On some propreties of orthogonal complements

Question

In my book the following propositions on orthogonal complements are given without any proof. However, I cannot figure out how to prove them, even though they must follow directly from the definition somehow. Could you show me what to do?

$\color{#a00}{\text{Proposition}}$

Let $V$ be a finite dimensional euclidean vector space with scalar product $g$. Let $S_1$ and $S_2$ be two vector subspaces of $V$; and let $S_1'$ and $S_2'$ be their orthogonal complements in $V$. Then, we have:

(a) $(S_1')'=S_1$;

(b) $S_1 \subset S_2 \implies S_2' \subset S_1'$;

(c) $(S_1 \cap S_2)'=S_1'+S_2'$;

(d) $(S_1+S_2)'=S_1' \cap S_2'$.

Answer 1

The orthogonal complement (with respect to the inner product $g$) of the subspace $S$ is $$ S'=\{v\in V:g(v,w)=0,\text{ for all $w\in S$}\} $$ First of all we can show that $S\cap S'=\{0\}$, because if $v\in S\cap S'$, then in particular $g(v,v)=0$, so $v=0$.

Two of the statements follow from the definition.

(b) If $S_1\subset S_2$ and $v\in S_2'$, then $g(v,w)=0$ for all $w\in S_2$; in particular $g(v,w)=0$ for all $w\in S_1$, which means $v\in S_1'$.

(d) From (b) we know that $(S_1+S_2)'\subset S_1'$ and $(S_1+S_2)'\subset S_2'$, so we just need to prove the converse inclusion. Suppose $v\in S_1'\cap S_2'$. If $w_1\in S_1$ and $w_2\in S_2$, then $$ g(v,w_1+w_2)=g(v,w_1)+g(v,w_2)=0+0=0 $$ Since $w_1+w_2$ is a generic element of $S_1+S_2$, we conclude.

For proving (a) we need the concept of orthogonal projection: every vector $v\in V$ can be written in one and only one way as $v=w+u$, where $w\in S_1$ and $u\in S_1'$ (you surely find this in your textbook). We also need the fact that, for every subspace $S$, $S\subset (S')'$ (which follows from the symmetry of $g$). Indeed, if $w\in S$ and $v\in S'$, we have $g(v,w)=0$ and so also $g(w,v)=0$; since $v$ is arbitrary, $g(w,v)=0$ for all $v\in S'$, which means that $w\in (S')'$.

In particular $V=S_1+S_1'$, so $$ \dim V=\dim S_1+\dim S_1'-\dim(S_1\cap S_1')=\dim S_1+\dim S_1'-0 $$ and therefore $\dim S_1'=\dim V-\dim S_1$. For the same reason, $\dim (S_1')'=\dim V-\dim S_1'$. Hence $$ \dim S_1=\dim V-\dim S_1'=\dim (S_1')' $$ and, from $S_1\subset(S_1')'$ we conclude $S_1=(S_1')'$.

Note that this part essentially requires that $V$ is finite dimensional, as it is false, in general, for infinite dimensional spaces (Hilbert spaces are introduced precisely for this purpose, as this property is valid in them).

It remains to see (c). Now, consider $T_1=S_1'$ and $T_2=S_2'$. From (d) we have $$ (T_1+T_2)'=T_1'\cap T_2' $$ which implies $$ (T_1'\cap T_2')'=((T_1+T_2)')'=T_1+T_2=S_1'+S_2' $$ But $(T_1'\cap T_2')'=((S_1')'\cap(S_2')')'=(S_1\cap S_2)'$.

Also this part requires finite dimension of $V$, as it uses (a).

Answer 2

Let's prove (a) and hopefully this will be enough to get you started on the others.

The orthogonal complement of a subspace $S$ of a finite-dimensional $\Bbb R$-inner-product space $V$ is defined as $$ S^{\perp}=\{\vec v\in V:\vec s\in S\Rightarrow\langle\vec v,\vec s\rangle=0\} $$

Lemma. $\dim S+\dim S^\perp=\dim V$

Proof. Let $\vec s_1,\dotsc,\vec s_m$ be a basis for $S$ and let $T:V\to V$ be the linear map $$ T(\vec v)=\langle\vec v,\vec s_1\rangle\,\vec s_1+\dotsb+\langle\vec v,\vec s_m\rangle\,\vec s_m $$ Then one shows that $\DeclareMathOperator{im}{im}\im T=S$ and $\ker T=S^\perp$. The Lemma then follows from the rank-nullity theorem. $\Box$

This gives us enough machinery to prove (a).

Proposition. $S^{\perp\perp}=S$

Proof. By the lemma we have $$ \dim S = \dim V-\dim S^\perp=\dim S^\perp+\dim S^{\perp\perp}-\dim S^\perp=\dim S^{\perp\perp}\tag{1} $$ Furthermore, note that $$ S^{\perp\perp}=\{\vec v\in V:\vec s\in S^\perp\Rightarrow\langle \vec v,\vec s\rangle=0\} $$ Now, to show that $S\subseteq S^\perp$, suppose $\vec v\in S$ and let $\vec s\in S^\perp$. Then, by definition $\langle\vec v,\vec s\rangle=0$ so $\vec v\in S$. This proves that $S\subseteq S^\perp$. This fact combined with (1) proves that $S$ is a subspace of $S^{\perp\perp}$ such that $\dim S=\dim S^{\perp\perp}$. A standard theorem from linear algebra then implies that $S=S^{\perp\perp}$ as advertised. $\Box$

As you can see, your the full answer to your question requires a decent amount of legwork. I suggest you read this answer over and see if you can prove the other parts of the proposition yourself.