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U and Uperpendicular

  1. Dec 15, 2009 #1
    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 isnt 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?
     
  2. jcsd
  3. Dec 15, 2009 #2
    You have confused union with direct sum. What you have said is that [tex]\mathbb{R}^3 = U \cup U^\perp[/tex], which is not true, as you have noted. The correct statement is that [tex]\mathbb{R}^3 = U \oplus U^\perp[/tex] Note that the direct sum in this case can be defined as [tex]A \oplus B = \{ a + b | a \in A, b \in B \}[/tex]. Try to show that every element of [tex]\mathbb{R}^3[/tex] may be written as a sum of an element in U and an element in [tex]U^\perp[/tex]. The decomposition for your specific vector is (1,1,2) = (1,0,2) + (0,1,0).
     
  4. Dec 15, 2009 #3
    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 :)
     
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