• anonymous
For a projection matrix P, how can I prove that N(P)=R(I-P) and also R(P)=N(I-P) ?
MIT 18.06 Linear Algebra, Spring 2010
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
For this type of problem it might be best to think in wide angle. We are thinking of matrices - and in given instances we are asked to create them ie. produce the entries to place in each row and column - but there is an 'operational' concept also. So one can think of the identity matrix as the one which does 'nothing', as when applied to some vector it returns the same vector. And that is true for any vector it operates on. However the identity also projects any vector onto itself! :-) However by projection matrix we ( usually ) intend to project any given vector onto/along some specific direction of interest. You already know how to do that along some 'special' directions - the coordinate axes - simply by plucking out one of the vector's components as expressed in that coordinate system. Or if you like take the dot/inner product of the vector of interest with a specific unit vector for some coordinate axis. For a general projection operation ( or operator ) you apply said operation to a vector to 'resolve' or establish it in terms of two other vectors - the sum of which gives the original vector : |dw:1359584340887:dw| then from that operational definition it is clear that once you have resolved to some direction, then resolving that component yet again gives that component. Thus \[R = RP^{n}\]for any number of projection operations n. Specifically : \[R=RP\]thus\[R-RP=0\]and \[R(I-P)=0\]now because the normal vector is exactly that ( perpendicular ) then : \[NP = 0\]Now your second equation doesn't look right. Do you mean \[R(N)=P(I-P)\]

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