Is it true that all matrices of this type are involutory?

Question

I came across a problem that suggested the following for a $3 \times 3$ matrix. $$\text{Let } H = I - \frac{2}{3}A$$ $$\text{Then } H^2 = I$$ For $H, A \in M_{3,3}$ and $A$ being a matrix of all ones. I played around with this and found that for $H, A \in M_{n,n}$ and $A$ being a matrix of all ones $$\text{Let } H = I - \frac{2}{n}A$$ $$\text{Then } H^2 = I$$ Works in the cases for $0 \leq n \leq 4$, which makes me think it's probably true for any $n$. Is that the case?

Answer 1

First of all, it is known that the $n\times n$ matrix of all ones is such that $A^2=nA$. To show this, see that every entry is going to be the sum of $n$ products of ones: \begin{align*} A^2&=\begin{bmatrix} 1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{bmatrix}\\ &=\begin{bmatrix} \sum_{i=1}^n1 & \sum_{i=1}^n1 & \cdots & \sum_{i=1}^n1\\ \sum_{i=1}^n1 & \sum_{i=1}^n1 & \cdots & \sum_{i=1}^n1\\ \vdots & \vdots & \ddots & \vdots\\ \sum_{i=1}^n1 & \sum_{i=1}^n1 & \cdots & \sum_{i=1}^n1 \end{bmatrix}=\begin{bmatrix} n & n & \cdots & n\\ n & n & \cdots & n\\ \vdots & \vdots & \ddots & \vdots\\ n & n & \cdots & n \end{bmatrix}=nA. \end{align*} Then, we get \begin{align} H^2=\left(I-\frac{2}{n}A\right)\left(I-\frac{2}{n}A\right)=I^2&-\frac{2}{n}IA-\frac{2}{n}AI+\frac{4}{n^2}A^2\\ &=I-\frac{4}{n}A+\frac{4}{n^2}(nA)=I-\frac{4}{n}A+\frac{4}{n}A=I. \end{align}