Proof there isn't a vector u such Su=u where S is the rotation transformation in R2

Question

We have the rotation matrix

\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}

Proof: there isn't exist a vector $u \in\ {\mathbb{R}^2}$ ($u\neq0$) such $Au=u$ where $A$ is the rotation matrix.

Thank you!

Answer 1

Hint:

First note that if $\theta=2k\pi$ the matrix is the identity matrix, so the statement in OP can be true only for $\theta\ne 2k\pi$.

Now note that $Ru=u$ means that the matrix $R$ has an eigenvalue $\lambda=1$, so find the eigenvalues of your matrix and show that they are different from $1$.

Answer 2

Rotation matrix is turning your vector. It is just a rotation. If you will rotate vector about axis, you can get your initial vector iff $\theta=2\pi k$ or your vector equals to zero.