- #1
franky2727
- 132
- 0
prove R3=Udirect sum V
u={(a,2a,3a)} aER
v={(x,y,z)} x,y,z ER x+y+z=0
i solved the first bit UnV=0 but I'm having problems with the R3=U+W bit, my notes say i need to be able to "solve" a+x=b,,,2a+y=c,,,3a+z=d but what does this mean? is this simply putting the b,c,d all equal to one of the a,x,y or z?
i have done 6a=b+c+d is this adequate?
u={(a,2a,3a)} aER
v={(x,y,z)} x,y,z ER x+y+z=0
i solved the first bit UnV=0 but I'm having problems with the R3=U+W bit, my notes say i need to be able to "solve" a+x=b,,,2a+y=c,,,3a+z=d but what does this mean? is this simply putting the b,c,d all equal to one of the a,x,y or z?
i have done 6a=b+c+d is this adequate?