Suppose that $e\_1, \cdots, e\_n$ are the standard vectors of the Euclidean space $\mathbb{R}^n$.


Let us consider the backward shift operator $T:\mathbb{R}^n\to \mathbb{R}^n$ given by $Te\_k=e\_{k-1}$ if $k\geq2$ and $Te\_1=e\_n$.


Let $A:\mathbb{R}^n\to \mathbb{R}^n$ be the operator given by $Ae\_3=Ae\_4=e\_1$ and $Ae\_k=0$ for any other $k$.


It is well-known that the columns of the discrete Fourier matrix are just the eigenvectors of $T$. What about the (analytical proof) eigenvectors of $T+A$?



The eigenvectors of adding a particular rank one matrix to the circulant matrix
$\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$For any $x=(x\_1,\dots,x\_n)\in\R^n$ we have $Tx=(x\_2,\dots,x\_n,x\_1)$ and $Ax=(x\_3+x\_4,0\dots,0)$, so that for $U:=T+A$ we have $Ux=(x\_2+x\_3+x\_4,x\_3,\dots,x\_n,x\_1)$.


Let now $(x\_1,\dots,x\_n)\in\R^n$ be an eigenvector of $U$ belonging to an eigenvalue $\la$ of $U$.
Then
\begin{equation}
\left\{
\begin{aligned}
&x\_k=\la x\_{k-1}\text{ for }k=3,\dots,n, \\ 
&x\_2+x\_3+x\_4=\la x\_1, \\ 
&x\_1=\la x\_n. 
\end{aligned}
\right. 
\end{equation}
So, $x\_k=\la^{k-2} x\_2$ for $k=3,\dots,n$, $x\_1=\la x\_n=\la^{n-1} x\_2$, $(1+\la+\la^2)x\_2=\la x\_1=\la^n x\_2$. So, $x\_2\ne0$ and without loss of generality (wlog) $x\_2=1$. So, $\la$ is one of the $n$ roots $\la\_1,\dots,\la\_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x\_{j,1},\dots,x\_{j,n})$ belonging to the eigenvalue $\la\_j$ is given by th formulas
\begin{equation}
 x\_{j,2}=1,\quad x\_{j,1}=\la\_j^{n-1},\quad 
 x\_{j,k}=\la\_j^{k-2}\text{ for }k=3,\dots,n. 
\end{equation}




---


It is now shown (see [this question and the answers to it](https://mathoverflow.net/q/447134/36721)) that the eigenvalues $\la\_1,\dots,\la\_n$ are pairwise distinct and hence the corresponding eigenvectors form a basis of $\R^n$.


