Isomorphism between quotient space $X/M$ and the orthogonal complement of $M$. Let $X$ be the Euclidean space $\mathbb{C}^n$ and let $M=\mathbb{C}^k$, for $1≤k<n$. Show that $X/M$ is isomorphic to the orthogonal complement of $\mathbb{C}^k$.
The orthogonal complement of $M$ to be a subspace of the dual of $X$ defined by $M^⊥=\{T∈X^*: \forall w∈M, T(w)=0\}$. Where $X^*$ is the collection of all linear transformations from $X$ to $\mathbb{C}$. Now I have to show that $X/M$ is isomorphic to $M^⊥$. But I am totally stuck that... What function I will take from $X/M$ to $M^⊥$ so that it will an Banach space isomorphism. 
 A: Elements in the space $X/M$ are cosets of the form $x+M$, with $x\in X$. For simlicity denote $[x]=x+M$.
Also denote the canonical basis of $X$ by $\{e_1,\dots,e_n\}$, where for each $e_j$ its $j-$th coordinate is $1$ and all other coordinates are $0$. Then it's not difficult to see that $\{[e_{k+1}],[e_{k+2}],\dots,[e_n]\}$ form a basis of $X/M$.
On the other hand, define $T_j\in X^*$ via:
$$T_j(w)=\begin{cases} \lambda, &w=\lambda e_j\in\text{span}\{e_j\}\\
0, &\text{else}\end{cases}.$$
Then it's not hard to see that $\{T_{k+1},T_{k+2},\dots,T_n\}$ form a basis of $M^\perp$. Hence
$$\text{dim}(X/M)=\text{dim}(M^\perp)=n-k.$$
This proves $X/M\cong M^\perp$.
A: Frank's answer is correct, but I wanted to provide some additional perspective. It's well-known you can set-theoretically identify the collection of cosets of $M$ with a subset of $X$ by picking a unique representative of each coset. This much is also true for cosets of groups, ideals of rings, and indeed for equivalence classes of any equivalence relation.
What's special about vector spaces is that you can pick this subset to be a subspace, so that the set of representatives is closed under addition and scalar multiplication. This is a choice of a complement for $M$, i.e. a subspace $M'$ such that $X = M + M'$ and $M \cap M' = 0$. Then it follows from $(x + M) + (x' + M) = (x + x') + M$ and $r(x + M) = (rx) + M$ that $X/M ≅ M'$. [1]
As usual in linear algebra, you can also identify subspace complements with projections. A choice of a complement for M is equivalent to a choice of projection $p : X \to X$ onto M, i.e. a linear operator such that $p^2 = p$ and $im(p) = M$. Then $M' = ker(p)$ and $q = 1-p$ is the complementary projection onto $M'$ corresponding to the canonical projection $\pi : X \to X/M$. Hence  $X/M = X/ker(q) ≅ im(q) = M'$.
In homological algebra, these observations are equivalent to the fact that every short exact sequence of vector spaces is split exact. This is not true for arbitrary modules because they don't always have complements: https://en.wikipedia.org/wiki/Splitting_lemma. The homological language also provides an alternative expression of the preceding ideas which dispenses with subspaces by talking about injections between different vector spaces and identifying projections onto subspaces with left inverses of such injections.
[1] This argument extends to the case where $X$ may be a non-abelian group and $X$ is the internal direct product of the subgroups $M$ and $M'$. The last condition implies that $M$ and $M'$ commute with each other even though multiplication in $X$ is generally non-commutative. Hence $x M = M x$ for all $x \in M'$ and so $(x M) (x' M) = (x x') M$ for all $x, x' \in M'$. This can also be generalized to the case where $X$ is the internal semidirect product of $M$ and $M'$ and they don't necessarily commute but only satisfy a twisted commutativity law.
As a geometric example, the group $E(n)$ of orientation-preserving rigid motions in $\mathbb{R}^n$ is the internal semidirect product of the subgroup $R(n, x) ≅ SO(n)$ of rotations around some arbitrary point $x \in \mathbb{R}^n$ and the subgroup $T(n) ≅ \mathbb{R}^n$ of translations. Rotations and translations don't commute but do follow a twisted commutativity law: $(r, 0) (1, v) = (1, rv) (r, 0)$ for all $r \in R(n, x), v \in T(n)$, where the rotations act automorphically on the subgroup of translations by rotating the translation vector. Hence $E(n)/T(n) ≅ R(n, x)$. But it's certainly not true that $E(n)/R(n, x) ≅ T(n)$ since $R(n, x)$ isn't even a normal subgroup: if you conjugate $R(n, x)$ by a nonzero translation $v \in T$ you get $R(n, x+v)$, not $R(n, x)$. 
