Given a vector space $V$ and a subspace $U$ of $V$.
$$ V \cong U \oplus(V/U) $$
Does the above equation always hold? Where $\oplus$ is external direct sum.
For finite dimensional vector space $V$, here is my attemp of prove:
Let dimension of $U$ be m, dimension of $V$ be $n$.
Find a basis of $U$ : $\{ \mathbf{ u_1, u_2, \cdots ,u_m}\}$ and extend it to a basis for $V$ : $\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$.
For every vector $\mathbf{x} \in V$, we can write $\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$ uniquely. Define a linear map $T$ as $$T(\mathbf{x})=(c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}])$$
,where $[]$ is used to express the equivalent class. We claim $T$ is an isomorphism.
Surjectivity is obvious. As for injectivity,
if $T(\mathbf{x})=(\mathbf{0},[\mathbf{0}])$, then $ c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0} $
$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$
$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$
Since $[d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$, we have
$ (d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$, which means $d_1=0, d_2=0, \cdots,d_{n-m}=0$
$\Rightarrow \mathbf{x}=\mathbf{0}$, so $T$ is injective.
Is the above proof correct? Does this mean $V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$ for any linear map $F$, because $\ker F$ is a subspace of $V$ ?
The final question is about how should I prove it when the dimension of $V$ is infinite?
Best Answer
It is true that for every finite dimensional vector space $V$ with $U$ a vector subspace that $$ V \cong U \oplus (V / U) $$ I think your proof is essentially correct. And yes it is true that for $T: V \rightarrow W$ any linear that we have $$ V \cong \operatorname{ker}(T) \oplus \operatorname{Im}(T) $$ The rank-nullity theorem is a direct consequence of this.
In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $k$ $\textit{splits}$. What this means is that if $T : U \rightarrow V$, $S : V \rightarrow W$ are linear maps such that $T$ is injective, $\operatorname{ker}(S) = \operatorname{Im}(T)$, and $S$ is surjective, then $V \cong U \oplus W$.
Note then that this directly gives us our result since if $U$ is a subspace of $V$, then the inclusion map $\iota : U \rightarrow V$ and projection map $\pi : V \rightarrow (V / U) $ set up exactly a short exact sequence.
In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.