R3: U & Uperpendicular Vector Logic

[sjeddie]

Let's say we are in R3, and U is the x,z plane, i think then all of Uperpendicular should be some translation of the span of the y axis. Now, since that U and Uperpendicular together form R3, then isn't it true that all vectors in R3 should be contained in either U or Uperpendicular? But given a vector, say (1,1,2), it is in neither U nor Uperpendicular. What's wrong with my logic?

 

[rochfor1]

You have confused union with direct sum. What you have said is that \mathbb{R}^3 = U \cup U^\perp, which is not true, as you have noted. The correct statement is that \mathbb{R}^3 = U \oplus U^\perp Note that the direct sum in this case can be defined as A \oplus B = \{ a + b | a \in A, b \in B \}. Try to show that every element of \mathbb{R}^3 may be written as a sum of an element in U and an element in U^\perp. The decomposition for your specific vector is (1,1,2) = (1,0,2) + (0,1,0).

 

[sjeddie]

I get it. (x,y,z)=(a,0,c) + (0,b,0) where (a,0,c) is in U and (0,b,0) is in Uperp. I didn't know what direct sum is, now I do! Thanks a lot rochfor1, you're awesome :)