Why can we treat systems of linear differential equations as systems of linear equations?

Question

The system of linear equations in vector/matrix form is given by $$\textbf{b=Ax}$$ Learning linear algebra I've always thought that both vectors and matrix in this equation are filled with numbers and all the theory about linear systems is based on this fact (am I wrong?).

It is clear for me that we indeed can write system of first order differential equations in vector/matrix form as $$\frac{d}{dt}\textbf{x=Ax}$$ I understand that if we unroll this into separate equations we would in fact get our original ODE's. But it just seems to me as a hacky notational trick mostly. I wouldn't think myself that we can now apply linear algebra simply because we can write it like that.

I mean this time x is vector of functions, and $\frac{d}{dt}\textbf{x}$ is vector of derivatives. It's not obvious for me at all that we can still treat this construction as linear algebra original equation and apply all the rules exactly the same way, even though there are now functions and operators in place of numbers. Doesn't it kind of changes everything / or may change is some cases? Why not?

P.S. I don't know if my confusion comes from differential equations or maybe I lack understanding of linear algebra in its full generality. Could something like that be correct and useful linear algebra construction as well? $$\int_{}^{}\textbf{b=Ax}$$ $$\sin(\textbf{b})=\textbf{Ax}$$ Integration is linear, sine is not, so sine would probably never go like that?

Answer 1

"I've always thought that both vectors and matrix in this equation are filled with numbers and all the theory about linear systems is based on this fact (am I wrong?)."

Yes, you are wrong. The theory of linear algebra is based on the fact that matrices act on column vectors linearly, that is, $A(x+cy)=Ax+cAy$ for any two vectors $x,y$ and any number $c$. If any mathematical object behaves the same way, the theory of linear algebra applies. The differential operator $\frac{\mathrm d}{\mathrm dx}$ behaves the same way in relation to functions:

$$\frac{\mathrm d(f+cg)}{\mathrm dx}=\frac{\mathrm df}{\mathrm dx}+c\frac{\mathrm dg}{\mathrm dx}$$

so the theory of linear algebra applies.

Answer 2

Write it as a differential form in $dt$

$$ d \mathbb x(t) \ = \ \mathbb A(t) \mathbb x(t) dt$$

This is the limit of the dicrete equation

$$ \mathbb x(t_i+ dt) \ = \ \mathbb x(t_i) + \mathbb A(t_i)\ \mathbb \x(t_i) dt$$

So each time step is indeed a small change by the image of a linear map on the current value of the vector.

Crucial is here that $\mathbb A(t)$ is time dependent. That implies, that the matrices at different times do not commute and sorting products produces extra terms.

If $\mathbb A(t)$ is a constant matrix, the solution is the exponential of the factor times $t$

$$ \mathbb x(t) == e^{(t-t0) \mathbb A} \mathbb x(t0)$$

Proof by differentiation of the power series. No problem with the exponential series, that converges in any norm on finite matrices. Convincing first argument: diagonalize $\mathbb A$. The exponential function maps onto the diagonal.

In the case of a time dependent matrix $\mathbb A(t)$, the solution is similar up to the delicate effect, that the powers at different times in the time discrete approximation are not time ordered with latest factor to the left, so it comes down as the first factor in

$$ e^{::\int_0^{t+dt}\ \mathbb A(s) ds }\ = e^{::\int_t^{t+tdt} \ \mathbb A(s) ds::} \ e^{\int_0^t \mathbb A(s) ds ::} = ( \mathbb A(t) dt) \ e^{\int_0^t \mathbb A(s) ds ::}$$

This sign $:: ::$ for time ordering of integrals over noncommuting entities is the central problem in quantum field theory (QFT).