$A=I-\frac{1}{n}J$ where $I$ is an $n\times n $ identity matrix, $J$ is an $n\times n $ matrix with all entries as $1$ Consider $$A=I-\frac{1}{n}J$$ where $I$ is an $n\times n $ idenity matrix, $J$ is an $n\times n $ matrix with all entries as $1$.
Which of the following is NOT true?


*

*$A^k=A$ for every positive integer.

*Trace of $A$ is $n-1$

*$\text{Rank $(A)$+Rank $(I-A)$}=n$

*$A$ is invertible.


My first concern was to conclude fourth option is false and it looks more natural for me to conclude fourth is false.
Reason is  $I-\frac{1}{n}J$ is invertible iff $\frac{1}{n}J$ is nilpotent i.e., $J$ is nilpotent and I am sure just by looking at it that $J$ would never be nilpotent. 
So, $A$ would never be invertible.
Trace of $A$ would be just $\text{ Trace of $A$ - $\frac{1}{n}$Trace of $J$}=n-\frac{1}{n}(n)=n-1$
I consider $A^2=(I-\frac{1}{n}J)(I-\frac{1}{n}J)=I-\frac{1}{n}J-\frac{1}{n}J+\frac{1}{n^2}J^2$
I see $-\frac{1}{n}J+\frac{1}{n^2}J^2=0$ but I am not sure if this implies $A^k=A$ for all $k$.
I can not say anything about third option.
Could some one help me to clear this.
Thank you.
 A: Notice that 
$$J^2=nJ$$
hence
$$A^2=\left(I-\frac 1 nJ\right)^2=I-\frac2nJ+\frac{1}{ n^2}J^2=A$$
hence we can answer the first point. The second point is also easy since $\mathrm{tr}$ is linear.
Now since the polynomial  $x^2-x=x(x-1)$ annihilates $A$ and $A\ne \alpha I_n$ then $0$ and $1$ are eigenvalues of $A$ and then $A$ isn't invertible, but since $A$ is diagonalizable then 
$$\dim E=n=\dim\ker A+\dim\ker(A-I)=\mathrm{Rank} A+\mathrm{Rank }(I-A)$$
by the rank-nullity theorem and this answers the third point.
A: A is the orthogonal projector along the direction of all ones, $e=(1,1,...,1)$, $A=I-\frac{e\,e^T}{e^Te}$. The usual identities for projectors apply.

The column-row product $e\cdot e^T$ is just the matrix where every entry is $1$, so it is $J$. $e^Te=n$ should be obvious. The general form of any corank-1-projector is $P=I-\frac{u\,v^T}{v^Tu}$. In the case $u=v$ this gives an orthogonal projector, the case $u=v=e$ gives the matrix $A$.
Now either you know the properties of projectors, esp. $P^2=P$, and that orthogonal projectors are additionally symmetric $P^T=P$, or you have to prove them in general or only for this example.
