SOLUTION: Let P be a projection, so {{{P^2 = P}}}. If {{{c<>1}}}, compute {{{(I-cP)^(-1)}}}.

Algebra ->  Vectors -> SOLUTION: Let P be a projection, so {{{P^2 = P}}}. If {{{c<>1}}}, compute {{{(I-cP)^(-1)}}}.      Log On


   

Question 1062030: Let P be a projection, so P%5E2+=+P. If c%3C%3E1, compute %28I-cP%29%5E%28-1%29.
Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
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Let P be a projection, so P%5E2+=+P. If c%3C%3E1, compute %28I-cP%29%5E%28-1%29.
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I can do it for  abs%28c%29 < 1.  Again,  let us assume that  |c| < 1. 


Then  %28I-cP%29%5E%28-1%29 = 1+%2B+cP+%2B+%28cP%29%5E2+%2B+%28cP%29%5E3+%2B+ellipsis = (since P%5E2+=+P+) = 1+%2B+cP+%2B+c%5E2%2AP+%2B+c%5E3%2AP+%2B+ellipsis = 


      ( apply the formula for the sum of an infinite geometric progression ) 


      = 1+%2B+c%2A%281+%2B+c+%2B+c%5E2+%2B+c%5E3+%2B+ellipsis%29%2AP%29 = 1+%2B+c%2A%281%2F%281-c%29%29%2AP = 


      = 1+%2B+%28c%2F%281-c%29%29%2AP = %28%281-c+%2B+c%29%2F%281-c%29%29%2AP = %281%2F%281-c%29%29%2AP.

Answer. %28I-cP%29%5E%28-1%29 = %281%2F%281-c%29%29%2AP.


You may check it by making direct calculations that   %28I+-+cP%29 * %281%2F%281-c%29%29%2AP = I   under the condition   P%5E2 = P.


And after checking it directly  (as I said)  you may to extend the formula for  |c| > 1.  It is valid for  |c| > 1,  too.