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I am trying to show that the matrix $\mathbf{(H-\frac{1}{n}J_n)}$ is idempotent where $\mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $n\times n$ matrix with $1$ in all its inputs. Taking :

$$\mathbf{(H-\frac{1}{n}J_n)(H-\frac{1}{n}J_n)= HH - H\frac{1}{n}J_n - \frac{1}{n}J_nH + \frac{1}{n}J_n\frac{1}{n}J_n}$$

Now, we know that $\mathbf{H}$ and $\mathbf{\frac{1}{n}J_n}$ are idempontent, thus :

$$\mathbf{(H-\frac{1}{n}J_n)(H-\frac{1}{n}J_n)=H-H\frac{1}{n}J_n - \frac{1}{n}J_nH +\frac{1}{n}J_n}$$

How would I continue now in order to show that the given matrix is idempontent ?

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  • $\begingroup$ The "hat matrix" should be redefined, It is not a usual name... $\endgroup$ – Jean Marie Nov 22 '18 at 19:08
  • $\begingroup$ @JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix $\endgroup$ – Rebellos Nov 22 '18 at 19:12
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    $\begingroup$ @JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $\hat y$, that is, it "puts a hat on $y$", figuratively. $\endgroup$ – Jean-Claude Arbaut Nov 22 '18 at 19:15
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    $\begingroup$ @Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP. $\endgroup$ – Jean Marie Nov 22 '18 at 19:19
  • $\begingroup$ Looks to me $H$ has to be the identity matrix for $H-\frac{1}{n}J_n$ to be idempotent. $\endgroup$ – StubbornAtom Nov 22 '18 at 19:26

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