An involutory matrix $\mathbf{A}$ is such that $\mathbf{A}^{2}=\mathbf{I}$.
An idempotent matrix $\mathbf{A}$ is such that $\mathbf{A}^{2}=\mathbf{A}$
So we'd have to satisfy both conditions simultaneously.
I do know the answer to the question is the identity matrix itself. But how can we go about proving it?
One way I thought of is to expose that the diagonal of $\mathbf{A}^{2}$ and condition that all other entries must be zero, but then I would already be biased by knowing the answer beforehand.
It'd be really nice if you guys gave me a hand with this one. In the meantime I'll carry on trying.
For idempotent matrix $A^2 = A$, it means the polynomial $p(x) = x^2 - x = x(x-1)$ is satisfied by matrix $A$. Hence its possible options for minimal polynomial for $A$ are $x$ , $(x-1)$ and $x(x-1)$. While for an involutary matrix $A^2 = I$ . So polynomial $q(x) = x^2 -1 = (x-1)(x+1)$ is satisfied by matrix A. So the possible cases for minimal polynomial are : (x-1), $(x+1)$ and $(x-1)(x+1)$. So when the matrix is both idempotent and involutary the only possible common option for minimal polynomial is : m(x) = (x-1). Since any matrix satisfies its minimal polynomial , we get A - I = O, the zero matrix. Hence $A = I$, the identity matrix .