A Proof that Orthogonal Complement is unique So our professor asked us to prove that considering any subspace $S$ of a vector space $V$, the orthogonal complement $S^{\perp}$ is unique. I have devised a proof and I am not sure whether this works. Please check my proof, which is as follows:

Consider that there are at least two different orthogonal complements $S^{\perp}_1$ and $S^{\perp}_2$ of $S$. Then because we know that $V=S+S^{\perp}_1=S+S^{\perp}_2$ and the sum is direct, $v=s_1+s^{\perp}_1=s_2+s^{\perp}_2$ where $v\in V,s_1,s_2\in S,s^{\perp}_1\in S^{\perp}_1,s^{\perp}_2\in S^{\perp}_2$. $v$ is arbitrary.
So that gives $s^{\perp}_1=(s_2-s_1)+s^{\perp}_2$. Now observing that the inner product $\left<s^{\perp}_1,s_2-s_1\right>=0$ by orthogonality, and then putting in $s^{\perp}_1=(s_2-s_1)+s^{\perp}_2$ in this inner product, we obtain $||s_2-s_1||=0$ implying that $s_1=s_2=s$(say).
So we have $v=s+s^{\perp}_1=s+s^{\perp}_2$ implying $s^{\perp}_1=s^{\perp}_2$. But $s^{\perp}_1\in S^{\perp}_1$ and $s^{\perp}_2\in S^{\perp}_2$ implying that $S^{\perp}_1\subset S^{\perp}_2$ and $S^{\perp}_2\subset S^{\perp}_1$ as $v$ was arbitrary (and hence $s^{\perp}_1,s^{\perp}_2$ are arbitrary). This shows that $S^{\perp}_1=S^{\perp}_2$ and the result is proved. Further, this shows that for any vector $v=s+s^{\perp}$ the complement $s^{\perp}$ is also unique.

 A: Assume $$S\oplus T_1=S\oplus T_2=V\ ,\tag{1}$$
where both $T_1$ and $T_2$ are orthogonal to $S$. Consider a $t\in T_1\ (\subset V)$. Then $V=S\oplus T_2$ implies that there is an $s\in S$ and a $t'\in T_2$
with
$$t=s+t'\ .$$
Taking the scalar product with $s$ on both sides one immediately gets $\langle s,s\rangle=0$, or $s=0$. This implies $t=t'\in T_2$, and as $t\in T_1$ was arbitrary we conclude that $T_1\subset T_2$. By symmetry, the reverse inclusion holds as well, whence $T_1=T_2$.
Note that this proof works even in the infinite-dimensional case. 
A: Towards the end of the proof, you show that $s_1^\perp=s_2^\perp$, and then claim that, *since $s_1^\perp,s_2^\perp$ are arbitrary, you can say that $S_1^\perp\subset S_2^\perp$ and $S_2^\perp\subset S_1^\perp$.  This is incorrect.  You cannot immediately assume that $s_1,s_2$ are arbitrary elements of $S_1,S_2$: they have been carefully chosen at the beginning so that $s_1+s_1^\perp=s_2+s_2^\perp=v$, where $v$ is an arbitrarily chosen vector that never ends up getting used.  
You've got the right idea: you want to show that an arbitrary element of $S_1^\perp$ is contained in $S_2^\perp$ and vice versa.  But tying them down with a vector $v$ at the beginning isn't going to help you.  Moreover, you never use $v$, only $s_1,s_2,s_1^\perp,s_2^\perp$.  Can you think of a way you might fix this without introducing the arbitrary vector $v$?
A: Let $S\subset V$ and suppose $S_1^\perp,S_2^\perp$ are two different orthogonal complements for $S$.  Let $s_1^\perp\in S_1^\perp$.  Then $s_1^\perp\in V$, so we can write $s_1^\perp=s+s_2^\perp$, where $s\in S,s_2^\perp\in S_2^\perp$.  Now observe that $s=s_1^\perp-s_2^\perp$, so $\|s\|^2=\left<s,s_1^\perp-s_2^\perp\right>=\left<s,s_1^\perp\right>-\left<s,s_2^\perp\right>=0-0=0$.  Therefore, $s=0$ and so $s_1^\perp=s_2^\perp\in S_2^\perp$.  Since $s_1^\perp$ was arbitrary, we conclude that $S_1^\perp\subset S_2^\perp$.  An identical argument tells us that $S_2^\perp\subset S_1^\perp$.  We conclude that $S_1^\perp=S_2^\perp$.
A: Theorem

At the very heart your problematic is about:
  $$V=U\oplus_\perp U'\implies U'=U^\perp$$

On the one hand, one has:
$$U\perp U':\quad u'\in U'\implies u'\perp U\implies u'\in U^\perp$$
On the other hand, one has:
$$V=U\oplus U':\quad u^\perp\in U^\perp\implies 0=\langle u,u^\perp\rangle=\langle u,u+u'\rangle=\|u\|^2\implies u^\perp\in U'$$
(That unveils orthogonal decompositions as orthogonal complements!)
Corollary
Especially, your assertion is a corollary then as:
$$V=U\oplus_\perp U'=U\oplus_\perp U''\implies U'=U^\perp=U''$$
(Note that uniqueness of the orthogonal complement is not an issue.)
