How to find an upper triangular with $\ U^2 = I $ which gives $\ U $ is its own inverse It's obvious that $\ I $(identity matrix) of any size $\ N $ satisfies$\ I^2 = I$  so that $\ I $ is its own inverse. 
However if we consider $\ N = 2$ and attempt to find such a triangular matrix $\ U \neq I $ we have this scenario:
$$\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} .\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} = I$$
$$\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} .\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
$$\begin{bmatrix} u_1^2 & u_1u_2 + u_2u_3 \\ 0 & u_3^2 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
After solving this we realize that $\ u_1 = 1,u_2, u_3 = -1$ is the form of the solution where $\ U \neq I $ for $\ N = 2 $.
But for $\ N \ge 3 $, what will be the form of such matrices?
 A: Observe that $U = \operatorname{diag}(\pm 1, \dots, \pm 1)$ is a diagonal (and in particular, upper triangular) matrix that satisfies $U^2 = I$. If $P$ is an invertible upper triangular matrix then $U' = P^{-1} U P$ is also upper triangular (as a product of upper triangular matrices) and satisfies 
$$ U'^2 = P^{-1} U P P^{-1} U P = P^{-1} U^2 P = P^{-1} I P = I. $$
In fact, the other direction is also true. That is, if $U$ is an upper triangular matrix that satisfies $U^2 = I$ then we can an upper triangular $P$ such that $P^{-1} U P = \operatorname{diag}(\pm 1, \dots, \pm 1)$. To see this, note that the equation $U^2 = I$ implies that the minimal polynomial of $U$ divides $(x^2 - 1) = (x - 1)(x + 1)$ and so $U$ is diagonalizable with eigenvalues $\pm 1$. Since $U$ is upper triangular, each subspace $\left< e_1, \dots, e_i \right>$ is $U$-invariant (where $e_i$ are the standard basis vectors of $\mathbb{F}^n$) and $U|_{\left< e_1, \dots, e_i \right>}$ is also diagonalizable. This implies that we can choose eigenvectors $v_i$ for $U$ such that $v_i \in  \left< e_1, \dots, e_i \right>$ for all $1 \leq i \leq n$. Taking the vectors $v_i$ to be the columns of $P$, we obtain the required result.
A: Here's one possible form. Let $a$ be any number. Define an $n \times n$ matrix as given below.
\begin{equation*}
A = \begin{bmatrix}
1 & a & \dfrac{a^2}{2} & \dfrac{a^3}{4}& \cdots & \dfrac{a^{n-1}}{2^{n-2}}\\
0 & -1 & -a & -\dfrac{a^2}{2} & \cdots & -\dfrac{a^{n-2}}{2^{n-3}}\\
0 & 0 & 1 & a & \cdots & \dfrac{a^{n-3}}{2^{n-4}}\\
0 & 0 & 0 & -1 & \cdots & -\dfrac{a^{n-4}}{2^{n-5}}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \cdots & (-1)^{n-1}
\end{bmatrix}
\end{equation*}
Then $A^2 = I$ (verify).
