How to construct a Hermitian matrix such that the magnitude of all the elements of its eigenvectors are same I need to construct a Hemintian Matrix with some constraint of its eigenvectors.
Suppose the eigenvectors of the matrix are $u_1=[u_{1,1},u_{1,2},u_{1,3}]$, $u_2=[u_{2,1},u_{2,2},u_{2,3}]$, and $u_3=[u_{3,1},u_{3,2},u_{3,3}]$, and I want $|u_{n,m}|=1$ for all $n=1,2,3$ and $m=1,2,3$.
How can I construct these kind of  matrices?
Thanks!
 A: 
Let $U=\begin{pmatrix}1&e^{-i\pi/3}&e^{i\pi/3}\\1&e^{i\pi}&-1\\e^{2\pi i/3}&e^{\pi i/3}&-1\end{pmatrix}$. Then, since $1+e^{2i\pi/3}=e^{i\pi/3}$, we find that $V\overline V^T=I_3$, where $V=U/\sqrt3$. Now let $A=V\begin{pmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{pmatrix}\overline V^T$, where $\lambda_1, \lambda_2, \lambda_3\in \mathbb C$.  

It is then immediate that $A$ is a Hermitian matrix, with eigen-vecotrs of the required type, since the eigen-spaces still contain the vectors in $U$.
Essentially it is to find a set of three orthonormal vectors with unit entries. And we can assign any eigen-value that we want to the matrix. And I just find three orthogonal such vectors.  

The following is about the verification of the properties of $V$. Skip it if needed.
  Let me show that $U$ is orthogonal:
  $(1,e^{-i\pi/3},e^{i\pi/3})\cdot(1,e^{i\pi},-1)=1+e^{2i\pi/3}-e^{i\pi/3}=0$, 
  $(1,e^{-i\pi/3},e^{i\pi/3})\cdot(e^{2\pi/3},e^{i\pi/3},-1)=e^{2i\pi/3}-e^{i\pi/3}+1=0$,
  $(1,e^{i\pi},-1)\cdot(e^{2i\pi/3},e^{i\pi/3},-1)=e^{2i\pi/3}-e^{i\pi/3}+1=0$.
  (Notice that $e^{i\pi}+1=0$)
  In the end, to see the claimed equality $e^{2i\pi/3}+1=e^{i\pi/3}$, see the following (rough) picture:  


Uniqueness 

In fact, the space of eigen-vectors of the required Hermitian matrix is unique, and we shall show it in the following:
  Suppose $u_1\cdot u_2=0$ where $u_1, u_2$ have unit entries. Then we obtain an identity $x_1+x_2+x_3=0$, where each $x_i$ is of the form $e^{i\theta}$. By factoring out a constant, which affects not the eigen-space, we might assume that $x_1=1$. Now it is easily verified that such an equality $1+e^{i\theta_1}+e^{i\theta_2}=0$ is of the above type.  It is now a simple but tedious matter to verify that the choice of three vectors which satisfy the identities right above the picture is unique, in the sense that the eigen-space is unchanged. Thus the requirement defines a unique eigen-space in $\mathbb C^3$.  

Hope the above is right, and that this helps.  
