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Define

$$ \mathbf{H}=\mathbf{X}\left(\mathbf{X}^{\prime}\mathbf{X}\right)^{-1}\mathbf{X}^{\prime} $$

where $\mathbf{X}$ is a design matrix of order $n \times k$

and

$$ \overline{\mathbf{J}}=\frac{1}{n}\mathbf{J}=\frac{1}{n}\mathbf{1}\mathbf{1}^{\prime} $$

where $\mathbf{1}$ is a unit vector of order $n \times 1$.

Now $$ \mathbf{H}\mathbf{H}=\mathbf{H} $$

and

$$ \overline{\mathbf{J}}\overline{\mathbf{J}}=\overline{\mathbf{J}} $$

Thus both $\mathbf{H}$ and $\overline{\mathbf{J}}$ are idempotent matrices.

My question is whether $\mathbf{H}-\overline{\mathbf{J}}$ would be idempotent. If so then

$$ \left(\mathbf{H}-\mathbf{\overline{\mathbf{J}}}\right)\mathbf{\left(\mathbf{H}-\mathbf{\overline{\mathbf{J}}}\right)}=\mathbf{H}-\mathbf{H\overline{\mathbf{J}}-\overline{\mathbf{J}}H}+\overline{\mathbf{J}}=\mathbf{H}-\mathbf{\overline{\mathbf{J}}-\overline{\mathbf{J}}}+\overline{\mathbf{J}}=\mathbf{H}-\overline{\mathbf{J}} $$

But I'm not able to show that

$$ \mathbf{H\overline{\mathbf{J}}=\overline{\mathbf{J}}H}=\overline{\mathbf{J}} $$

I'd highly appreciate if you guide me to figure this. Thanks for your time and help.

Edited

$\mathbf{X}$ is a design matrix, has a widespread use in Statistics, see here.

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    What you need is ${\bf H\bar J}+{\bf \bar JH}=2{\bf\bar J}$, but I think it is rarely going to hold, and much depends on ${\bf X}$.2012-11-12
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    Might it have been intended that one of the columns of $\mathbf{X}$ is a column of $1$s?2012-11-12
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    @MichaelHardy: You right, the first column of $\mathbf{X}$ is a a column of $\mathbf{1}$s.2012-11-12

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