I want to prove that every matrix in $M_{2\times 2}(\mathbb{R})$ whose characteristic polynomial does not split is similar to a positive multiple of a rotation matrix, $$ s\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}. $$
I know that two matrices $A$ and $B$ are similar if there's an (invertible) matrix $Q$ such that $B = Q^{-1}AQ$. I think my issue is that I don't know how to operationalize the fact that the polynomial does not split, and furthermore how this is connected to the rotation matrix.
Any tips on how to proceed?
One proof of this is as follows. First, note that every matrix of the form $$ M = \pmatrix{a&-b\\b&a} $$ (where $a,b$ are not both zero) is a positive multiple of some rotation matrix. Now, the characteristic polynomial of $A$ can be written in the form $$ p(x) = (x-a)^2 + b^2 $$ for some constants $a,b$ with $b$ non-zero. Let $B$ denote the matrix $B = A - aI$. The characteristic polynomial of $B$ must be of the form $q(x) = x^2 + b^2$.
Let $x \in \Bbb R^2$ be any non-zero vector. We note that $Bx = \lambda x$ cannot hold for any $\lambda \in \Bbb R$; otherwise, $(x-\lambda)$ would be factor of the characteristic polynomial $p(x)$. Thus, $x$ and $Bx$ are linearly independent. Consider the basis of $\Bbb R^2$ given by $$ \mathcal B = \{v_1,v_2\}, \qquad v_1 = x, \quad v_2 = \frac 1{b} Bx. $$ By the Cayley Hamilton theorem, we must have $q(B) = 0$, and hence $q(B)x = 0$, so that $$ (B^2 + b^2 I)x = 0 \implies B \left(\frac 1b x\right) = -bx. $$ Thus, we have $Bv_1 = bv_2$ and $Bv_2 = -bv_1$. Conclude that the matrix of $B$ relative to the basis $\mathcal B$ is given by $$ [B]_{\mathcal B} = \pmatrix{0&-b\\b&0}. $$ Conclude that the matrix of $A = aI + B$ is given by $$ [A]_{\mathcal B} = [aI + B]_{\mathcal B} = aI + [B]_{\mathcal B} = \pmatrix{a&-b\\b&a}. $$ Equivalently, if $Q$ is the matrix whose columns are $v_1,v_2$, then $$ Q^{-1}AQ = \pmatrix{a&-b\\b&a}, $$ which was what we wanted.