Finding the eigenvalues of an $n \times n$ matrix Compute the eigenvalues of the $n\times n$ matrix M defined by
$$(m_{ij}) = \begin{cases}
1 - 1/n & \text{ if } i = j  \\
-1/n & \text{ otherwise.}
\end{cases} $$
Using Wolfram Alpha, I found a pattern: for an $n \times n$ matrix, the eigenvalues are $0$ (with multiplicity $1$) and $1$ (with multiplicity $ n - 1)$. But how can I prove this fact? I didn't get anywhere with induction.
For example, in the $n = 2$ case, we have that $\text{trace M } = 1$, and $\text{det} M = 0$. So the product of the eigenvalues is $0$ (which means one eigenvalue is $0$) and the sum is $1$, so the other one must be $1$. This doesn't really work for higher orders though.   
 A: Since $M-\operatorname{Id}$ has rank $1$ (all columns are equal and different from $0$), $1$ is an eigenvalue with multiplicity $n-1$. And since $(1,1,\ldots,1)$ is an eigenvector with eigenvalue $0$, $0$ is the remaining eigenvalue.
A: The matrix is actually 
$$A = I - \frac{1}{n} \pmb{1}\pmb{1}^T$$
where $\pmb{1}$ is a column vector of size $n$ of all ones.
Notice that 
$$A\pmb{1} = \pmb{1}- \frac{1}{n} \pmb{1}\pmb{1}^T \pmb{1} =  \pmb{1}- \pmb{1} = 0$$
So $ \pmb{1}$ is an eigenvector of eigenvalue zero. 
Now, the other $n-1$ eigenvalues/vectors come from the following observation. Let $(\lambda,x)$ be eigenvector/value pairs, so
$$Ax = x - \frac{\sum x_i}{n} \pmb{1} = x - \bar{x} \pmb{1}$$
where $\bar{x}$ is the sample mean of the entries of $x$.
The above equation suggests that $(1, x)$ is an eigenvector/value pair if $\bar{x} = 0$. You could build an $n-1$ linearly independent span of vectors satisfying $\sum x_i = 0$. So 
\begin{align}
\lambda_1 &= 1 \\
& \vdots \\
\lambda_{n-1} &= 1 \\
\lambda_{n} &= 0
\end{align}
For $n = 1$ (base case), the only eigenvalue is $0$.
