# Show that a matrix is nonsingular [closed]

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, JMPApr 6 '18 at 2:53

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• You probably need a stronger assumption than distinctness of the $\theta_i$. At least distinctness $\mod 2 \pi$ is required. – Hans Engler Mar 29 '18 at 21:47
• @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 – krms Mar 29 '18 at 21:49
• For what it's worth, for the case $n=3$ the determinant is $-64 \sin \theta_1 \sin \theta_2 \sin \theta_3 (\cos \theta_1 -\cos \theta_2)^2 (\cos \theta_1 -\cos \theta_3)^2 (\cos \theta_2 -\cos \theta_3)^2$ – Hans Engler Mar 30 '18 at 0:21
• @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 – krms 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 $e^{i\theta_1},\ e^{-i\theta_1},\ \ldots,\ e^{i\theta_n},\ e^{-i\theta_n}$.
• @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. – user1551 Mar 31 '18 at 2:55