I have two vectors, $\mathbf{u}_{m1} = M^{-1/2} [ 1 \ e^{\jmath m_1\beta} \ e^{\jmath 2 m_1 \beta} \cdots e^{\jmath (M-1) m_1 \beta}]^T$ and
$\mathbf{u}_{m2} = M^{-1/2} [ 1 \ e^{\jmath m_2\beta} \ e^{\jmath 2 m_2 \beta} \cdots e^{\jmath (M-1) m_2 \beta}]^T$,
where $M$ is a positive natural number, $m_1,m_2 \in \{1,2,\cdots, M\}$, $m_1 \neq m_2$, $\beta = 2\pi/M$, $^T$ is the transpose operator, and $\jmath$ is the imaginary unit with $\jmath^2 = -1$.
Based on the simulation result, I found that the matrix $\mathbf{u}_{m1}\mathbf{u}_{m2}^H + \mathbf{u}_{m2}\mathbf{u}_{m1}^H$ always have eigenvalues
$\lambda_1 = 1$ and $\lambda_2 = -1$.
I would like to know is there anything special about this matrix???
Best Answer
No this is not a special property of the vectors $u_1,u_2$, since this result can be derived by just imposing orthonormality conditions on them. It is a simple matter to see that the linear combinations $u_1\pm u_2$ are eigenvectors of the proposed matrix since
$$(u_1u_2^\dagger+ u_2u_1^\dagger)(u_1\pm u_2)=\pm (u_1\pm u_2)$$
where here we have used the fact that $|u_1|^2=|u_2|^2=1, ~u_1^H u_2=0$.