Diagonalizability via self-inverse matrix

Question

Suppose $A\in K^{n\times n}$ is diagonalizable via an involutory (or self-inverse) matrix, i.e. there exist a diagonal $D\in K^{n\times n}$ and an involutory transformation $T\in K^{n\times n}$ (i.e. $T^{-1}=T$) such that $$A=TDT^{-1}(=TDT)$$

What would be a characterization of such matrices, in particular for $K=\mathbb{C}$ (and $K=\mathbb{R}$)?

Answer 1

Clearly, $A$ is diagonalizable via involutary (denoted by DVI) IFF $A$ admits a basis of eigenvectors $(e_i)$ s.t. $[e_1,\cdots,e_n]$ is involutary.

Assume $K=\mathbb{C}$.

When $n=2$, every $A\in M_2$ that is diagonalizable is DVI and we have the same result if $K=\mathbb{R}$.

When $n=3$, the algebraic set $Z_3=\{A\in M_3; A\text{ is DVI }\}$ has dimension $7$ ($7$ degrees of freedom). In particular, a randomly chosen matrix in $M_3$ is not in $Z_3$ with probability $1$.

More generally, I think that $dim(Z_n)=dim(Inv_n)+n$, where $Inv_n=\{A\in M_n; A^2=I_n\}$ is an algebraic set of dimension $\approx n^2/2$.