Let $K$ be a field of characteristic$\ne 2$ and $u$ be an invertible element of $K$.
Show that $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $\begin{pmatrix}u&0\\0&-u\end{pmatrix}$ are congruent.
I tried to find a suitable transition matrix but I'm stuck, any ideas please?
On
If $u^{1/2}$ exists (e.g. $u \in \mathbb{R}$) and $(u^{1/2})^2=u$ then you can take $P = u^{1/2} I_2$, so that $$P^T\begin{pmatrix}1&0\\0&-1\end{pmatrix} P= u^{1/2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}u^{1/2}= \begin{pmatrix}u&0\\0&-u\end{pmatrix}$$
On
As a path towards Rene's answer: Pick some matrix $A$ with unknown coefficients and substitute this into the conjugacy definition. This will give you three constraints, from which you can persuade yourself that it's enough to find the two upper elements. To solve the last equation left, what's a factorization of $u$ that always exists?
Use the matrix
$$A=\begin{pmatrix} \frac{u+1}{2}& \frac{u-1}{2}\\ \frac{u-1}{2}& \frac{u+1}{2}\\ \end{pmatrix}$$