# Show that these matrices are congruent.

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 '14 at 19:50
@PraphullaKoushik No, I mean congruent. –  user142836 Jul 16 '14 at 19:52
What do you mean by congruent? –  Brian Fitzpatrick Jul 16 '14 at 19:52
Have you tried it with $K=\mathbb R$? –  Ragnar Jul 16 '14 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 '14 at 20:05

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 '14 at 23:19
Try $A^T\begin{pmatrix}1&0\\0&-1\\\end{pmatrix}A$ –  Rene Schipperus Jul 16 '14 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 '14 at 20:22
It being obvious to you doesn't make it obvious to the asker. -@ReneSchipperus –  Semiclassical Jul 16 '14 at 20:22
It wasnt obvious to me, I had to work it out. –  Rene Schipperus Jul 16 '14 at 20:24
@ReneSchipperus: I meant the route to it, not the matrix itself. –  Semiclassical Jul 16 '14 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 '14 at 20:00