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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 $\vec u=0$

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

2 Answers 2

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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.

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    $\begingroup$ This wont work for infinite dimension V. $\endgroup$ Jun 9, 2015 at 17:58
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    $\begingroup$ @BhaskarVashishth The “recursion” method doesn't work, we need Zorn's lemma (or the equivalent axiom of choice). $\endgroup$
    – egreg
    Jun 9, 2015 at 18:00
  • $\begingroup$ @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 $K$ $\endgroup$ Jun 9, 2015 at 18:01
  • $\begingroup$ @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. $\endgroup$
    – egreg
    Jun 9, 2015 at 18:03
  • $\begingroup$ 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 $V$? $\endgroup$
    – JJ1993
    May 1, 2017 at 10:17
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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 $K=\text{span}{\{\beta'-\beta\}}$

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  • $\begingroup$ 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 $\endgroup$ Jun 9, 2015 at 17:48
  • $\begingroup$ Which book of linear algebra are you studying? $\endgroup$ Jun 9, 2015 at 17:49
  • $\begingroup$ Gilbert Strang's $\endgroup$ Jun 9, 2015 at 17:50
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    $\begingroup$ see page 109, Heading 2L $\endgroup$ Jun 9, 2015 at 17:59
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    $\begingroup$ 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. $\endgroup$ Jun 9, 2015 at 18:11

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