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The short explanation of the DFT that I like the most is that it projects the signal vector on to the vector room with the exponential functions as basis vectors. Then you can see how much of the signal that each basis vector, corresponding to a frequency, can explain.

It's intuitive to see if you start with a 2D vector space (the regular euclidian plane) and a 2D vector, and then you can expand the definition of vector spaces to include orthogonal functions as bases.



> vector room

Is Swedish, "rum" means both space, and a room. In English, "vector space" is used.


Minor point, it's not a projection because you don't lose dimensions.


nit: the identity matrix is a projection

P^2 = P is the definition. not losing dimensions


Nit to your nit, which is incorrect w.r.t. parent reply:

A Fourier transform is not a projection, it's a change of basis represented by a unitary transformation.


The w Fourier coefficient F(w) is the dot product of f with an exponential function, `e_w • f`, and is in that sense a projection. The inverse Fourier transform writes the original function as a sum of the projected components: `f = sum_w (e_w • f) e_w = sum_w F(w) e_w`. This is exactly how writing an "arrow" style 2- or 3-D vector as a sum of orthogonal projections works.


True, but missing the point as you would not normally refer to the identity matrix as a projection.


What was your point? An incorrect characterization of projections?


Is it a math terminology? Because the way I think it's still a projection, just with 0 residuals?




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