[Math] Positive definite matrices diagonalised by orthogonal matrices that are also involutions

Question

Let $A$ be a positive definite matrix. Then, $A$ is diagonalized by an orthogonal matrix $P$.

I want to know when this matrix is also an involution, i.e., $P^2 = I$.

If there is any characterization of such $A$, please kindly share. Thank you.

Answer 1

Let $n \times n$ matrix $\rm A$ be symmetric and positive definite. Since $\rm A$ is symmetric, it is diagonalizable. Hence, there exists a (non-singular) matrix $\rm P$ such that $\mathrm A = \mathrm P \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm P^{-1}$, where $\lambda_1, \lambda_2, \dots, \lambda_n > 0$. Suppose $\rm P$ is orthogonal and involutory — and, thus, symmetric. Hence,

$$\mathrm A = \mathrm P \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm P$$

Since $\rm P$ is symmetric and involutory, it has a spectral decomposition and its eigenvalues are $\pm 1$

$$\mathrm P = \mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top$$

where $\mathrm V$ is an orthogonal matrix and $\sigma_1, \sigma_2, \dots, \sigma_n = \pm 1$. Thus, we have the parameterization

$$\mathrm A = \mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top$$