Is P = P² the definition of a projection operator?

  • Thread starter Thread starter Stephan Hoyer
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the mathematical definition of a projection operator, specifically addressing the equation P = P². Participants confirm that this equation characterizes projection operators, which can be represented by matrices that replace certain elements with zeros. The conclusion drawn is that for any operator P satisfying P² = P, the vector space V can be expressed as the direct sum of the null space of P and the range of P. An example of a non-standard projection operator in R² is also provided, illustrating that not all projections conform to the simple identity matrix form.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly vector spaces and operators.
  • Familiarity with the definition and properties of projection operators.
  • Knowledge of null spaces and ranges of linear transformations.
  • Basic proficiency in matrix representation and manipulation.
NEXT STEPS
  • Study the properties of projection operators in linear algebra.
  • Learn about null spaces and ranges of linear transformations in depth.
  • Explore examples of non-standard projection operators in various dimensions.
  • Investigate the implications of the equation P² = P in different mathematical contexts.
USEFUL FOR

Mathematics students, educators, and professionals in fields requiring linear algebra knowledge, particularly those focused on operator theory and vector space analysis.

Stephan Hoyer
Messages
105
Reaction score
0
Ignore this post (it was irrecoverable due to bad LaTeX markup). See the one below.
 
Last edited:
Physics news on Phys.org
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.
 
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.
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]
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
5K
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
4
Views
4K