Let $\mathbf{F}$ be a discrete Fourier transform (DFT) matrix such that
\begin{align}
F\_{m,n}=e^{-j2\pi(m-1)(n-1)/N},\quad m,n=1,\ldots,N.
\end{align}
What we can say about the singular value decomposition (SVD) of truncated DFT matrix (the following matrix)?
\begin{align}
\tilde{\mathbf{F}} = 
\begin{bmatrix}
\mathbf{I}\_k,\mathbf{0}
\end{bmatrix}
\mathbf{F} 
\begin{bmatrix}
\mathbf{I}\_k\\\mathbf{0}
\end{bmatrix},
\end{align}
where $\mathbf{I}\_k$ is the identity matrix of size $k<N$.



Singular value decomposition of truncated discrete Fourier transform matrix
Let me insert a factor $N^{-1/2}$, so that the Fourier transform is unitary:
$$U\_{mn}=N^{-1/2}e^{-2\pi i(m-1)(n-1)/N},\quad m,n=1,\ldots,N.$$
We truncate the $N\times N$ matrix $U$ to the $k\times k$ upper left corner,
$$U^{(k)}\_{mn}= N^{-1/2}e^{-2\pi i(m-1)(n-1)/N},\quad m,n=1,\ldots,k\leq N.$$


A characterisation of the singular values of $U^{(k)}$ is given in [The Eigenvalue Distribution of Discrete Periodic Time-Frequency Limiting Operators](https://arxiv.org/abs/1707.05344) and in [The Future Fast Fourier Transform?](https://epubs.siam.org/doi/pdf/10.1137/S1064827597316266).



> 
> **Of order $k^2/n$ of the singular values are close to unity and
> $k-k^2/n$ are close to 0.**
> 
> 
> 


![](https://ilorentz.org/beenakker/MO/truncatedDFT1024.png)
singular values squared of $U^{(k)}$ for $N=1024$, $k=256$.


**Comment:** Actually a total of $\max(0,2k-N)$ of the singular values of $U^{(k)}$ are *precisely* equal to 1, as Noam Elkies was kind enough to explain to me [here.](https://mathoverflow.net/q/416100/11260)


