I have been learning about the DFT:
$$ X(k) = \sum\limits^{N-1}_{n=1} x(n) e^{-i(2\pi/N)nk}$$
$$ x(n) = \frac{1}{N}\sum\limits^{N-1}_{k=0} X(k) e^{i(2\pi/N)nk} $$
Which allows for a discrete Signal of length N to be expressed as the sum of N complex sinousoids of frequency 0,1,..,N-1
It makes intuative sense that much fewer sinousoids should be required as the signal is not continous, and therefor does not have to deal with issues due to continuity in any order and only has to be equal to the function at discrete points.
I however saw no justification of why exactly N sinousoids are required.
Why does a N length DFT only require N sinousoids?
One of the reasons this makes sense is because the DFT is a linear map that is (or should be!) invertible, the map has to be a linear operator $A:\mathbb C^N\to\mathbb C^M$ where $N=M$.
The dimension $N$ of the first vector spaces is given by the number of sample points, the dimension $M$ of the second vector space is given by the number of basis function, ie. sinuoids. With the argument above, they have to be equal.