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Next: Άσκηση 54 Up: Άσκηση 53 Previous: Υπόδειξη

Λύση

Έστω $x_1+y_1i, x_2+y_2i\in \mathbb C$ και $c\in \mathbb R$ τότε

\begin{displaymath}F((x_1+y_1i)+(x_2+y_2i))=F((x_1+x_2)+(y_1+y_2)i)=
\left(\beg...
...
x_1+x_2 & y_1+y_2\\
-y_1-y_2 & x_1+x_2
\end{array}\right)=\end{displaymath}


\begin{displaymath}\left(\begin{array}{cc}
x_1 & y_1\\
-y_1 & x_1
\end{array...
...y_2\\
-y_2 & x_2
\end{array}\right)=F(x_1+y_1i)+F(x_2+y_2i).\end{displaymath}

Επίσης,

\begin{displaymath}F(c(x_1+y_1i))=F(cx_1+cy_1i)
\left(\begin{array}{cc}
cx_1 & cy_1\\
-cy_1 & cx_1
\end{array}\right)=\end{displaymath}


\begin{displaymath}c\left(\begin{array}{cc}
x_1 & y_1\\
-y_1 & x_1
\end{array}\right)=cF(x_1+y_1i).\end{displaymath}

Εύκολα βλέπουμε ότι ${\rm Ker}F={\bf0}$ και άρα η $F$ είναι 1-1. Τέλος,

\begin{displaymath}F(z_1z_2)=F((x_1+y_1i)(x_2+y_2i))=F((x_1x_2-y_1y_2)+(x_1y_2+x_2y_1)i)=\end{displaymath}


\begin{displaymath}\left(\begin{array}{cc} x_1x_2-y_1y_2 & x_1y_2+x_2y_1\\
-x_...
...begin{array}{cc}
x_2 & y_2\\
-y_2 & x_2
\end{array}\right)=\end{displaymath}


\begin{displaymath}F(x_1+y_1i)F(x_2+y_2i)=F(z_1)F(z_2).\end{displaymath}



Vassilis Metaftsis
1999-09-15