What is an efficient way of showing that the matrix $$\begin{align} P\triangleq \begin{bmatrix}\cos\theta_1&\sin\theta_1&...&\cos\theta_n&\sin\theta_n\\ \cos2\theta_1&\sin2\theta_1&...&\cos2\theta_n&\sin2\theta_n\\ \vdots&\vdots&~&\vdots&\vdots\\ \cos2n\theta_1&\sin 2n\theta_1&...&\cos2n\theta_n&\sin2n\theta_n\end{bmatrix}\in\mathbb{R}^{2n\times 2n} \end{align}$$ is nonsingular for distinct $\theta_i\in(0,\pi)$ (or similar conditions on $\theta_i$).

I have seen this similar post but I cannot do the same here. Any help is appreciated. Thanks


closed as off-topic by Namaste, Saad, Xander Henderson, Xam, JMP Apr 6 '18 at 2:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Saad, Xander Henderson, Xam, JMP
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ You probably need a stronger assumption than distinctness of the $\theta_i$. At least distinctness $\mod 2 \pi$ is required. $\endgroup$ – Hans Engler Mar 29 '18 at 21:47
  • $\begingroup$ @HansEngler I agree. The main problem is that, when I want to calculate the determinant using Chebyshev polynomials, I cannot think of any helpful row operations $\endgroup$ – Yasi Mar 29 '18 at 21:49
  • 1
    $\begingroup$ For what it's worth, for the case $n=3$ the determinant is $-64 \sin \theta_1 \sin \theta_2 \sin \theta_3 (\cos \theta_1 -\cos \theta_2)^2 (\cos \theta_1 -\cos \theta_3)^2 (\cos \theta_2 -\cos \theta_3)^2$ $\endgroup$ – Hans Engler Mar 30 '18 at 0:21
  • $\begingroup$ @HansEngler Someone told me that if I use Chebyshev polynomials, then by some row operations, I can get a Vandermonde matrix. But I don't see how to do it $\endgroup$ – Yasi Mar 30 '18 at 0:27

Let $D=\pmatrix{1&1\\ i&-i}$. Then $$ P(D\oplus D\oplus\cdots\oplus D) =\pmatrix{ e^{i\theta_1}&e^{-i\theta_1}&\cdots&e^{i\theta_n}&e^{-i\theta_n}\\ e^{2i\theta_1}&e^{-2i\theta_1}&\cdots&e^{2i\theta_n}&e^{-2i\theta_n}\\ \vdots&\vdots&&\vdots&\vdots\\ e^{(2n-1)i\theta_1}&e^{-(2n-1)i\theta_1}&\cdots&e^{(2n-1)i\theta_n}&e^{-(2n-1)i\theta_n}\\ e^{2ni\theta_1}&e^{-2ni\theta_1}&\cdots&e^{2ni\theta_n}&e^{-2ni\theta_n}}, $$ and in turn $P(D\oplus D\oplus\cdots\oplus D)\operatorname{diag}(e^{-i\theta_1},e^{i\theta_1},\cdots,e^{-i\theta_n},e^{i\theta_n})$ is the Vandermonde matrix for $e^{i\theta_1},\ e^{-i\theta_1},\ \ldots,\ e^{i\theta_n},\ e^{-i\theta_n}$.

  • $\begingroup$ Thanks, nicely done. I have a question if you don't mind. Do you think it's possible to get a closed-form expression for the inverse of P in a reasonable amount of time? $\endgroup$ – Yasi Mar 30 '18 at 20:56
  • 1
    $\begingroup$ @Mohkam7 There is an explicit formula for the inverse of a Vandermonde matrix. So, $P^{-1}$ is just the Vandermode matrix inverse times the diagonal matrix inverse times the block diagonal matrix inverse. $\endgroup$ – user1551 Mar 31 '18 at 2:55
  • $\begingroup$ Thanks, I see what you mean. I think now I have the formula for the inverse but it looks complicated and not very explicit. I suspect there is another approach to finding the inverse, similar to what you did for the determinant, by guessing some other multiplication matrices. $\endgroup$ – Yasi Mar 31 '18 at 15:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.