Isomorphisms between $V/U$, $U^0$ and $U^\perp$ I am working through Axler's LADR on my own and was wondering about what the "natural" isomorphisms between these spaces would be. If $V$ is finite-dimensional all 3 of these spaces have the same dimension ($dimV-dimU$) and hence should be isomorphic, correct? 
An isomorphism between $U^\perp$ and $U^0$ seems the most natural: for each vector $u$ of $U^\perp$, $u\mapsto\varphi_u$, where $\varphi$ is an element of $U^0$ given by $\langle,u\rangle$. This mapping is obviously injective and surjective by the functional representation theorem (called the Riesz representation theorem in the book).
Next I tried coming up with one between $U^0$ and $V/U$ (and $U^\perp$ and $V/U$) in pretty much the exact way; for each $\varphi_v$ of $U^0$, $\varphi_v\mapsto v+U$, where a unique $v$ exists by the representation theorem ($\forall w\in U^\perp,w\mapsto w+U$). I seem to have managed to prove that both are isomorphisms. For the first one's injective part, suppose that the kernel is non-empty. Then there is a non-zero functional $\varphi_v$ which is mapped to the identity of $V/U$, i.e. $0+U$, hence $v\in U$. But since $\varphi_v(w)=\langle w,v\rangle\in U^0$ is $0$ for any $w\in U$, $\langle v,v\rangle=0$ and $v=0$, which is a contradiction. This felt a bit counter-intuitive to me, because at first I thought that there should be no reason why 2 different functionals shouldn't possibly map to the same affine subset, but the proof seems to be correct. For the other isomorphism proof of injection seems to be just as simple: suppose the kernel was non-empty, then $w\neq0$ maps to $0+U$ and hence $w\in U$. But if it were it would have to be orthogonal to every element of $U$, including itself, therefore $w=0$, which is a contradiction. For the surjective part of the 2nd one I thought it could be argued as follows: we know that $V=U\oplus U^\perp$, i.e. each vector $v$ can be uniquely represented as $u+u^\perp$. Then any element $v+U$ of $V/U$ can be written as $(u+u^\perp)+U$ or simply $u^\perp+U$. That is, every affine subset has a representative that is orthogonal to $U$ and hence a pre-image under this mapping. Argue surjectivity for the first one the same way, since $\varphi_{u^\perp}$ is in the annihilator. 
My question I suppose would simply be whether my general line of reasoning is correct or not. I understand that all this is utterly trivial, I was just toying around by myself and was looking for validation for my direction of thinking. Apologies for the wall of text.
 A: I assume that $V$ is a Hilbert space (e.g. a finite dimensional vector space with an inner product), $U\subseteq V$ is a sub-vector space and that by $U^0$ you denote the subspace of the dual of $V$ which evaluates to zero on $U$.
You have a canonical isomorphism $V\cong V^*$ given by sending $v$ to $\langle-,v\rangle$. This immediately implies that $U^\perp\cong U^0$ via this isomorphism (which makes the statement completely trivial).
The canonical isomorphism $V\cong U\oplus U^\perp$ implies the statement that $U^\perp\cong V/U$, as you noticed. However, a maybe more natural way to see this (that doesn't involve splitting $V$) is to consider the short exact sequence
$$0\longrightarrow U\longrightarrow V\longrightarrow V/U\longrightarrow0\ .$$
Taking duals, we obtain the short exact sequence
$$0\longrightarrow (V/U)^*\longrightarrow V^*\longrightarrow U^*\longrightarrow0\ .$$
But the dual $V^*\to U^*$ of the inclusion $U\to V$ is given by the restriction of  an element of $V^*$ (acting on $V$) to $U$. Now since the sequence is exact, we have
$$\ker(V^*\to U^*)\cong(V/U)^*,$$
and this kernel is tautologically $U^0$. Thus, the most natural statement is that
$$U^0\cong (V/U)^*,$$
and this becomes the statement that you want once again by the isomorphism of Hilbert spaces and their duals.
