# When is this matrix singular?

Consider matrix $A$ with $(j,k)$′th entry $A_{j,k}=\sin(\omega_j t_k+\phi_j),\,\forall j,k\in\{1,2,...,n\}$, where $\omega_j,t_k,\phi_j\in \mathbf R$.

1) For $t_k=k$, what is the condition on $\omega_j$, aside from the trivial one $\omega_j=q_j\pi$ for some $q_j\in\mathbf Z$, for A to be singular? What can we say about the eigenvalues and eigenvectors of A? Particularly, I am interested in whether it is true that, when $q_j\pm q_k \neq q\pi, \forall j,k\in\{1,2,...,n\}$ and $\forall q\in\mathbf Z$, $A$ is nonsingular.

2) The same question as in 1) but for general $t_k\in\mathbf R$.

Thoughts: For 1), the complex variable version of the problem, i.e. $A_{j,k}=e^{i\omega_j k}$, is much simpler. $A$ is a Vandermonde matrix and is singular iff $\exists j,k, \omega_j-\omega_k=2q\pi$ for some $q\in\mathbf Z$. However, I am not clear whether and how I can convert between the complex variable version and the above real variable one. I am not aware of and can not find any results regarding the eigenvalues and eigenvectors of Vandermonde matrix.

Problem 2) seems even harder.

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Have you tried $2x2$ and $3x3$ cases? This may give you a sense if this is feasible as stated. –  Lev Borisov Feb 23 '14 at 22:06
If $\omega_j=1$, $\phi_j=-t_j$, then $A_{jk}=\sin(t_k-t_j)$, which means that $A$ is a rank-2 matrix, and hence singular (for arbitrary $t_k$). I guess you are trying to get necessary and sufficient conditions which will be much harder... –  Suvrit Feb 24 '14 at 0:20
@LevBorisov: Yes, I have tried $2\times 2$ case with $\phi_1=\phi_2=0$. The singularity condition is the trivial $\omega_2=\pm\omega_1+q\pi$ for some integer $q$, similar to the ones listed in the question. Perhaps I should find a way to use some symbolic manipulation software to simplify the determinant of the general $3\times 3$ cases. –  Hans Feb 24 '14 at 0:56
@Suvrit: This condition falls in the simple scenarios mentioned in question 1). Still, in this case, why is $A$ always rank 2? –  Hans Feb 24 '14 at 1:36
$\sin(x-y)=s_xc_y-c_xs_y$, which is at most rank-2. –  Suvrit Feb 24 '14 at 1:58

## 1 Answer

For $\phi_j=0$ and $t_k=k$ you have (up to a constant) the matrix $A_{jk}= \lambda_j^k - \lambda_j^{-k}$ where $\lambda_j=\exp( i w_j)$. You can factor out $\lambda_j-\lambda_j^{-1}$ to get a matrix $B_{jk}$ with the entries that are symmetric under transformation $\lambda_j\leftrightarrow\lambda_j^{-1}$. Moreover, you have $$B_{jk}=P_k(\lambda_j+\lambda_j^{-1})$$ where $P_k$ is a monic polynomial of degree $k-1$. By subtracting columns (note that $P_k$ depends on $k$ only, not on $j$) you get Vandermonde matrix $(\lambda_j+\lambda_j^{-1})^{k-1}$ and you can go from there.

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Fabulous! Do you have any idea how to deal with $\phi_j\neq 0$? –  Hans Feb 24 '14 at 4:42
Not at the moment. I again suggest trying $2x2$ case. But it may well be something more complicated. The same goes for more complicated $t_k$. –  Lev Borisov Feb 24 '14 at 12:04