Briefly, the answer to your question is “No.”
Cropping an image in the spatial domain (cropping the bitmap) is not the same as cropping an image in the frequency domain (cropping the DCT). The latter corresponds more closely to resizing the image to one that's rasterized with a larger pixel.
When you crop an image in the spacial domain, then compute its DCT, you're using an entirely different set of basis functions to represent the image in terms of. I can't think of a way to describe how the DCT changes, and I doubt there's anything much recognizable to spot.
If you think of this in one dimension, it might help.
Suppose you have a signal $(X_1,X_2,\dots,X_{20})$ that's 20 “pixels” long. The DCT of that signal is a set of coefficients $(x_1,x_2,\dots,x_{20})$ that (roughly) represent the relative power of the signal at frequencies (cycles per pixel) $0,\frac{1}{40},\frac{2}{40},\frac{3}{40}\,$, and so on up to $\frac{19}{40}$.
If you crop the image to 18 pixels wide, the DCT gives you the power at frequencies $0,\frac{1}{36},\frac{2}{36},\frac{3}{36}\,$, and so on up to $\frac{17}{36}$.
Except for the power at frequency zero, none of the DCT coefficients for the cropped image represent power at the same frequency as a DCT coefficient for the uncropped image. (Note that the coefficients of the DCT are scaled by $\sqrt{2/N}$, where $N$ is the width of the image, so the frequency-0 coefficient will be slightly different.)
The situation where you would expect to see a DCT that looked like a crop of the DCT of your original image (up to the scaling factor) is if you made your image smaller not by cropping, but by resizing.
Best Answer
It depends what you are asking for exactly. A matrix represents a linear map, in this case I will assume you mean a map $\Bbb R^2\to\Bbb R^2$.
The answer is no because linear maps must preserve the origin. Consider when $\square ABCD$ is the unit square $[0,1]\times[0,1]$ and $\square EFGH$ is any rectangle which doesn't contain the origin.
Yes. Any rectangle can be translated so that one corner is the origin. The two sides of the rectangle can be considered as vectors $u$ and $v$ (which you can visualize as arrows pointing away from the origin). Then the matrix taking $[0,1]\times[0,1]$ to the rectangle with one corner at the origin and sides formed by $u$ and $v$ is just $[u\ v]$ (here $u$ and $v$ are column vectors, so this is a $2\times 2$ matrix). We can remember where the original rectangle's corner was using another vector.
There won't be any way to avoid needing more data than just one matrix here, because in my interpretation cropping is not a linear map (but rather affine).