I am working through a multi-part proof of how orthogonal projection matrices give specific results from their properties. I read through the Gauss-Markov model theory to get a start. This is only a part of the proof, but if I can get help on this part, then I will be better equipped to try and tackle the rest.

Part (a) Prove $X'XA = X'XB$ iff $XA = XB$

Part (b) Use result of (a) to prove $X(X'X)^-X'X = X$ for any generalized inverse of $X'X$.

Part (c) Prove if $A$ is symmetric and $G$ is a generalized inverse of $A$, then it must be true that $G'$ is also a generalized inverse of $A$.

Part (d) Use these to show $X'X(X'X)^-X' = X'.$

(There's a lot more, but this is the stuff I'm stuck on, and I honestly am struggling with how these matrices prove these types of properties.)

Introductory (a) Proof

$XA=XB$ therefore $X'XA = X'XB$ holds only trivially

Must prove that $X'XA = X'XB$ therefore $XA =XB.$

Rewrite $X'XA=X'XB$ as $X'XA-X'XB = 0.$

Factor $X'X(A-B) = 0$

Multiply both sides $(A'-B') X'X(A-B) = (A'-B') 0$


I know I should Somehow show $A'A = 0$, then use this.

  • $\begingroup$ Hi, welcome to CV! I took the liberty of editing your post to make it more readable, and perhaps get an answer. Let me know if I made any mistakes. Also, I find these types of questions more likely to be answered in SE Mathematics. $\endgroup$ – Antoni Parellada Jan 17 '16 at 17:00
  • $\begingroup$ In part (a) can we assume that $\mathbf{X}$ has full column rank as in the Gauss-Markov conditions? $\endgroup$ – JohnK Jan 17 '16 at 21:02
  • $\begingroup$ I hadn't thought of this (shows my ignorance) but yes, I can state that as an assumption. $\endgroup$ – Jennifer Cooke Jan 17 '16 at 21:08
  • $\begingroup$ @Antoni and JohnK - Is there a collection you know of, for proofs of different uses of the GMM, like...when having to prove the estimate for mean, having to prove the estimate for variance, having to prove...etc.? $\endgroup$ – Jennifer Cooke Jan 19 '16 at 20:02

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