0
$\begingroup$

Think this question is easy but I'm having much trouble to resolve it.

Let $\mathbb{R}(A/B)$ be the field of the racional funcions generated by $\frac{A}{B}$ over $\mathbb{R}$, where $A$ and $B$ are polynomials in $\mathbb{R}$. Then for all $\alpha,$ $\beta$, $\gamma$, $\delta$, satisfying $ \alpha \delta − \beta \gamma \neq 0$, we have $\mathbb{R}(A/B)$ $=$ $\mathbb{R}\left(\frac{\alpha A+\beta B}{\gamma A + \delta B}\right)$

Can someone help me or at least give a hint?

$\endgroup$
1
$\begingroup$

Hint: Write $A/B$ in the form \begin{bmatrix}A\\B\end{bmatrix} and now you have $$\begin{bmatrix}\alpha&\beta\\ \gamma&\delta\end{bmatrix}\begin{bmatrix}A\\B\end{bmatrix}=\begin{bmatrix}\alpha A+\beta B\\ \gamma A+\delta B\end{bmatrix}$$

If only the square matrix on the left were invertible, we could have so much fun..

$\endgroup$
  • $\begingroup$ Nice hint, but how I transform the racional funtion $\frac{A}{B}$ in the vector $\left[\begin{array}{l}A\\ B \end{array}\right]$? $\endgroup$ – Matheus Manzatto Jul 14 '17 at 2:06
  • $\begingroup$ We are just using the matrix notation for convenience. What is really going on behind the scenes is that we are solving a linear equation. Once you invert the matrix, you will have a recipe for $A/B$ in terms of $A'/B'$ where $A'=\alpha A+\beta B$ and $B'=\gamma A +\delta B $. $\endgroup$ – Ravi Jul 14 '17 at 2:09
  • $\begingroup$ ooooh! now I undertand, very clever, thx bro $\endgroup$ – Matheus Manzatto Jul 14 '17 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.