# Is this acceptable

1. Jan 7, 2009

### 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?

2. Jan 7, 2009

### tiny-tim

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)

3. Jan 7, 2009

### Staff: Mentor

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.

4. Jan 8, 2009

### 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 isnt that just a+x=b 2a+y=c 3a+z=d

5. Jan 8, 2009

### tiny-tim

Yup!

… but now you know why!

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