# 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 $\vec u=0$

• Whatever $V-H$ means here, I don't think that approach will work. Jun 9, 2015 at 17:40
• $0 \notin K=V-H$ Jun 9, 2015 at 17:40
• We define $U+W=\{\,u+w\mid u\in U,w\in W\,\}$, but how do you define $V-H$? Jun 9, 2015 at 17:40

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.

• This wont work for infinite dimension V. Jun 9, 2015 at 17:58
• @BhaskarVashishth The “recursion” method doesn't work, we need Zorn's lemma (or the equivalent axiom of choice). Jun 9, 2015 at 18:00
• @egreg yes I know that because of finite-dimensionality I can assume basis for $V$ and $H$ but how can I conclude from that the subspaces $H=Span(v_1,\dots,v_h)$ and $K =Span(v_{h+1},\dots,v_n$ have the properties mentioned or in fact how can I say that the basis of $V$ is the union of the basis of $H$ and $K$ Jun 9, 2015 at 18:01
• @sepideh That $H=\operatorname{Span}(v_1,\dots,v_h)$ is true by assumption. Just apply the definitions for the rest, I'll add something. Jun 9, 2015 at 18:03
• Hello @egreg and sorry for bringing you back to this post. I wonder, in the case that $V$ is finite-dimensional, can you use the subspaces $H$ of $V$ and the corresponding $K_r$'s to reconstruct the dual $V^*$ of $V$? May 1, 2017 at 10:17

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 $$K=\text{span}{\{\beta'-\beta\}}$$

• would you please explain more? I'm totally a beginner. I don't understand what you mean by extending the basis of $H$ to the basis of $V$ and then how can I conclude from that the subspace $K$ exists Jun 9, 2015 at 17:48
• Which book of linear algebra are you studying? Jun 9, 2015 at 17:49
• Gilbert Strang's Jun 9, 2015 at 17:50
• see page 109, Heading 2L Jun 9, 2015 at 17:59
• I personally do not like strang's book . There are many other interesting books from beginners to advanced level. You should at least explore in some library and keep one more book by your side while doing strang. Jun 9, 2015 at 18:11