Start from easy and low dimensional matrix:

mv=fv=m1fm\vec v=\vec f \longrightarrow \vec v=m^{-1}\vec f

v={m11m12m21m22}{f1f2}=(m11f1+m12f2)e^1+(m21f1+m22f2)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=Σjmijfjv_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(xx)ddxqdx\frac{d}{dx}=\int \delta(x-x')\frac{d}{dx'}q\cdot dx'

\downarrow

2 indexes

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

\downarrow

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


Conclusion:

Just the result of an infinite dimensional inverse problem.