Every subspace of a vector space has a complement I want to see if my proof is true or I thought very trivially?  
If $H$ is a subspace of a finite dimensional vector space $V$, show there is a subspace $K$ such that $H\cap K=0$ and $H+K=V$  
So far I have tried :  
$H\subseteq V$ is a subspace $\Rightarrow\,\exists\;K=(V-H)\subseteq V$
$K$ is a subspace because it's the sum of two subspace $V$ and $(-H)$
Now we should prove that $H\cap K=0$
We assume $H\cap K\neq0\Rightarrow\exists\,\vec u\neq0\,,\,\vec u\in H\cap K$
$\vec u\in H\cap K\Rightarrow \begin{cases}\vec u\in H\\\vec u\in K\,\Rightarrow\vec u\in V-H\Rightarrow\vec u\in V,\,\vec u\notin H\end{cases}$
That's a contradiction so our hypothesis was wrong and $\vec u=0$
 A: You should consider basis of $H$, say $\beta=\{b_i\}_{i\in I}$, extend it to a basis for $V$, say $\beta'=\{a_j\}_{j\in J}$ and take $K=\text{span}{\{\beta'-\beta\}}$
A: You're misled by the symbol $V-H$.
If it is $V-H=\{v\in V:v\notin H\}$ then it is not a subspace, because $0$ is not in this set; if it is $V-H=\{v-x:v\in V,x\in H\}$ then it's equal to $V$ so generally not contained in $H$.
The proof is by repeated selection of elements. If $H=V$, we select $K=\{0\}$ and we're done.
Otherwise there is $w_1\in V$, $w_1\notin H$. So, setting $K_1=\operatorname{Span}(w_1)$, we have $K_1\cap H=\{0\}$. If $H+K_1=V$, we're done.
Otherwise there is $w_2\in V$, $w_2\notin H+K_1$. Set $K_2=\operatorname{Span}(w_1,w_2)$ and prove


*

*$\{w_1,w_2\}$ is linearly independent

*$H\cap K_2=\{0\}$


This starts a recursion, so, suppose we have selected $w_1,\dots,w_{r}$ in such a way that


*

*$\{w_1,\dots,w_r\}$ is linearly independent

*$H\cap K_r=\{0\}$, where $K_r=\operatorname{Span}(w_1,\dots,w_r)$


If $H+K_r=V$ we're done, otherwise we can select $w_{r+1}\in V$, $w_{r+1}\notin H+K_r$ and the set $\{w_1,\dots,w_r,w_{r+1}\}$ has the properties above.
The recursion must stop, because a linearly independent set cannot have more than $\dim V$ elements.

If you already know that every linearly independent set can be extended to a basis, it's simpler, of course. Take $\{v_1,\dots,v_h\}$ a basis of $H$ and extend it to $\{v_1,\dots,v_h,v_{h+1},\dots,v_n\}$, a basis of $V$. Then $K=\operatorname{Span}(v_{h+1},\dots,v_n)$ is the subspace you're looking for.
Indeed, any element of $v$ can be written as
$$
v=\alpha_1v_1+\dots+\alpha_hv_h+\alpha_{h+1}+\dots+\alpha_nv_n
$$
and so
$$
v=x+y\in H+K
$$
where
$$
x=\alpha_1v_1+\dots+\alpha_hv_h\in H\qquad
\text{and}\qquad
y=\alpha_{h+1}+\dots+\alpha_nv_n\in K
$$
If $v\in H\cap K$, then
$$
v=\beta_1v_1+\dots+\beta_hv_h=\gamma_{h+1}v_{h+1}+\dots+\gamma_nv_n
$$
for some scalars. Now
$$
\beta_1v_1+\dots+\beta_hv_h+(-\gamma_{h+1})v_{h+1}+\dots+(-\gamma_n)v_n=0
$$
and so, by the linear independence,
$$
\beta_1=\dots=\beta_h=0\\
\gamma_{h+1}=\dots=\gamma_n=0
$$
and $v=0$, which implies $H\cap K = \{0\}$.

If the space $V$ is not finite dimensional, the result is true provided you accept Zorn's lemma. Consider the set $\mathcal{F}$ of all subspaces $L$ of $V$ satisfying $H\cap L=\{0\}$.
The set $\mathcal{F}$ can be ordered by inclusion and it's not empty, because $\{0\}\in\mathcal{F}$. Let $\mathcal{C}$ be a chain in $\mathcal{F}$; then
$$
L_0=\bigcup\{L:L\in\mathcal{C}\}
$$
is easily seen to be a subspace of $V$ and
$$
H\cap L_0=\bigcup\{H\cap L:L\in\mathcal{C}\}=\{0\}
$$
so $L_0\in\mathcal{F}$ and is an upper bound for $\mathcal{C}$. By Zorn's lemma we can select a maximal element $K\in\mathcal{F}$. If $H+K\ne V$, there is $w\in V$, $w\notin H+K$. But then
$$
H\cap (K+\operatorname{Span}(w))=\{0\}
$$
(proof?) and $K\subsetneq K+\operatorname{Span}(w)$, contradicting the maximality of $K$. Thus $H+K=V$ and we're done.
