Start from easy and low dimensional matrix:

mv⃗=f⃗⟶v⃗=m−1f⃗m\vec v=\vec f \longrightarrow \vec v=m^{-1}\vec f

v⃗={m‾11m‾12m‾21m‾22}⋅{f1f2}=(m‾11f1+m‾12f2)e^1+(m‾21f1+m‾22f2)e^2\vec v= \left\{ \begin{matrix} \overline m_{11}&\overline m_{12}\\ \overline m_{21}&\overline m_{22} \end{matrix} \right\} \cdot \left\{ \begin{matrix} f_1\\ f_2 \end{matrix} \right\} =(\overline m_{11}f_1+\overline m_{12}f_2)\hat e_1+(\overline m_{21}f_1+\overline m_{22}f_2)\hat e_2

Thus,

vi=Σjm‾ijfjv_i=\Sigma_j\overline m_{ij}f_j


For function:

d2ydx2+k2y=f(x)\frac{d^2y}{dx^2}+k^2y=f(x)

operator:

O^=d2dx2+k2\hat O=\frac{d^2}{dx^2}+k^2

O^y(x)=f(x)⟶y(x)=O^−1f(x)?\hat Oy(x)=f(x)\longrightarrow y(x)=\hat O^{-1}f(x) ?


ddx=∫δ(x−x′)ddx′q⋅dx′\frac{d}{dx}=\int \delta(x-x')\frac{d}{dx'}q\cdot dx'

↓\downarrow

2 indexes

y(x)=∫G(x,x′)f(x′)dx′y(x)=\int G(x,x')f(x')dx'

↓\downarrow

O^G(x,x′)=δ(x−x′)\hat OG(x,x')=\delta(x-x')


Conclusion:

Just the result of an infinite dimensional inverse problem.