Reducing a matrix of the form J-n*In, where J is the all-ones matrix. How can I reduce a matrix like this? The goal is to prove that the dimension of the generated space is 1. It's the matrix of the homogeneous system.
$\left( \begin{array}{cccccc}1-n&1&\dots &\dots&\dots&1 \\ 1 & 1-n & 1 & \dots & \dots & 1 \\ \vdots & 1 & \ddots & \ddots & & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \vdots && 1 & 1-n & 1 \\ 1 &1 & \dots & \dots & 1 & 1-n\end{array} \right)$.

The matrix diagonalizes to $$\begin{pmatrix}
0&0&0&0\\
0&-n&0&0\\
0&0&-n&0\\
0&0&0&-n\\
\end{pmatrix}$$
Let $A=\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
\end{pmatrix}$
And if $X=\begin{pmatrix}
x_1\\
\vdots \\
x_n
\end{pmatrix}$ let $\sum X=x_1+\cdots x_n$.
Then we have $$AX=\sum X \begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$$
Now consider the eigenvalue problem
$$(A-nI)X=\lambda X$$
or equivalently,
$$AX=(n+\lambda)  X=\sum X \begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$$
Now if $\lambda +n \neq 0$ then $X$ is a multiple of $\begin{pmatrix}
1\\
\vdots \\
1
\end{pmatrix}$ and we have that $\lambda +n=n$ so $\lambda$ is zero with an eigenspace of dimension one.
Otherwise $\lambda=-n$ and this has an eigenspace of dimension $n-1$
as can be easily checked the vectors $$\begin{pmatrix}
1\\
0\\
\vdots \\
0\\-1\\
0\\
\vdots\\
0\\
\end{pmatrix}$$ are eigen vectors. 
Finally it would be nice to check the minimal polynomial which should be $x(x+n)$
