Question 1062030
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Let P be a projection, so  {{{P^2 = P}}}.  If {{{c<>1}}}, compute {{{(I-cP)^(-1)}}}.
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I can do it for &nbsp;{{{abs(c)}}} < 1.  &nbsp;Again, &nbsp;let us assume that &nbsp;|c| < 1.



<pre>
Then  {{{(I-cP)^(-1)}}} = {{{1 + cP + (cP)^2 + (cP)^3 + ellipsis}}} = (since {{{P^2 = P }}}) = {{{1 + cP + c^2*P + c^3*P + ellipsis}}} = 


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


      = {{{1 + c*(1 + c + c^2 + c^3 + ellipsis)*P)}}} = {{{1 + c*(1/(1-c))*P}}} = 


      = {{{1 + (c/(1-c))*P}}} = {{{((1-c + c)/(1-c))*P}}} = {{{(1/(1-c))*P}}}.
</pre>

<U>Answer</U>.  {{{(I-cP)^(-1)}}} = {{{(1/(1-c))*P}}}.



You may check it by making direct calculations that &nbsp;&nbsp;{{{(I - cP)}}} * {{{(1/(1-c))*P}}} = I  &nbsp;&nbsp;under the condition &nbsp;&nbsp;{{{P^2}}} = {{{P}}}.



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