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?


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..

  • $\begingroup$ 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

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