I understand that the discrete Fourier transform simply changes basis to the discrete Fourier basis, which is an orthonormal basis of eigenvectors for any shift-invariant linear operator on $\mathbb C^n$. The discrete Fourier basis can be easily discovered by computing the eigenvectors of the shift operator $S:\mathbb C^n \to \mathbb C^n$, which maps $\begin{bmatrix} x_0 & x_1 & x_2 & \cdots & x_{n-1} \end{bmatrix}^T$ to $\begin{bmatrix} x_{n-1} & x_0 & x_1 & \cdots & x_{n-2} \end{bmatrix}^T$.

Is there an analogous way understand the discrete cosine transform? Does the discrete cosine basis diagonalize some particular class of linear operators (as the discrete Fourier basis diagonalizes any shift-invariant linear operator)? Is there some simple and fundamental linear operator, analogous to $S$, which can be used to discover the discrete cosine basis?

How can I expand my understanding of the DFT to include the DCT? How does DCT fit into the picture?


I stumbled upon this paper by Gilbert Strang while searching for something related to the DCT: http://www-math.mit.edu/~gs/papers/dct.pdf.

On page 3 it says: "The basis vectors of cosines are actually eigenvectors of symmetric second-difference matrices." Might be what you are looking for.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.