This was asked in Math Stackexchange [here](https://math.stackexchange.com/questions/1009004/dirichlet-characters-as-eigenvectors) but generated no comments or answers. I have slightly edited the original question with the comment in the fourth paragraph and the explicit matrix example at the end.


I have a very concrete question. First an example: If one considers the additive group $Z/(n)$ which is cyclic, the corresponding group characters are the rows of the discrete fourier transform (DFT)matrix $D$.


Take any $n\times n$ circulant matrix $C$. Obviously this matrix is generated by the action of the cyclic group on the first row. Its eigenvectors are the columns of $D$ (which is symmetric) and the matrix of eigenvectors diagonalizes any circulant matrix. Corresponding to each root of unity $\omega$ of order $n,$ its consecutive powers make up the eigenvector and the inner product of the first row of $C$ is clearly the corresponding eigenvalue.


Now comes the question. For simplicity, assume that $n$ is a prime. Consider the multiplicative group of $Z/(n)$ which has $n-1$ elements. The Dirichlet characters of this group form an $(n-1)\times(n-1)$ matrix $D'$. Now take a ``multiplicative matrix'', $M$ where first one specifies the first row $(a\_1,\ldots,a\_{n-1})$ and then obtains the $k^{th}$ row as $(a\_{k},a\_{2k},\ldots,a\_{(n-1)k})$ where the indices are computed modulo $n.


In fact, if $n$ is assumed to be prime, I'd expect that the eigenvectors to be obtained by a coordinate permutation of the eigenvectors which are the columns of the DFT matrix since both groups in question are cyclic.


I'd have assumed that such a matrix would have exactly the Dirichlet characters, of which there are enough when $n$ is a prime, as its eigenvectors. A few basic computations on Mathematica seemed to contradict this hope. For example taking $n=7,$ and the first row as
$(1,1,1,0,0,0)$ yielding the matrix
$$
\left(
\begin{array}{cccccc}
1 &1 &1 & 0& 0& 0\\
0 &1 &0 & 1& 0& 1\\
0 &1 &1 & 0& 0& 1\\
1 &0 &0 & 1& 1& 0\\
1 &0 &1 & 0& 1& 0\\
0 &0 &0 & 1& 1& 1
\end{array}
\right)
$$
the expected eigenvectors materialized. However,
taking the first row as $(1,1,0,0,0,0)$ yielding the matrix
$$
\left(
\begin{array}{cccccc}
1 &1 &0 & 0& 0& 0\\
0 &1 &0 & 1& 0& 0\\
0 &0 &1 & 0& 0& 1\\
1 &0 &0 & 1& 0& 0\\
0 &0 &1 & 0& 1& 0\\
0 &0 &0 & 0& 1& 1
\end{array}
\right)
$$
did not.


Why is this the case? References, pointers welcome.



Dirichlet Characters as Eigenvectors
I believe the eigenvectors are the ones you guessed, but in your second example, the dimensions of some of the eigenspaces are larger than one. I would guess that Mathematica chose a basis for those eigenspaces different than the eigenvectors coming from the multiplicative Dirichlet characters. By the way, if you view the second example as the adjacency matrix of a directed graph, the graph is two disjoint directed 3-cycles (with loops), and from this point of view the eigenvectors might be more transparent. 


