1,\vec{v}_2,...,\vec{v}_{n}\right\}$ is linearly dependent
$\Rightarrow$ there exists some $\vec{v}_i$ such that it is a linear combination of the preceding vectors (let $\vec{v}_i = \vec{v}_n$ by reordering)
Thus, $\vec{v}_n \in Span(\left\{ \vec{y}_1, \vec{v}_1,\vec{v}_2,...,\vec{v}_{n-1} \right\} )$
Since $\left\{ \vec{y}_1, ..., \vec{y}_{n-1} \right\}$ is linearly independent, we can continue with this method until we get
$Span(\left\{ \vec{y}_1, \vec{y}_2, ..., \vec{y}_{n-1}, \vec{v} \right\} ) = V$ where $\vec{v} \in \left\{ \vec{v}_1, \vec{v}_2, ... , \vec{v}_n \right\}$
Let $\vec{z} \in V/Ker(f)$
$\Rightarrow \vec{z} = \left[ b_1\vec{y}_1 + b_2\vec{y}_2 + ... + b_{n-1}\vec{y}_{n-1} + b_n\vec{v} \right]_{Ker(f)} = [b_n\vec{v}]_{Ker(f)}$
Thus, $\vec{v}$ spans $V/Ker(f)$
$\Rightarrow Dim(V/Ker(f))=1$
Thus, $V/Ker(f)$ and $F$ are isomorphic.

2. Sep 26, 2012

### jbunniii

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. Sep 27, 2012

### Ninty64

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.

It is "the collection of cosets of $V$ modulo $Ker(f)$".
I quoted that from the book because I would have gotten that wrong. I understand that the coset is $\vec{v} + Ker(f)$ where $\vec{v} \in V$. I also understand that it relates to the modulus because if
$\vec{v} = \vec{x} + \vec{y}, \vec{x} \in V-Ker(f), \vec{y} \in Ker(f)$
then $\vec{v} + Ker(f) = \vec{x} + \vec{y} + Ker(f) = \vec{x} + Ker(f)$
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.
I know that every element in $Ker(f)$ is $[ \vec{0} ]_{V/Ker(f)}$ and that all the other elements in $V-Ker(f)$ 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.

Last edited: Sep 27, 2012
4. Sep 27, 2012

### jbunniii

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. Sep 27, 2012

### Ninty64

What I take away from the First Isomorphism Theorem is that if two vectors $\vec{v}, \vec{u} \in V$, then for any Linear transformation $T:V\rightarrow W$,
if $T(\vec{u}) = T(\vec{v})$ then $\vec{u} \equiv \vec{v}$ modulus $Ker(T)$
So if you only take the cosets of $V$ mod $Ker(T)$, then it follows that
$\overline{T}:V/Ker(T) \rightarrow W$ is 1-to-1 since any elements that mapped to the same element in $W$ were in the same coset.
Furthermore, if you restrict $W$ to $Range(T)$, then the new transformation,
$\hat{T}:V/Ker(T) \rightarrow Range(T)$ 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 $f \in L(V,F)$
Since $f \neq \vec{0}_{V \rightarrow F}$, then $Dim(Range(f)) \geq 1$
Since $Dim(F) = 1$, then it follows that $Rank(f) = 1$ and, furthermore, $Range(f) = F$
By the first isomorphism theorem, the linear transformation $\hat{T}:V/Ker(f) \rightarrow Range(f)$ is isomorphic.
Since $Range(f) = F$, then $V/Ker(f)$ and $F$ are isomorphic.

6. Sep 27, 2012

### jbunniii

Yes, it looks right to me.

7. Sep 27, 2012

### Ninty64

Thank you so much for your help!