aplicacions lineals

f(x,y) = ax+by


ker(f):
f(x,y)=0
ax+by=0
<x,y>=k·<(-b),a>=(-k)·<b,(-a)>=ak·<(-b)/a,1>
<x,y>=k·<(-b)/a,1>
<x,y>=k·<b,(-a)>=(-k)·<(-b),a>=bk·<1,(-a)/b>
<x,y>=k·<1,(-a)/b>


Im(f):
f(x,y)=v
ax+by=v
<x,y>=(v/2)·<(1/a),(1/b)>


f(x,y,z) = ax+by+cz


ker(f):
f(x,y,z)=0
ax+by+cz=0
<x,y,z>=k·<(-c),(-c),(a+b)>=(-k)·<c,c,(-1)(a+b)>=(-c)k·<(1,0,(-a)/c>+(-c)k·<(0,1,(-b)/c>
si ( x=1 & y=0 ) ==> <x,y,z>=k·<(1,0,(-a)/c>
si ( x=0 & y=1 ) ==> <x,y,z>=k·<(0,1,(-b)/c>
<x,y,z>=k·<(-b),(a+c),(-b)>=(-k)·<b,(-1)(a+c),b>=(-b)k·<(1,(-a)/b,0>+(-b)k·<(0,(-c)/b,1>
si ( x=1 & z=0 ) ==> <x,y,z>=k·<(1,(-a)/b,0>
si ( x=0 & z=1 ) ==> <x,y,z>=k·<(0,(-c)/b,1>
<x,y,z>=k·<(b+c),(-a),(-a)>=(-k)·<(-1)(b+c),a,a>=(-a)k·<((-c)/a),0,1>+(-a)k·<((-b)/a),1,0>
si ( y=0 & z=1 ) ==> <x,y,z>=k·<((-c)/a),0,1>
si ( y=1 & z=0 ) ==> <x,y,z>=k·<((-b)/a),1,0>


k·<(1,(-a)/b,0>=k·<(1,0,(-a)/c>+((-a)/b)k·<(0,1,(-b)/c>
k·<(0,(-c)/b,1>=0·k·<(1,0,(-a)/c>+((-c)/b)k·<(0,1,(-b)/c>
k·<((-c)/a),0,1>=((-c)/a)·k·<(1,0,(-a)/c>+0·k·<(0,1,(-b)/c>
k·<((-b)/a),1,0>=((-b)/a)·k·<(1,0,(-a)/c>+k·<(0,1,(-b)/c>


k·<(1,0,(-a)/c>=k·<(1,(-a)/b,0>+((-a)/c)k·<(0,(-c)/b,1>
k·<(0,1,(-b)/c>=0·k·<(1,(-a)/b,0>+((-b)/c)k·<(0,(-c)/b,1>
k·<((-c)/a),0,1>=((-c)/a)k·<(1,(-a)/b,0>+k·<(0,(-c)/b,1>
k·<((-b)/a),1,0>=((-b)/a)k·<(1,(-a)/b,0>+0·k·<(0,(-c)/b,1>


k·<(1,0,(-a)/c>=((-a)/c)k·<((-c)/a,0,1>+0·k·<((-b)/a,1,0>
k·<(0,1,(-b)/c>=((-b)/c)k·<((-c)/a,0,1>+k·<((-b)/a,1,0>
k·<(1,(-a)/b,0>=0·k·<((-c)/a,0,1>+((-a)/b)k·<((-b)/a,1,0>
k·<(0,(-c)/b),1>=k·<((-c)/a,0,1>+((-c)/b)k·<((-b)/a,1,0>






dim( ker(f) )=2


Im(f):
f(x,y,z)=v
ax+by+cz=v
<x,y,z>=(v/3)·<(1/a),(1/b),(1/c)>