Show that a given matrix always has an eigenvector in $\mathbb{R}$ Can somebody give a hint? The given exercise is, for all $\theta$ in $\mathbb{R}$, show that the matrix always has an eigenvector in $\mathbb{R^2}$ $$
A = \begin{pmatrix}
\cos(\theta) & \sin(\theta) \\
\sin(\theta) & -\cos(\theta)
\end{pmatrix}
$$
The use of determinants isn't allowed, it is just allowed that I should use de definition of eigenvector and eigenvalue, and the fact that the eigenvectors are linearly independent if they are distinct (if needed). 
I don't fully understand the problem because I can't see how is possible that this matrix can give me a multiple scalar of some vector.
 A: the matrix $\pmatrix{\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta}$ represents  the reflection on a mirror along the line $y = \tan (\frac{\theta}{2})\  x.$ therefore $\pmatrix{\cos (\theta/2) \\\sin(\theta/2)}$ is an eigenvector corresponding to the eigenvalue $1$  and $\pmatrix{\sin (\theta/2) \\-\cos(\theta/2)}$ is an eigenvector corresponding to the eigenvalue $-1$.
A: AFTERTHOUGHT: it occurs to me that there is something that does not need determinant, although the concept is implicit. As you can easily confirm, we have a matrix $A$ such that $\color{red}{A^2 = I}.$ Now, if you are willing to accept the proposition that every square matrix has an eigenvalue (possibly complex) then we can write
$$ Av = \lambda v  $$ for $v$ a nonzero column matrix, possibly complex as well. Then 
$$ v = Iv =  A^2 v = \lambda Av = \lambda^2 v.  $$
So, in fact, $\lambda = \pm 1.$ This argument holds for things that are not reflections or symmetric; consider
$$ A =
\left(
\begin{array}{cc}
0 & 7 \\
\frac{1}{7} & 0
\end{array}
\right)   
$$
Also, note that  possession of an eigenvalue is not automatic in infinite dimension.  From what I can see, in the space of one-sided infinite sequences with, say, complex entries,  the right-shift operator does not have any non-zero eigenvector. See http://en.wikipedia.org/wiki/Shift_operator#Sequences
ORIGINAL: People seem confused. This matrix gives a REFLECTION. Determinant is $-1,$ trace is $0,$ so, no matter what $\theta$ might be, the characteristic polynomial is
$$ \lambda^2 - 1  $$
and the eigenvalues are $1$ and $-1.$ there is no counterexample.
It saves a little writing if you write the matrix as
$$
\left(
\begin{array}{cc}
a & b \\
b & -a
\end{array}
\right)   
$$
with the understanding that
$$ a^2 + b^2 = 1.  $$
So, the $+1$ eigenvector is a column vector sent to the zero vector by
$$
\left(
\begin{array}{cc}
a -1 & b \\
b & -a -1
\end{array}
\right).   
$$
Think about it. With $a^2 + b^2 = 1,$ why is the matrix immediately above singular? What is the actual eigenvector in terms of $a,b?$
Similar, the $-1$ eigenvector is a column vector sent to the zero vector by
$$
\left(
\begin{array}{cc}
a +1 & b \\
b & -a +1
\end{array}
\right).   
$$
Again, why is this singular? Oh, symmetric matrix, the eigenvectors have different eigenvalues and are perpendicular to each other.
You can tell that a matrix is a reflection if it can be written as $\color{red}{I - 2 v v^T},$ where the letter $v$ refers to a column vector of length $1.$ Oh, in the other order, $v^T v$ is the (squared) length of $v,$ it is a 1 by 1 matrix with entry $v \cdot v.$ In the order used above, $v v^T$ is a symmetric, rank one, positive semidefinite matrix, furthermore its trace is exactly $1.$ So, the determinant of $\color{red}{I - 2 v v^T}$ is $-1$ and its trace is $n-2,$ where $n$ is the dimension. In dimension 2, you need only check the trace and determinant to know for sure. 
A: Suppose $Av=\lambda v$.
We can scale $v$ to have unit norm, so that $v=(\cos\alpha,\sin\alpha)$.
Writing out $Av=\lambda v$ gives
$$
(\cos\theta\cos\alpha+\sin\theta\sin\alpha,\sin\theta\cos\alpha-\cos\theta\sin\alpha)
=
\lambda(\cos\alpha,\sin\alpha).
$$
We can simplify the LHS:
$$
(\cos(\theta-\alpha),\sin(\theta-\alpha))
=
\lambda(\cos\alpha,\sin\alpha).
$$
If we calculate the norm of both sides, we get
$$
1=\lambda^2\cdot1,
$$
so $\lambda=\pm1$.
If $\lambda=1$, we have $\cos(\theta-\alpha)=\cos\alpha$ which implies $\theta-\alpha=\pm\alpha+2\pi n$ for some $n\in\mathbb Z$.
The negative sign is only possible if $\theta$ is a multiple of $2\pi$ (this case is simple to solve separately).
For the positive sign we get $\alpha=\theta/2-\pi n$ for some $n\in\mathbb Z$.
Shifting $\alpha$ by $\pi$ changes the sign of $v$.
Therefore, up to sign, the normalized eigenvector corresponding to $\lambda=1$ is $(\cos(\theta/2),\sin(\theta/2))$.
A similar calculation for $\lambda=-1$ gives $(\sin(\theta/2),-\cos(\theta/2))$.
A: That transformation $A$ is a reflection $P_x$ along the $x$-axis followed by a rotation $R_\theta$ by an angle $\theta$ :
$$
A =
\left(
\begin{matrix}
\cos\theta & \sin\theta \\
\sin\theta & -\cos\theta
\end{matrix}
\right)
=
\left(
\begin{matrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 \\
0 & -1
\end{matrix}
\right)
=
R_\theta \, P_x 
$$
The length of the vector will be preserved by the transformation, because reflections and rotations do.
We want $A x = \lambda x$ for some non-zero vector $x$ and real number $\lambda$. Thus
$$
\lVert x \rVert = \lVert A x \rVert = \lvert \lambda \rvert \, \lVert x \rVert
$$
The preserved length enforces $\lambda = \pm 1$.
So a vector $u_\phi$ with angle $\phi$ and non-zero length $r$
$$
u_\phi
=
\left(
\begin{matrix}
r \cos \phi \\
r \sin \phi
\end{matrix}
\right)
\quad (*)
$$
and 
$$
A u_\phi 
= 
R_\theta \, P_x u_\phi
=
R_\theta u_{-\phi}
=
u_{-\phi + \theta}
= 
u_\phi
\iff \\
-\phi + \theta = \phi \iff
\phi = \frac{\theta}{2}
$$ 
will have an eigenvalue $1$. For the eigenvalue $-1$, 
$$
A u_\phi = - u_\phi = u_{\phi + \pi}
$$
one needs a vector with
$$
-\phi + \theta = \phi + \pi \iff \phi = \frac{\theta-\pi}{2}
$$
Both vectors exist for any $\theta$, just use the above determined  $\phi$ for any non-zero $r$ in equation $(*)$.
