## What does this blog do

We are treating linear ODE problems as an example of quotient space in this blog post. You are assumed to be able to solve linear ODEs without theoretical problems, with basic understanding of normal subgroups.

## General theories

### Quotient Space

Let $X$ be a vector space, and $N$ a subspace of it. Naturally $N$ is normal in $X$ since $X$ is abelian. Define

for $x \in X$, then the collection of sets ${\pi(x):x \in X}$ is the *quotient space* of $X$ *modulo* $N$, in which case we write $X/N$. Addition and scalar multiplication are defined by

This is well-defined since $N$ is a vector space. The kernel of $\pi$ or the origin of $X/N$ has to be $N=0+N$.

### Linear ODE

If one solves a linear ODE problem of order $n$ on an interval $(a,b)$, namely

they will find that the solution can be

where $\mathbf{c}$ is a given constant vector and $\mathbf{\Phi}(x)=\begin{bmatrix}\mathbf{\varphi_1},\mathbf{\varphi_2},\cdots,\mathbf{\varphi_n}\end{bmatrix}$ and ${\mathbf{\varphi_i}}$ are the fundamental solutions to $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}$. We’ll translate this into the language of quotient space.

## Steps to quotient space

So where is the $X$? It suffices to pick $\mathcal{C}^n$, the space of all functions $\mathbf{y}=(y_1,y_2,\cdots,y_n)^T$ such that $y_k$ is $n$-times differentiable. The crux therefore becomes finding $N$. And we’ll show that it’s denoted by $\mathbf{\Phi}(x)\mathbf{c}$.

### The solutions of $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}$ form a vector space

Indeed, it’s trivially verified since $\frac{d}{dx}$ and matrix multiplication are linear. The question is, how does this vector space *look* like? Why the fundamental solutions to this equation has and only has $n$ elements? Does that mean this space (denoted by $N$), has dimension $n$? Fortunately, the following isomorphism answers this question in the affirmative.

TheoremThe vector space $N$ is isomorphic to $\mathbb{R}^n$

Pick and fix $x_0 \in (a,b)$. Picard’s existence and uniqueness theorem ensures that, the initial value problem

has a unique solution. Therefore we have a bijection

It suffices to prove that $H$ is linear. Namely, we need to show that

which is trivial, since

### Quotient space and coset are there

Let’s see this solution again

For $\mathbf{c}=(c_1,c_2,\cdots,c_n)^T$, we have

We want to state that ${\mathbf{\varphi_k}}$ is a **basis** of $N$. Yes, it has $n$ elements, but it should also be linear independent. We have no worry about that, since the existence of $\mathbf{\Phi}^{-1}(x)$ forces $|\mathbf{\Phi}| \neq 0$, which is equivalent to the linear independence.

There we have it. $\mathbf{c}$ can be any element of $\mathbb{R}^n$, hence $\mathbf{\Phi}(x)\mathbf{c}$ goes through all elements of $N$. Also we know, the function

is a special solution to $\frac{d\mathbf{y}}{dx}=\mathbf{Ay}+\mathbf{f}$. Thus we have

to be the set of all solutions where $\pi(\mathbf{z}) \in X/N$.