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Linear Algebra

  1. Nov 1, 2005 #1
    Ignore this post (it was irrecoverable due to bad LaTeX markup). See the one below.
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2
    Here's the question I'm stuck in on my latest problem set:

    Suppose P is an operator on V and [itex]P^2=P[/itex]. Prove that [itex]V = null P \oplus range P[/itex].

    I'm pretty sure that P is simply the projection operator (consisting of the identity matrix replacing some 1's with 0's), in which case the conclusion follows easilly. But I've been looking at this one for a while and I can't see how I can prove that P must be the projection operator.

    Thanks for your help.
  4. Nov 2, 2005 #3


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    If I recall correctly, P² = P is the definition of a projection operator!

    Not all projections have the form you gave: for example, consider in R² the orthogonal projection onto the line y = x given by [x, y] -> (1/2)[x+y, x+y]
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