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When working on a problem, I encountered the following statement.

Let $x,y,t \in \mathbb R$

$\left( \begin{smallmatrix} x & y \\ y & t \\ \end{smallmatrix} \right)$ is orthogonally similar to $\left( \begin{smallmatrix} a & b \\ -b & a \\ \end{smallmatrix} \right)$ for some $a,b \in \mathbb R$

In order to prove it, I considered multiplying by conjugate rotation matrices with unknown angle $h$

$$ \begin{pmatrix} \cos (h) & \sin (h) \\ -\sin (h) & \cos (h) \\ \end{pmatrix}\cdot\begin{pmatrix} x & y \\ y & t \\ \end{pmatrix}\cdot\begin{pmatrix} \cos (h) & -\sin (h) \\ \sin (h) & \cos (h) \\ \end{pmatrix} =\begin{pmatrix} \scriptstyle{ \sin (h) (y \cos (h)+t \sin (h))+\cos (h) (x \cos (h)+y \sin (h))} & \scriptstyle{\cos (h) (y \cos (h)+t \sin (h))-\sin (h) (x \cos (h)+y \sin (h))} \\ \scriptstyle{\cos (h) (y \cos (h)-x \sin (h))+\sin (h) (t \cos (h)-y \sin (h))} & \scriptstyle{\cos (h) (t \cos (h)-y \sin (h))-\sin (h) (y \cos (h)-x \sin (h))} \end{pmatrix}$$ But finding a suitable $h$ is actually not trivial ...

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This is false. Take $x=t=0$, $y\ne0$. The eigenvalues of the first matrix are real $(\pm y)$. The eigenvalues of the second matrix are $\lambda_1=\lambda_2=a$ if $b=0$ and they are not real if $b\ne0$. Thus, these two matrices cannot be similar, because similar matrices have the same eigenvalues.

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