Show that a given matrix always has an eigenvector in $\mathbb{R}$ Can somebody give a hint?

Question

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.

Answer 1

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.

Answer 2

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 $(\sin(\theta/2),-\cos(\theta/2))$.

Answer 3

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 $(*)$.

Answer 4

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 $-1$.