Start from easy and low dimensional matrix:
mv=f⟶v=m−1f
v={m11m21m12m22}⋅{f1f2}=(m11f1+m12f2)e^1+(m21f1+m22f2)e^2
Thus,
vi=Σjmijfj
For function:
dx2d2y+k2y=f(x)
operator:
O^=dx2d2+k2
O^y(x)=f(x)⟶y(x)=O^−1f(x)?
dxd=∫δ(x−x′)dx′dq⋅dx′
↓
2 indexes
y(x)=∫G(x,x′)f(x′)dx′
↓
O^G(x,x′)=δ(x−x′)
Conclusion:
Just the result of an infinite dimensional inverse problem.