Show that the rank is 1 I am given a matrix in $\mathbb{R}^{2 \times 1}$, such that $y \in (1, \infty)$:
\begin{pmatrix}
-\frac{1}{2y^2}cos(2y)-\frac{1+y}{y}sin(2y)\\
-\frac{1}{2y^2}sin(2y)+\frac{1+y}{y}cos(2y)
\end{pmatrix}
and I am supposed to show that it has rank $1$.
My attempt: I tried showing that the two can't be $0$ at the same time. I also tried showing that the two don't intersect in a point that's a root. Both without success. Any hints maybe?
Thanks in advance!
 A: The vector you give doesn't describe a vector space but a spiral curve, whose asymptotic limit is the unit circle.

Therefore, I don't see how it can make any sense to speak about a "rank" because the word is reserved to a vector space.
Further properties could be obtained by writing the vector under the form:
$$\underbrace{\begin{pmatrix}
\cos(2y)&\ \sin(2y)\\
\sin(2y)&-\cos(2y)
\end{pmatrix}}_{S_{y}}\underbrace{\begin{pmatrix}
-\frac{1}{2y^2}\\ -\frac{1+y}{y}
\end{pmatrix}}_{V_y}$$
where $S_{y}$ is an invertible matrix in general (its determinant is $-1$ : it is a symmetry matrix).
A: The only $(2\times 1)$-matrix which does not have rank $1$ is $\binom00$, so assume that
\begin{cases}-\frac1{2y^2}\cos(2y)-\frac{1+y}y\sin(2y)=0,\\-\frac1{2y^2}\sin(2y)+\frac{1+y}{y}\cos(2y)=0.\end{cases}
Multiply both equations with $-2y^2$ to obtain
\begin{cases}\cos(2y)+2y(1+y)\sin(2y)=0,\\\sin(2y)-2y(1+y)\cos(2y)=0.\end{cases} To make the expressions more readable, let $c=\cos(2y),s=\sin(2y)$. Now square both equations above and add them up to get \begin{align*}0&=c^2+4y(1+y)cs+4y^2(1+y)^2s^2+s^2-4y(1+y)sc+4y^2(1+y)^2c^2\\&=(c^2+s^2)+4y^2(1+y)^2(c^2+s^2)\\&=1+4y^2(1+y)^2\geq 1,\end{align*}
where the identity $\cos^2x+\sin^2x=1$ has been used. Since this clearly is a contradiction, the assumption was wrong and both entries of the matrix cannot simultaniously be $0$. Hence the matrix has rank $1$ for all $y$.
