I was thinking about matrices and then why arent there matrices with uncountable many values? (Probably this conecpt already exists for a very long time, but i don't know it)
Assume there are matrices $A$ and $B$ with the size $N\times N$. Then the product $C=A*B$ is defined as: $$ C_{y,x}=\sum_{k=1}^N{A_{y,k}*B_{k,x}} $$
Now this could to be done with an uncountable matrix (which is just a function):
$$ A(x,y)=x^2+y, B(x,y)=x*y $$
for
$$ x, y\in[0,N] $$
Then the value of $C=A*B$ ($*$ for the "special matrix multiplication") could be defined as: $$ C(x,y)=\int\limits_0^N{A(k,y)*B(x,k)}\mathrm{d}k $$
Now we can multiply two matrices with uncountable many values. There is also a neutral element which can be defined as $e(x,y)=\delta(x-y)$ (dirac delta function).
Of course also vector with uncountable many values could be defined:
$$ b(y)=exp(y) $$
It is also no problem to define a multiplication of such a matrix and such a vector:
$$ A(x,y)=x^2+y, b(y)=exp(y) $$ for $$ x,y\in[0,N] $$
$$ C(y)=A*b=\int\limits_0^N{A(y,k)*b(k)}\mathrm{d}k $$
And now another thing could be defined. A simple equation ($A,b$ are known, $x$ is unknown):
$$ A(x,y)*x(y)=b(y) $$
Or simpler: $$ Ax=b $$
Which is equal to: $$ \int\limits_0^N{A(y,k)*x(k)}\mathrm{d}k=b(y) $$
Now it is not obvious if such an equation has a solution or not. Is it possible to generalize the classic determinant to a function $f(x,y)$, where $x,y\in[0,N]$? So that it is possible to see if such an equation has a unique solution.
All things here were just copied from the classic linear algebra, so i think something like a determinant should also be available?
Thank you very much
best regards
Kevin