Linear algebra — similarity of reflection matrices

Question

The question is as follows:

Let $A,B$ be $2\times2$ reflection matrices.
Are $A$ and $B$ similar?

What I’ve tried:

It did seem like a proof to me:

1. I have calculated the characteristic polynomial and found that it is $x^2 - 1$ for every $2\times2$ reflection matrix but it did not help me much to prove similarity.

2. I have tried showing that a general reflection matrix is similar to a specific reflection matrix and use the association property, did not lead me anywhere as well.

3. Another shot was to take two general reflection matrices and to try finding an invertible matrix $P$ that fulfills $AP=PB$. but I got stuck.

Would appreciate any help.

Answer 1

If $A$ is a $2\times2$ matrix which is the matrix of a reflection with respect to the standard basis of $\Bbb R^2$, then since its characteristic polynomial is $x^2-1$, $A$ is similar to $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$. Therefore, if $B$ is also the matrix of a reflection with respect to the same basis, $B$ is similar to $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$, and so $A$ and $B$ are similar.

Answer 2

Here's a more geometric approach. I assume you're talking about lines through the origin in $\mathbb R^2$.I'll denote reflection about the line $\ell$ by $\omega_{\ell}$ Whenever I say "line" I mean "line through the origin,"(i) Write the matrix that corresponds to reflection about the line $ax+by=0. $ Show that if $\ell_1$ is the line $$Ax+By=0$$ and $\ell_2$ is the line $$A^{\prime}x+B^{\prime}y=0$$ then $\omega_{l_1}(\ell_2)$ is a line $$A^{\prime \prime}x+B^{\prime \prime}y=0$$ and find $A^{\prime \prime}$ and $B^{\prime \prime}$ in terms of $A,B,A^{\prime}$ and $B^{\prime}$.(ii) Show that if $\ell$ is the line $y=(\tan(\theta))x$ and $\ell_*$ is the line $y=(\tan(\theta + \phi))x$ then $\omega_*(\ell)$ is the line $y=(\tan(\theta + 2\phi))x$. It follows that for any two lines $\ell_1$ and $\ell_2$ there exists a line $\ell_3$ such that $\omega_{\ell_3}(\ell_1)=\ell_2$. (iii) Show that for any two lines $L_1$ and $L_2$ $$\omega_{\omega_{L_1}(L_2)}=\omega_{L_1}\circ\omega_{L_2}\circ\omega_{L_1}$$ (iv)Let $\ell$ and $\ell^{\prime}$ be any two lines Let $\ell ^{\prime \prime}$ be a line such that $\ell^{\prime}=\omega_{\ell^{\prime \prime}}(\ell)$. Then $$\omega_{\ell^{\prime}}=\omega_{\ell^{\prime \prime}}\circ\omega_{\ell}\circ \omega_{\ell^{\prime \prime}}$$ $$=\omega_{\ell^{\prime \prime}}^{-1}\circ\omega_{\ell}\circ \omega_{\ell^{\prime \prime}}$$ because a reflection is idempotent.Thus not only are any two reflection matrices similar, we can find a similarizing matrix that is idempotent.