Prove that $V$ is isomorphic to $W \times V/W$.

Question

I have been asked to prove the following:

Let $V$ be a finite dimensional vector space of a field $K$ and $W$ be a subspace of $V$. Prove that $V$ is isomorphic to $W \times V/W$ (the direct product of $W$ and $V/W$).

Here is my proof thus far:

Define $\pi: V \rightarrow V/W$ by $\pi(v) = [v]$.

We need to show that $\pi$ is a linear map and that it is surjective and injective.

To show that $\pi$ is a linear map we must show that $\pi(a+b) = \pi(a) + \pi(b)$ and that $\pi(ka) = k\pi(a)$.

1. Take some $a, b \in V$. $\pi(a+b) = [a + b] = [a] + [b] = \pi(a) + \pi(b)$ by our definition of addition of equivalence classes.

2. Take some $a \in V$ and some $k \in \mathbb{F}$. Now, $\pi(ka) = [ka] = k \cdot [a] = k\pi(a)$ by our definition of multiplication of equivalence classes.

So, we have show that $\pi$ is a linear map and now we must show that it is injective and surjective.

To show that $\pi$ is surjective:

1. Take $[a] \in V/W$, where $[a]$ := {$v \in V | a-v \in W$}

2. Let $v = a$, and so, $a-a \in W$ which implies that $0 \in W$. Since $W$ is a subspace, we know by definition that $0 \in W$, and it follows that $\pi$ is surjective.

To show that $\pi$ is injective:

1. We know by definition that $\pi$ is injective if $\pi(a) = \pi(b)$ implies that $a=b$.

2. Take $a,b \in V$ and assume that $\pi(a) = \pi(b)$. By definition we know this implies that $[a]=[b]$. This then implies that $a \sim b$, so we are done.

I feel confident that I showed that $\pi$ is a linear map correctly but I am worried about the part where I illustrate that $\pi$ is a linear map. Any help would be appreciated.

Answer 1

Hint: consider a basis $\{w_1,\dots,w_m\}$ of $W$ and extend it to a basis $$ \{w_1,\dots,w_m,v_1,\dots,v_n\} $$ of $V$. Now note that $\{[v_1],\dots,[v_n]\}$ is a basis for $V/W$.

Some more details. A vector $v\in V$ can be uniquely written as $$ v=a_1w_1+\dots+a_mw_m+b_1v_1+\dots+b_nv_n $$ Define $f(v)= a_1w_1+\dots+a_mw_m$ and prove this defines a linear map $f\colon V\to W$. Now consider also the projection map $\pi\colon V\to V/W$, $\pi(v)=[v]$ and prove that $F\colon V\to W\times V/W$, $$ F(v)=(f(vj,\pi(v)) $$ Is the required isomorphism.