How could I approach determining if this 2D system represented as a 2D summation formula is linear? I have a given 2D system:
$$y(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} x(k_1,k_2)$$
My usual approach to determining if a system is linear is to test if it is homogeneous and additive. First I would input an arbitrary point: (0,0), (1,1), (0,1), (1,0), calculate its output. Then I would scale the input and scale the output and see if the scaling amount matches. Next I would test the additive property in a similar manner. This is fine for a simple 2D system formula, for example: $y(m,n) = x(m,n) + 3$
With a 2D system represented as summation, I am not sure how to apply this test.
Searching the web, it seems this system looks like a 2D convolution:
$$y[i,j]= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} h[m,n]⋅x[i−m,j−n] $$
However, I am still stuck thinking of how to prove this system is linear. I would assume the process is fundamentally the same (prove it is homogeneous and additive or not). I am just not sure of the steps.
I'm actually starting this out not knowing the answer, although I have my suspicions.  I think it'll be a linear system, but I'm not sure.  Here's your starting point: you define the system $y = h(x)$ as:
$$y = h(x) \mid y(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} x(k_1,k_2)$$
So let
$$\begin{aligned}
y_1(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} A\,x_1(k_1,k_2) = h\left(A\, x_1 \right) = A h(x_1) \\
y_2(m,n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} B\,x_2(k_1,k_2) = h\left(A\, x_2 \right) = B h(x_2)
\end{aligned}$$
Then
$$
y_1(m,n) + y_2(m, n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} A\,x_1(k_1,k_2) + \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} B\,x_2(k_1,k_2)
$$
Because the summations have identical limits, we can bring everything under one pair of summations:
$$
y_1(m,n) + y_2(m, n) = \sum_{k_1=-\infty}^{m} \sum_{k_2=-\infty}^{n} \left ( A\,x_1(k_1,k_2) + B\,x_2(k_1,k_2) \right)
$$
but this is just $y_1 + y_2 = h(A\,x_1 + B\,x_2)$.  You already know that $y_1 = A h(x_1)$ and $y_2 = B h(x_2)$, so linearity is proved.
(Which is what I thought -- I was going to just dismiss your original problem as &quot;summations are linear, go away!&quot; -- then I realized I wasn't sure).