Prove that the rank of $(1-I)$ is $n$ The rank of $(1-I_n)$, where $1$ is the $n \times n$ all-1 matrix and $I_n$ the $n \times n$ identity matrix, seems to be $n$.
How to prove this concisely?
 A: Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
\det(J-XI_n)
$$
which means that $\det(J-I_n)\ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
A: If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$\sum_{\overset{i=1}{i \neq j}}^n x_i = 0$$ for all $j \in \{1,2,\ldots,n\}$. Can this be true for $x \neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- \left(I + \dfrac{ee^T}{1-e^Te}\right) = - \left(I  - \dfrac{ee^T}{n-1}\right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u \neq 0$. In your case, this corresponds to the condition that $n-1 \neq 0$, i.e., $n \neq 1$.
A: $\newcommand{\rank}{\operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{n\times n}$is full column rank ($\rank(A)=n$)if and only if it's invertible.
$$\det(1-I_n)=(n-1)(-1)^{n-1}\neq0$$ 
use  row echelon operations to Compute $\det(B)$$$\forall i, ~~ 2\le i\le n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column.  Now;$$\det(B)=\det\begin{pmatrix}
    r & a & a & \cdots &a \\
    a & r& a & \cdots & a \\
     \vdots  & \vdots& \vdots & \ddots & \vdots \\
     a & a & a & \cdots & r    
     \end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$ 
A: If $A$ is any $n\times n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $\def\tr{\operatorname{tr}}\tr A$ (as that is the sum of the eigenvalues). Therefore $A-\lambda I$ with $\lambda\neq0$ is invertible unless $\lambda=\tr A$, in which case $\def\rk{\operatorname{rk}}\rk(A-\lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $\tr A=n$, so for $\lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
