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.
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1$\begingroup$ Geometrically, this occurs precisely when there is a reflection (through some subspace) taking the eigenspaces of A to coordinate-subspaces. $\endgroup$– JamesCommented Jun 28, 2018 at 2:23
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$\begingroup$ @RodrigodeAzevedo the $P$ is the question has to be symmetric as it is orthogonal and involution. That much only I can understand. If you have further insight please share with me. $\endgroup$– GA316Commented Jun 28, 2018 at 5:06
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$\begingroup$ @James But can you say anything more algebraically which would be help me more? $\endgroup$– GA316Commented Jun 28, 2018 at 5:07
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$\begingroup$ It's better to start with a positive definite diagonal matrix $B.$ Then for any orthogonal involution $Q,$ you see that $A = QBQ$ is still a positive definite matrix, and $QAQ = B,$ so that $A$ is a positive definite matrix diagonalized by the orthogonal involution $Q.$ Conversely, every positive definite matrix $A$ which may be diagonalized by an orthogonal involution $P$ has such a form, since $PAP$ is positive definite diagonal, and is inverted by the orthogonal involution $P.$ $\endgroup$– Geoff RobinsonCommented Jun 28, 2018 at 11:03
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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
$$\mathrm A = \mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top$$