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Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and strictly positive eigenvalues (note that $A$ is not required to be symmetric).

My question. Do there exist an orthogonal matrix $T\in\mathbb{R}^{n\times n}$ and a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ such that $$ TAPT^\top = D+S, $$ where $D\in\mathbb{R}^{n\times n}$ is a diagonal matrix with positive diagonal entries and $S\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix?

Of course, if the diagonal entries of $D$ are not required to be positive then the answer is in the affirmative (see, e.g., this related question).

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  • $\begingroup$ Question 2 seems trivial - scaling $P$ scales $tr(D)$. $\endgroup$ – user44191 Sep 25 '18 at 17:24
  • $\begingroup$ @user44191: Absolutely right! I will remove my second question right away. $\endgroup$ – Ludwig Sep 25 '18 at 17:27
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Choose any positive definite matrix $Q$. Since $A$ has eigenvalues with positive real part, the Lyapunov equation $$AP + PA^\top = Q$$ is solvable, and its solution $P$ is symmetric and positive definite.

Now decompose $AP = H+\hat{S}$, where $H$ is symmetric and $\hat{S}$ is skew-symmetric. Plugging this decomposition into the Lyapunov equation, we see that $H=\frac12 Q$. Then, take an eigendecomposition $\frac12 Q = TDT^\top$, with orthogonal $T$; since we chose $Q$ positive definite, $D$ has positive diagonal entries.

Hence we have $$ AP = \frac12 Q + S = TDT^\top + \hat{S}, $$ with $\hat{S}$ skew-symmetric, or $$ T^\top AP T = D + T^\top \hat{S} T, $$ where $S=T^\top \hat{S} T$ is skew-symmetric.

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Rewrite your equation: $A = (T^{-1} (D+S) T) (T^{-1} P T)$.

Choose $B, C$ symmetric positive definite with $A = BC$ (https://pure.tue.nl/ws/files/1810587/Metis198781.pdf ). Then choose $D$ similar to $B$, choose $T$ such that $B = T^{-1} D T$, and $P = T C T^{-1}$. Also choose $S = 0$.

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