Show that this matrix is unitary I'm currently working on quantum computing. 
In a literature source that I have, it is said that the following matrix $U$ is unitary. 
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=\begin{pmatrix} -1+\frac{2}{N} & \frac{2}{N} & ...& \frac{2}{N} \\ 
\frac{2}{N} & -1+\frac{2}{N} & ...& \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ...& -1+\frac{2}{N}  \end{pmatrix}  $
my first thought would be to pull the 1 out:
$U=\begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ 
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N}  \end{pmatrix}  - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
 A: Note that
$$
U=\frac{2}{N}vv^T -I
$$ where $v=(1,1,\ldots,1)^T\in\Bbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-\frac{4}{N}vv^T +\frac{4}{N^2}(v^Tv)vv^T=I-\frac{4}{N}vv^T +\frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
A: $ U^{*} = U = \begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ 
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N}  \end{pmatrix}  - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= \begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ 
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N}  \end{pmatrix} $ 
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$ 
$ A^2= \begin{pmatrix} \frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N} \\ 
\frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N}  \end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$ 
