- #1

- 46

- 0

## 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.