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I need to prove that $X^T e$ = 0 where $e$ is the residual in multiple linear regression model in matrix algebra?

I need some guidance on how to do it. Is there any good pdf for the proofs for multiple linear regression model for matrix alegbra?

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1 Answer 1

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$$\mathbf X'\mathbf e = \mathbf X'(\mathbf y -\mathbf {\hat y})= \mathbf X'(\mathbf y -\mathbf X\hat \beta) =...$$

ADDENDUM

$$=\mathbf X'\left(\mathbf y -\mathbf X (\mathbf X'\mathbf X)^{-1}\mathbf X' \mathbf y\right) =\mathbf X'\mathbf y -\mathbf X'\mathbf X (\mathbf X'\mathbf X)^{-1}\mathbf X' \mathbf y$$

$$\mathbf X'\mathbf y -\mathbf X' \mathbf y = \mathbf 0$$

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    $\begingroup$ could you kindly finish the proof? $\endgroup$ Commented Oct 31, 2013 at 7:19
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    $\begingroup$ I support @garciaj. Answers are not a place for hints or haze, even transparent. Reserve those for comments, please. $\endgroup$
    – ttnphns
    Commented Oct 31, 2013 at 9:25
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    $\begingroup$ @ttnphns I disagree: "hint-answers" can be the appropriate response when the OP just asks for a hint ("need some guidance", the OP wrote), or when the OP doesn't, but the question is related to homework/self-study. But now that the OP indicated that the hint-answer has covered the OP's needs, I am completing it so that the answer stands on its own. $\endgroup$ Commented Oct 31, 2013 at 9:49
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    $\begingroup$ You would be right if it were a cabinet conversation between you and the OP. But this site is being seen by many frequenters and passers-by who would be bewildered by an interrupted response. Once again: if you want to stop at a hint level, please do it as a comment, not an answer. $\endgroup$
    – ttnphns
    Commented Oct 31, 2013 at 10:11
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    $\begingroup$ @ttnphns Once again: I disagree. This site is both a conversation between the OP and the answerer, and a public place where people come to find ready-made complete Q&A - so we should always try to find ways to service both...which is what I did with the way I structured my answer. And in general, my answers tend to be rather ...over-complete. $\endgroup$ Commented Oct 31, 2013 at 10:21

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