When is the rank of this matrix equal to $2$?
$\begin{bmatrix}-\sin u + \sin (u+v) & \sin(u+v) \\ \cos u - \cos(u+v) & -\cos(u+v)\\ 1 & 0 \end{bmatrix}$
I’ve computed the two $2 \times 2$ determinants, which give $\sin v$ and $-\cos(u+v)$. However, if I equal them to $0$, I obtain $v=\pi n$ for the first case and $v+u=\pi/2 +\pi n$ for the second one (with $n\in \mathbb{N}$). How should I conclude?
On
I think it is easier to compute the kernel of the matrix. So, let $v_1 = \begin{pmatrix} -\sin u + \sin(u+v) \\ \cos u - \cos(u+v) \\ 1 \end{pmatrix}$ and $v_2 = \begin{pmatrix} \sin(u+v) \\ -\cos(u+v) \\ 0 \end{pmatrix}$ and consider the equation $$\lambda_1v_1 + \lambda_2v_2 = 0.$$ The matrix has maximal rank (rank $2$ in this case) if and only if this equation only has the trivial solution $\lambda_1 = \lambda_2 = 0$. From the third row we get $$\lambda_1 + 0 = 0,$$ hence $\lambda_1 = 0$ and $\lambda_2v_2 = 0$. As $v_2$ is never $0$, this implies that $\lambda_2 = 0$ as well. Thus your matrix always has rank $2$.
On
$$\begin{bmatrix} -\sin u + \sin (u+v) & \sin(u+v) \\ \cos u - \cos(u+v) & -\cos(u+v)\\ 1 & 0 \end{bmatrix} \overset1\sim \begin{bmatrix} -\sin u & \sin(u+v) \\ \cos u & -\cos(u+v)\\ 1 & 0 \end{bmatrix} \overset2\sim \begin{bmatrix} -\sin u & \sin u \cos v + \cos u \sin v \\ \cos u & \sin u \sin v - \cos u \cos v\\ 1 & 0 \end{bmatrix} \overset3\sim \begin{bmatrix} -\sin u & \cos u \sin v \\ \cos u & \sin u \sin v\\ 1 & 0 \end{bmatrix} \overset4\sim \begin{bmatrix} -\sin u & \cos u \sin v \\ \cos u - \sin u \tan v & 0\\ 1 & 0 \end{bmatrix}$$ Explanation:
Note that rank of this matrix is at least one (last row) and at most two (because its dimension is 3×2).
Now, you need to check when matrix rank is 1. If it is one, then first and second rows has to be linearly dependent on the last row, therefore $\sin(u+v) = -\cos(u+v) = 0$, but this can never happen as $\sin^2 x + \cos^2 x = 1$.
Therefore, this matrix has rank 2 for all values of $u$ and $v$