Consider $m$ by $m$ matrix $\bf A$ such that all entries equal to $\frac{1}{m}$. Prove that $\mathbf{A}^{n} = \mathbf A$ for $n ≥ 2$ 
Proposition. Consider $m$ by $m$ matrix $\bf A$ such that $a_{ij} = \frac{1}{m}$ for all $i$ and $j$. Then $\mathbf{A}^{n} = \mathbf{A}$, for $n ≥ 2$

My attempt:
By induction.
Base case: 
$k=2$.
Consider arbitrary entry of $\bf A$, say $[\mathbf{A}^2]_{ij}$
$$[\mathbf{A}^2]_{ij} = \sum_{k=1}^{m}\frac{1}{m^2} = \frac{1}{m}$$
Which implies that every entry of $\mathbf{A}^{2}$ equals to $\frac{1}{m}$ and therefore $\mathbf{A}^2 = \bf A$.
Induction step:
Suppose $\mathbf {A}^{n} = \bf A$ for some $n \in \mathbb Z^{+}$. 
$$\mathbf{A}^{n+1} = \mathbf{A}^{n}\mathbf{A} = \mathbf{A}^2 = \mathbf{A}$$
$\blacksquare$



*

*Is it correct?

*Is there a better way?
 A: The matrix has rank $1$, and has $1$ as an eigenvalue (with eigenvector $(1,1,\ldots,1)$. The characteristic polynomial is then $x^{m-1}(x-1)$. Since the matrix is diagonalizable (nullspace is of dimension $m-1$, and $E_1$ is of dimension $1$), the minimal polynomial is $x(x-1) = x^2-x$.
That means that $A^2-A=0$, hence $A^2=A$. Once you have $A^2=A$, you get that for every $m\geq 2$, $A^m = A$ as well.

Alternatively, $A=\frac{1}{m}B$, where $B$ is the matrix of all $1$s. It is easy to compute $B^2 = mB$. So $A^2 = \frac{1}{m^2}(mB) = \frac{1}{m}B = A$.
A: You may skip the $A^2=A$ step by expressing $A$ directly as a rank-one matrix. Let $e=(1,\ldots,1)^T$. Then $A=\frac1mee^T$ and $e^Te=m$. Therefore
\begin{aligned}
A^n
&=\frac{1}{m^n}\underbrace{(ee^T)(ee^T)\cdots(ee^T)}_{m \text{ times}}\\
&=\frac{1}{m^n}e\,\underbrace{(e^Te)(e^Te)\cdots(e^Te)}_{m-1 \text{ times}}\,e^T\\
&=\frac{m^{n-1}}{m^n}ee^T
=\frac1mee^T=A.
\end{aligned}
A: Note sure if I would call it better, but it is shorter:
Note that $Ax = \bar{x}$ for any $x$ where $\bar{x} = ({1 \over m}\sum_k x_k )(1,....,1)^T$.
 Since $\bar{\bar{x}} = \bar{x}$, we see that $A^n x = A x$ for all $n \ge 1$ and so $A^n = A$.
