1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find P1 and P2 such that P1, P2, and P1+P2 idempotent

  1. Oct 5, 2009 #1
    Hi everyone,

    I am working on the following question:
    I would need to find a non trivial example of two 3by3 matrices P1 and P2 such that
    P1 is idempotent
    P2 is idempotent
    P1+P2 is idempotent

    How I should go about attempting to find a solution?

    First I thought I should solve three equations
    P1^n= P1
    P2^n= P2
    (P1+P2)^n= P1+P2

    But I quickly convince myself that this would not work, and that I should try a trial and error avenue. I would appreciate your ideas.
     
  2. jcsd
  3. Oct 5, 2009 #2
    P1n= P1, P2n= P2 and (P1+P2)n= P1+P2 can be replaced by P12= P1, P22= P2 and (P1+P2)2= P1+P2 because matrix multiplication is associative so you can take the power one multiplication at a time.

    The second set equations are just the definition of a projection matrix applied to P1, P2 and P1 + P2. Do you know what that is?
     
  4. Oct 5, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I really don't think that's right. If P is the 2x2 matrix [[0,1],[1,0]] then P^3=P but P^2 is not equal to P, it's I. On the other hand, I don't think the question really requires you solve for anything. You only want examples of P1 and P2 that satisfy those properties. Just try and fish around for examples, as the OP said, trial and error. Are nonzero diagonal matrices non-trivial enough?
     
  5. Oct 6, 2009 #4
    But that matrix isn't idempotent. Doesn't an idempotent matrix have to satisfy Pn = P for all n?
     
    Last edited: Oct 6, 2009
  6. Oct 6, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, you're right. I was just thinking about the implication 'P^n=P implies P^2=P' which is false. But, yes, idempotent means P^2=P. Which does imply P^n=P. Sorry.
     
  7. Oct 7, 2009 #6
    >>The second set equations are just the definition of a projection matrix applied to P1, P2 >>and P1 + P2. Do you know what that is?

    No. P1, P2 and (P1+P2) are projection matrices?
     
  8. Oct 7, 2009 #7
    Nonzero diagonal matrices are non-trivial. Do you have one?
     
  9. Oct 7, 2009 #8
    cook up a couple of matrices of the that have 1's on some of the diagonals and 0's elsewhere. Play around with those. You should be able to get what you want. Those are projection operators by the by. They take a vector and return the vector that is zero except on the axes that the operator projects onto where it just returns the coordinates. The key to a projection operator is that it is the identity on its image (this allows idempotence).
     
  10. Oct 7, 2009 #9
    Here is the example I found
    A1=
    1,0,0;
    0,0,0;
    1,0,0;

    A2=
    0,0,0;
    0,1,0;
    0,0,0;


    A1+A2=
    1,0,0;
    0,1,0;
    1,0,0;

    All three are idempotent. Is there a way to make those matrices a little less trivial.
     
  11. Oct 7, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Idempotent matrices, by their nature, tend to look somewhat trivial. That's already more non-trivial than I was thinking of. I was thinking of just using the diagonal parts. The zero matrix I would think of as trivial.
     
  12. Oct 7, 2009 #11
    Change basis? The grader might hate you though.
     
  13. Oct 7, 2009 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I agree. About the hating.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to find P1 and P2 such that P1, P2, and P1+P2 idempotent
  1. Basis for P2 (Replies: 1)

Loading...