# Column space equivalent for systems of differential equations?

I apologize if this question is rather general, but I was looking for an answer for a while but couldn't find much.

Is there something like a "column space" equivalent for systems of linear/non-linear differential equations? That is, some function space of all linear combinations of functions.

If so, what would be its basis?

If not, what are the problems of generalizing the concept of a column space to differential equations?

• For linear systems of ODEs, there is a so-called fundamental matrix (just googled University of Washington lecture notes). For a nonlinear system, a linear combination of solutions need not be a solution. – user539887 May 2 '18 at 20:27

There is something like a "column space" for any map, and an operator - of any type - is a map. The "something" is the image of the map.

That is, if $f : A \to B$ is a map with domain $A$ and codomain $B$, then the set $f(A) := \{f(a)\mid a\in A\}$ is called the image of $f$.

In the particular case of an $n \times m$ real matrix $M$, the matrix can be considered to be a map $M : \Bbb R^m \to \Bbb R^n$ which carries $v \in \Bbb R^m$, considered as a column vector, to $Mv \in \Bbb R^n$. The "column space" of $M$ is exactly the image of $M$ as a map.

Now, this particular example has useful properties: We can multiply column vectors $v \Bbb R^k$ by elements of $\Bbb R$ and we can add two column vectors together. And further, $M$ respects these operations: if $a, b \in \Bbb R$ and $u,v \in \Bbb R^m$, then $M(au + bv) = a(Mu) + b(Mv)$.

These concepts generalize: A set $V$ with an addition operation defined on it, and a "scalar" multiplication between elements of $\Bbb R$ and elements of $V$ also defined which distributes over both the addition in $V$ and the addition in $\Bbb R$, then set $V$ is called a vector space over $\Bbb R$. You can generalize this to other fields than $\Bbb R$ ($\Bbb C$ in particular), but I'll stick with real vector spaces.

Vector spaces abound. If $A$ is any non-empty set, then the set $\mathscr F(A) = \{f \mid f\text{ is a function from }A\text{ into }\Bbb R\}$ is a vector space, where addition and scalar multiplication are defined by $(f + g)(x) := f(x) + g(x)$ and $(af)(x) := a(f(x))$ for all $f,g \in V, a \in \Bbb R$. Any subset of $\mathscr F(A)$ which is closed under the addition and scalar multiplication forms a vector subspace of $\mathscr F(A)$. For example, when $A$ is some open interval of $\Bbb R$, then the sets $$C^n(A) := \{f \in \mathscr F(A) \mid f\text{ is }n\text{-times differentiable and }f^{(n)}\text{ is continuous}\}$$ are vector spaces for any non-negative integer $n$ (where $f^{(0)}$ is taken to be $f$ itself). The space $C^0(A)$ of all continuous functions on $A$ is also denoted by $C(A)$.

If $V, W$ are vector spaces, and $T: V \to W$ is a map from $V$ to $W$ which satisfies for all $u, v \in V$ and $a, b \in \Bbb R$ that $T(au + bv) = aT(u) + bT(v)$, then $T$ is called a "linear map". (If $W = V$, then $T$ is called a "linear operator on $V$".) $T$ behaves just like a matrix $M$, and $V$ behaves like $\Bbb R^n$, except that $n$ need not be finite (to be sure, there are other differences - most particularly, $V$ may not have a "standard basis" of vectors $(1, 0, ...), (0, 1, ...), ...$ like $\Bbb R^n$, but the most important aspects are the same).

As before with $M$, the image $T(V)$ of $T$ is its "column space". Since $T$ is linear, it can be shown that $T(V)$ is a vector subspace of $W$ - that is, a subset of $W$ which is closed under addition and multiplication, and therefore can be considered a vector space in its own right.

An $n$=th order differential operator $\scr L$ over the set $A$ can be considered to be a map from $C^n(A) \to C(A)$. If $\scr L$ is linear, then its image $\text{Im}(\mathscr L) := \mathscr L(C^n(A))$ is a vector subspace of $C(A)$, and has most of the properties you expect of column spaces. If $\scr L$ is not linear, then there is no reason to expect that $\text{Im}(\scr L)$ will be a vector space. So adding functions in it or multiplying them by scalars will not necessarily provide other functions within it. But it still satisfies some of the properties of a column space.

• Thank you very much Paul :)! – holistic May 3 '18 at 19:14