TITLE: calculate kernels of matrices with angles
QUESTION [5 upvotes]: So my professor gave me this question:
I have to find the basis of the eigenvalues of this matrix
\begin{pmatrix}
  \cos(q) & \sin(q)\\
  \sin(q) & -\cos(q)\\
  \end{pmatrix}
so I calculate the eigenvalues and I found it is 1 and -1.
so now I need to find the basis of the kernel of those matrices
\begin{pmatrix}
  1-\cos(q) & -\sin(q)\\
  -\sin(q) & 1+\cos(q)\\
  \end{pmatrix}
\begin{pmatrix}
  -1-\cos(q) & -\sin(q)\\
  -\sin(q) & -1+\cos(q)\\
  \end{pmatrix}
so actually I need to find 
\begin{pmatrix}
  a \\ b\\
  \end{pmatrix}
that will be function as basis for the kernel of each one of those matrices.
but how do I do it ?
I know how to do it when there is no angles meaning I would just compare it to zero.
for example in the first matrix I get these two equations :
$x-xcos(q)-ysin(q)=0$
$-xsin(q)+y+ycos(q)=0$
so I get this equation:
$x(-sin(y)+((1-cos(y))/(sin(y)))+((cos(y)-cos(y)cos(y))/(sin(y))))=0$
which is true for every $x$ and then I can assign some value to $x$ for example zero and then $y$ equal zero also but after I am checking it, it is not true.
could u please help me ? 
so how can I do it ?
All he told us is that $q\neq{0}$ and $q\neq\pi$ and $q\neq{2}\pi $
and nothing more.

REPLY [2 votes]: Observe that $\cos(q-\pi/2)=\sin q$, and $\sin(q-\pi/2)=-\cos q$. So, the linear transformation corresponding to the matrix sends ${\bf i}\,(=\pmatrix{1\\0})$ to $\bf u$ in the unit circle at angle $q$ (the first column) and sends ${\bf j}\,(=\pmatrix{0\\1})$ to $\bf v$ in the unit circle at angle $q-\pi/2$.
If you draw it, it could be clear that the reflection through the line $e$ at angle $q/2$ also does the same to the basis vectors $\bf i$ and $\bf j$. As both are linear and coincide on a basis, we get that this is the matrix of reflection through $e$. From this, we easily get the eigenvectors: vectors parallel to $e$ will be fixed (i.e. they correspond to eigenvalue $1$) and vectors orthogonal to $e$ will be reversed (i.e. they correspond to eigenvalue $-1$).