Is the matrix $A = I - \frac{J}{n+1}$ idempotent? I am supposed to figure out if the following statement is false or true.

If $I$ is the $n \times n$ identity matrix, and $J$ is an  $n \times n$ matrix consisting entirely of ones, then the matrix $$A = I - \frac{J}{n+1}$$ is idempotent (i.e., $A^{2} = A$).


I understand obviously what $I$ and $J$ are, my issue is with the $A = I - \frac{J}{n+1}$. I searched my textbook and found no reference to it. What does it mean?
 A: The expression $A = I - \frac{J}{n+1}$ is matrix addition, where $\frac{J}{n+1}$ represents scalar multiplication of the matrix $J$ by $\frac{1}{n+1}$. Since the matrix $J$ consists completely of $1$s, each entry of the scaled matrix $\frac{J}{n+1}$ is $\frac{1}{n+1}$. Here is the matrix addition:
$$ I - \frac{J}{n+1} =
\left[\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{array}\right]
-
\left[\begin{array}{cccc}
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\end{array}\right]$$
$$ \implies A = \left[\begin{array}{cccc}
\frac{n}{n+1} & \frac{-1}{n+1} & \cdots & \frac{-1}{n+1} \\
\frac{-1}{n+1} & \frac{n}{n+1} & \cdots & \frac{-1}{n+1} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{-1}{n+1} & \frac{-1}{n+1} & \cdots & \frac{n}{n+1} \\
\end{array}\right] $$
An interesting matrix. You could describe it by $A_{ij}=\frac{n}{n+1}$ if $i=j$ and $A_{ij}=\frac{-1}{n+1}$ if $i\neq j$. 
A: It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:
$$A^2=(I-\frac{1}{n+1}J)^2=I^2-\frac{1}{n+1}IJ-\frac{1}{n+1}JI+\frac{1}{(n+1)^2}J^2 = \\
= I - \frac{2}{n+1}J + \frac{n}{(n+1)^2}J = I + \frac{n-2(n+1)}{(n+1)^2}J = \\
= I + \frac{-n-2}{(n+1)^2}J = I - \frac{1}{n+1}J - \frac{1}{(n+1)^2}J = A - \frac{1}{(n+1)^2}J \neq A$$
As you can see, $A$ is not idempotent.
