While solving an exercise I arrived to this simple matrix and I want to get its rank It is $$\begin{pmatrix}2\cos 2t\\ \cos t \end{pmatrix}$$
It is written in the solution that the rank is 1, they took 2 cases if $\cos t=0$ which is easy and the other case when $\cos t \ne 0$, and this second case is the one I didn't understand.
Help please by showing steps and by giving the precise definition of a rank, it is the number of nonzero rows in a matrix?
If $\cos(t)=0$ then $t=\pi n-\frac{\pi }{2}$ with $n\in \mathbb{Z}$ so $2\cos(2t)=2\cos\left(2\left(\pi n-\frac{\pi }{2}\right)\right)=2\cos\left(-\pi\right)=-2$ and your matrix is $\left(-2,0\right)$.
If $\cos(t)\neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.