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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?

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congruent? I think you mean similar... ? –  Praphulla Koushik Jul 16 at 19:50
@PraphullaKoushik No, I mean congruent. –  user142836 Jul 16 at 19:52
What do you mean by congruent? –  Brian Fitzpatrick Jul 16 at 19:52
Have you tried it with $K=\mathbb R$? –  Ragnar Jul 16 at 19:59
Hint: what you're after is the square root of $u$. You should consider the characteristic $\infty$ and the characteristic finite, $\not =2$ separately.. –  Matt B. Jul 16 at 20:05

3 Answers 3

up vote 6 down vote accepted

Use the matrix

$$A=\begin{pmatrix} \frac{u+1}{2}& \frac{u-1}{2}\\ \frac{u-1}{2}& \frac{u+1}{2}\\ \end{pmatrix}$$

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$$\begin{bmatrix} \frac{u + 1}2 & \frac{u - 1}2 \\ \frac{u - 1}2 & \frac{u + 1}2\end{bmatrix}^T \begin{bmatrix} u & 0 \\ 0 & -u \end{bmatrix} \begin{bmatrix} \frac{u + 1}2 & \frac{u - 1}2 \\ \frac{u - 1}2 & \frac{u + 1}2\end{bmatrix} = \begin{bmatrix} u^2 & 0 \\ 0 & -u^2\end{bmatrix} \ne \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$$ –  DanielV Jul 16 at 23:19
Try $A^T\begin{pmatrix}1&0\\0&-1\\\end{pmatrix}A$ –  Rene Schipperus Jul 16 at 23:22

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?

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well how do you think I found it in the first place ? –  Rene Schipperus Jul 16 at 20:22
It being obvious to you doesn't make it obvious to the asker. -@ReneSchipperus –  Semiclassical Jul 16 at 20:22
It wasnt obvious to me, I had to work it out. –  Rene Schipperus Jul 16 at 20:24
@ReneSchipperus: I meant the route to it, not the matrix itself. –  Semiclassical Jul 16 at 20:25

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 $$P^T\begin{pmatrix}1&0\\0&-1\end{pmatrix} P= u^{1/2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}u^{1/2}= \begin{pmatrix}u&0\\0&-u\end{pmatrix}$$

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Why does $u^{1/2}$ exist? Oh I overlooked the "positive" ... the question is about a general field. –  martini Jul 16 at 20:00

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