Proving R3=U+V: Solving a+x=b,2a+y=c,3a+z=d

[franky2727]

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?

 

[tiny-tim]

prove R³=Udirect sum V

u={(a,2a,3a)} aER
v={(x,y,z)} x,y,z ER x+y+z=0
…
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?

Hi franky2727!

I think they mean that, to prove that a general point (b,c,d) in R³ is in U + V, you have to find an (a,2a,3a) in U and an (x,y,z) in V which add to (b,c,d)

 

[Mark44]

It might be helpful to think about the geometry here. The set {a, 2a, 3a} is a line in R3, while the equation x + y + z = 0 is a plane that passes through the origin, and whose normal is the vector (1, 1, 1).

What you're doing is showing that any arbitrary vector in R3 can be written as the sum of a vector along the line plus a vector that extends from the origin out to the plane somewhere.

 

[franky2727]

you have to find an (a,2a,3a) in U and an (x,y,z) in V which add to (b,c,d)

well isn't that just a+x=b 2a+y=c 3a+z=d

 

[tiny-tim]

you have to find an (a,2a,3a) in U and an (x,y,z) in V which add to (b,c,d)

well isn't that just a+x=b 2a+y=c 3a+z=d

Yup! ​

… but now you know why!