aplicacions lineals

f(x,y) = <ax+by,cx+dy>


ker(f):
ax+by=0
cx+dy=0


x+(b/a)y=0
(c/d)x+y=0


x+(b/a)y=0
0·x+(1+(-1)(c/d)(b/a))y=0


Si ( a=b & c=d )  ==> x+y=0 ==> ( ( x=1 & y=(-1) ) or ( x=(-1) & y=1 ) )
Si ( ad=cb )  ==> x+(b/a)y=0 ==> ( ( x=1 & y=( (-a)/b ) ) or ( x=( (-b)/a) ) &  y=1 )
Si ( ad!=cb )  ==> ( x=0 & y=0 )
( ad=cb ) <==> ((-d)/c) = ((-b)/a)
Im(f):
ax+by=1
cx+dy=1
si ad!=cb ==> ( y=(a+(-c))/(ad+(-1)cb) & x=(d+(-b))/(ad+(-1)cb) )
ax+by=n
cx+dy=n
si ad!=cb ==> ( y=n·(a+(-c))/(ad+(-1)cb) & x=n·(d+(-b))/(ad+(-1)cb) )
ax+by=m
cx+dy=n
si ad!=cb ==> ( y=(an+(-c)m)/(ad+(-1)cb) & x=(dm+(-b)n)/(ad+(-1)cb) )


f(x,y) = <ax+by,cx+dy,ux+vy>


ker(f):
ax+by=0
cx+dy=0
ux+vy=0


x+(b/a)y=0
(c/d)x+y=0
(u/v)x+y=0


x+(b/a)y=0
0·x+(1+(-1)(c/d)(b/a))y=0
0·x+(1+(-1)(u/v)(b/a))y=0


Si ( a=b & c=d & u=v )  ==> x+y=0 ==> ( ( x=1 & y=(-1) ) or ( x=(-1) & y=1 ) )
Si ( ad=cb & av=ub)  ==> x+(b/a)y=0 ==> ( ( x=1 & y=( (-a)/b ) ) or ( x=( (-b)/a) ) &  y=1 )
Si ( ad!=cb or av!=ub )  ==> ( x=0 & y=0 )
( ad=cb & av=ub) <==> ( ((-d)/c) = ((-b)/a) & ((-v)/u) = ((-b)/a) )


Im(f):
ax+by=1+(-1)( (ud+(-1)cv)/( (ad+(-1)bc)+(ud+(-1)cv)+(bu+(-1)av) ) )
cx+dy=1+(-1)( (bu+(-1)av)/( (ad+(-1)bc)+(ud+(-1)cv)+(bu+(-1)av) ) )
ux+vy=1+(-1)( (ad+(-1)bc)/( (ad+(-1)bc)+(ud+(-1)cv)+(bu+(-1)av) ) )
si ( ad!=cb or ud!=cv or av!=bu ) ==>...
... ( y=(a+(-c)+u)/( (ad+(-1)bc)+(ud+(-1)cv)+(bu+(-1)av) ) &...
.... x=((-b)+d+(-v))/( (ad+(-1)bc)+(ud+(-1)cv)+(bu+(-1)av) ) )