Idempotency of difference of two idempotent matrices 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.
 A: In view of the comments I'll construe the word "design" in the phrase "design matrix" to mean that one of the columns of $\mathbf{X}$ is a column of $1$s, although that usage is not universal.
Then the column space of $\mathbf{J}$ is a subspace of the column space of $\mathbf{X}$.  The matrix $\mathbf{H}$ represents the projection onto the column space of $\mathbf{X}$.  If $\mathbf{Y}\in\mathbb{R}^{n\times 1}$ then $\mathbf{HY}$ is the projection of $\mathbf{Y}$ onto the column space of $\mathbf{X}$, and $\mathbf{\overline{J}Y}$ is the projection of $\mathbf{Y}$ onto the column of $1$s.
Since the column space of $\mathbf{J}$ is a subspace of that of $\mathbf{X}$, the column space of $\mathbf{\overline{J}}$ is also a subspace of that of $\mathbf{X}$, and we have $\mathbf{H\overline{J}}=\mathbf{\overline{J}}$, i.e. the projection of a column of $\mathbf{\overline{J}}$ onto the column space of $\mathbf{X}$ is itself.  Since $\mathbf{H}$ and $\mathbf{\overline{J}}$ are symmetric, we have
$$
\mathbf{H\overline{J}}=\mathbf{\overline{J}}\text{ and } \mathbf{\overline{J}H}=\mathbf{\overline{J}}.
$$
Consequently you get an affirmative answer.
