In general, $\operatorname{QFT}_m$ is used to denote the QFT defined on basis states $|0\rangle, |1\rangle, \dotsc, |m-1\rangle$ according to $$ \operatorname{QFT}_m \colon |x\rangle \mapsto \frac{1}{\sqrt{m}} \sum_{y=0}^{m-1} e^{2 \pi i \frac{x}{m} y} |y\rangle. $$
This is the definition of QFT. I think the eigenvectors of $\operatorname{QFT}_m$ must be some linear combinations of $|0\rangle, |1\rangle, \dotsc, |m-1\rangle$. However I can't see what those eigenvectors exactly are.
Could anyone explain to me what the eigenvectors and eigenvalues of QFT are?
This is asking for the eigenvalues and eigenvectors of the following matrix (with $\newcommand{\ze}{\zeta}\ze=\exp(2\pi i/m)$) $$ A=\frac1{\sqrt m}\pmatrix{1&1&1&\cdots&1\\1&\ze&\ze^2&\cdots&\ze^{m-1} \\1&\ze^2&\ze^4&\cdots&\ze^{2(m-1)}\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&\ze^{m-1}&\ze^{2(m-1)}&\cdots&\ze^{(m-1)^2}}.$$ It is both a Vandermonde matrix and unitary. Then $$A^2=\pmatrix{1&0&\cdots&0&0\\0&0&\cdots&0&1 \\0&0&\cdots&1&0\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&1&0&\cdots&0}.$$ Thus in general, $A^4=I$ and the eigenvalues are $\pm1$ and $\pm i$. Each eigenspace will have dimension approximately $m/4$. The exact figures will depend on the congruence class of $m$ modulo $4$.