# Can this matrix group be found?

I am trying to find the matrix group $G_D \le GL(n,\mathbb{R})$ consisting of all matrices $P$ which satisfy: $$P^T D P = D$$ where $D$ is a given positive definite diagonal $n \times n$ matrix.

For the special case $D = \lambda I$ the group is $G_D = O(n)$. But I am interested in the case of a general diagonal matrix $D$. Note that the solution $P = \pm I$ is valid for any $D$, but I am hoping there are nontrivial solutions as well.

My approach so far is to write $D$ as $$D = \textrm{diag}(d_1,\dots,d_n) = \sum_i d_iE_i$$ where $E_i$ is a matrix with 1 in position $(i,i)$ and zero elsewhere, i.e.: $$E_i = \left( \begin{array}{ccccc} 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & 0 & \cdots & 0 \end{array} \right)$$ Then the defining equation becomes: $$\sum_i d_i P^T E_i P^T = D$$ Now I write $P$ in block form as $$P = \left( \begin{array}{c|c} P_{1i} & P_{2i} \\ \hline P_{3i} & P_{4i} \end{array} \right)$$ where $P_{1i}$ is $i \times i$, $P_{2i}$ is $i \times (n-i)$, $P_{3i}$ is $(n-i) \times i$ and $P_{4i}$ is $(n-i) \times (n-i)$.

Putting this into the summation above gives: $$\sum_i d_i \left( \begin{array}{cc} P_{1i}^T \hat{E}_i P_{1i} & P_{1i}^T \hat{E}_i P_{2i} \\ P_{2i}^T \hat{E}_i P_{1i} & P_{2i}^T \hat{E}_i P_{2i} \end{array} \right) = D$$ Here I have defined $\hat{E}_i$ has the $i \times i$ matrix with 1 in position $(i,i)$ and 0 elsewhere: $$\hat{E}_i = \left( \begin{array}{ccc} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{array} \right)$$ At this point I'm stuck and am not sure how to make progress on solving for $P$ in the above equation. I'd appreciate if anyone has advice or possibly other insights into how to attack this problem.

• $P = \sqrt{D}^{-1} O \sqrt{D}$ for any $O \in O(n, \mathbb(R))$. your group is isomorphic to $O(n)$ under an inner automorphism. – achille hui May 27 '18 at 15:33
• Thanks! This is great. If you make an answer I'll give you credit – vibe May 27 '18 at 16:07

Since $D$ is positive definite, it has a positive definite square root. Notice
\begin{align} P^T D P = D \iff & (\sqrt{D}^{-1} P^T \sqrt{D})(\sqrt{D} P \sqrt{D}^{-1}) = I_n\\ \iff & (\sqrt{D}^T P \sqrt{D}^{-T})^T (\sqrt{D} P \sqrt{D}^{-1}) = I_n\\ \iff & (\sqrt{D} P \sqrt{D}^{-1})^T (\sqrt{D} P \sqrt{D}^{-1}) = I_n \end{align} If we let $O = \sqrt{D}P\sqrt{D}^{-1}$, then $O^T O = I_n \implies O \in O(n,\mathbb{R})$.
This means the collection of $P$ can be parameterized as $\sqrt{D}^{-1}O\sqrt{D}$ by $O \in O(n,\mathbb{R})$. Such a collection forms a group which is isomorphic to $O(n,\mathbb{R})$ by an inner automorphism induced by $\sqrt{D}$.
• I have a followup question, since the set of all $P$ form a group isomorphic to $O(n)$, shouldn't that mean there is a matrix representation of the group with is composed of orthogonal matrices? However in the form $P = D^{-1/2} O D^{1/2}$, these matrices are not orthogonal even if $O$ is. – vibe May 27 '18 at 23:05
• @vibe If you don't view $P$ as matrix but some linear transform $\tilde{P}$. and scale your basis according to $\sqrt{D}$, then in the new basis, the matrix corr to $\tilde{P}$ will be orthogonal matrices. – achille hui May 28 '18 at 3:08